Chapter 11Gases
2006 Prentice Hall
2
CHAPTER OUTLINE
Properties of Gases Pressure amp Its Measurement The Gas Laws Vapor Pressure amp Boiling Point Combined Gas Law Avogadrorsquos Law STP amp Molar Volume Ideal Gas Law Partial Pressures
3
PROPERTIESOF GASES
Gases are the least dense and most mobile of the three phases of matter
Particles of matter in the gas phase are spaced far apart from one another and move rapidly and collide with each other often
Gases occupy much greater space than the same amount of liquid or solid
This is because the gas particles are spaced apart from one another and are therefore compressible
Solid or liquid particles are spaced much closer and cannot be compressed
4
PROPERTIESOF GASES
Gases are characterized by four properties These are
1 Pressure (P)
2 Volume (V)
3 Temperature (T)
4 Amount (n)
5
KINETIC-MOLECULARTHEORY
Scientists use the kinetic-molecular theory (KMT) to describe the behavior of gases
The KMT consists of several postulates
1 Gases consist of small particles (atoms or molecules) that move randomly with rapid velocities
3 The distance between the particles is large compared to their size Therefore the volume occupied by gas molecules is small compared to the volume of the gas
2 Gas particles have little attraction for one another Therefore attractive forces between gas molecules can be ignored
6
KINETIC-MOLECULARTHEORY
5 The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin)
Animation
4 Gas particles move in straight lines and collide with each other and the container frequently The force of collision of the gas particles with the walls of the container causes pressure
Tros Introductory Chemistry Chapter
7
Kinetic Molecular Theory
8
PRESSURE ampITS MEASUREMENT
Pressure is the result of collision of gas particles with the sides of the container Pressure is defined as the force per unit area
Pressure is measured in units of atmosphere (atm) or mmHg or torr The SI unit of pressure is pascal (Pa) or kilopascal (kPa)
1 atm = 760 mmHg = 101325 kPa
1 mmHg = 1 torr
Tros Introductory Chemistry Chapter
9
Common Units of Pressure
Unit Average Air Pressure at Sea Level
pascal (Pa) 101325
kilopascal (kPa) 101325
atmosphere (atm) 1 (exactly)
millimeters of mercury (mmHg) 760 (exactly)
inches of mercury (inHg) 2992
torr (torr) 760 (exactly)
pounds per square inch (psi lbsin2) 147
10
PRESSURE ampITS MEASUREMENT
Atmospheric pressure can be measured with the use of a barometer
Mercury is used in a barometer due to its high density
At sea level the mercury stands at 760 mm above its base
Tros Introductory Chemistry Chapter
11
Atmospheric Pressure amp Altitude
bull the higher up in the atmosphere you go the lower the atmospheric pressure is around youat the surface the atmospheric pressure is
147 psi but at 10000 ft is is only 100 psibull rapid changes in atmospheric pressure
may cause your ears to ldquopoprdquo due to an imbalance in pressure on either side of your ear drum
Tros Introductory Chemistry Chapter
12
Pressure Imbalance in Ear
If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo
13
Example 1
The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm
1 mmHg = 1 torr
740 mmHg = 740 torr
740 mmHg atm
x mmHg760
1= 0974 atm
14
Example 2
The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr
1 mmHg = 1 torr
851 mmHg = 851 torr
112 atm mmHg
x atm1760
= 851 mmHg
15
PRESSURE amp MOLES OF A GAS
The pressure of a gas is directly proportional to the number of particles (moles) present
The greater the moles of the gas the greater the pressure
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
2
CHAPTER OUTLINE
Properties of Gases Pressure amp Its Measurement The Gas Laws Vapor Pressure amp Boiling Point Combined Gas Law Avogadrorsquos Law STP amp Molar Volume Ideal Gas Law Partial Pressures
3
PROPERTIESOF GASES
Gases are the least dense and most mobile of the three phases of matter
Particles of matter in the gas phase are spaced far apart from one another and move rapidly and collide with each other often
Gases occupy much greater space than the same amount of liquid or solid
This is because the gas particles are spaced apart from one another and are therefore compressible
Solid or liquid particles are spaced much closer and cannot be compressed
4
PROPERTIESOF GASES
Gases are characterized by four properties These are
1 Pressure (P)
2 Volume (V)
3 Temperature (T)
4 Amount (n)
5
KINETIC-MOLECULARTHEORY
Scientists use the kinetic-molecular theory (KMT) to describe the behavior of gases
The KMT consists of several postulates
1 Gases consist of small particles (atoms or molecules) that move randomly with rapid velocities
3 The distance between the particles is large compared to their size Therefore the volume occupied by gas molecules is small compared to the volume of the gas
2 Gas particles have little attraction for one another Therefore attractive forces between gas molecules can be ignored
6
KINETIC-MOLECULARTHEORY
5 The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin)
Animation
4 Gas particles move in straight lines and collide with each other and the container frequently The force of collision of the gas particles with the walls of the container causes pressure
Tros Introductory Chemistry Chapter
7
Kinetic Molecular Theory
8
PRESSURE ampITS MEASUREMENT
Pressure is the result of collision of gas particles with the sides of the container Pressure is defined as the force per unit area
Pressure is measured in units of atmosphere (atm) or mmHg or torr The SI unit of pressure is pascal (Pa) or kilopascal (kPa)
1 atm = 760 mmHg = 101325 kPa
1 mmHg = 1 torr
Tros Introductory Chemistry Chapter
9
Common Units of Pressure
Unit Average Air Pressure at Sea Level
pascal (Pa) 101325
kilopascal (kPa) 101325
atmosphere (atm) 1 (exactly)
millimeters of mercury (mmHg) 760 (exactly)
inches of mercury (inHg) 2992
torr (torr) 760 (exactly)
pounds per square inch (psi lbsin2) 147
10
PRESSURE ampITS MEASUREMENT
Atmospheric pressure can be measured with the use of a barometer
Mercury is used in a barometer due to its high density
At sea level the mercury stands at 760 mm above its base
Tros Introductory Chemistry Chapter
11
Atmospheric Pressure amp Altitude
bull the higher up in the atmosphere you go the lower the atmospheric pressure is around youat the surface the atmospheric pressure is
147 psi but at 10000 ft is is only 100 psibull rapid changes in atmospheric pressure
may cause your ears to ldquopoprdquo due to an imbalance in pressure on either side of your ear drum
Tros Introductory Chemistry Chapter
12
Pressure Imbalance in Ear
If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo
13
Example 1
The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm
1 mmHg = 1 torr
740 mmHg = 740 torr
740 mmHg atm
x mmHg760
1= 0974 atm
14
Example 2
The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr
1 mmHg = 1 torr
851 mmHg = 851 torr
112 atm mmHg
x atm1760
= 851 mmHg
15
PRESSURE amp MOLES OF A GAS
The pressure of a gas is directly proportional to the number of particles (moles) present
The greater the moles of the gas the greater the pressure
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
3
PROPERTIESOF GASES
Gases are the least dense and most mobile of the three phases of matter
Particles of matter in the gas phase are spaced far apart from one another and move rapidly and collide with each other often
Gases occupy much greater space than the same amount of liquid or solid
This is because the gas particles are spaced apart from one another and are therefore compressible
Solid or liquid particles are spaced much closer and cannot be compressed
4
PROPERTIESOF GASES
Gases are characterized by four properties These are
1 Pressure (P)
2 Volume (V)
3 Temperature (T)
4 Amount (n)
5
KINETIC-MOLECULARTHEORY
Scientists use the kinetic-molecular theory (KMT) to describe the behavior of gases
The KMT consists of several postulates
1 Gases consist of small particles (atoms or molecules) that move randomly with rapid velocities
3 The distance between the particles is large compared to their size Therefore the volume occupied by gas molecules is small compared to the volume of the gas
2 Gas particles have little attraction for one another Therefore attractive forces between gas molecules can be ignored
6
KINETIC-MOLECULARTHEORY
5 The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin)
Animation
4 Gas particles move in straight lines and collide with each other and the container frequently The force of collision of the gas particles with the walls of the container causes pressure
Tros Introductory Chemistry Chapter
7
Kinetic Molecular Theory
8
PRESSURE ampITS MEASUREMENT
Pressure is the result of collision of gas particles with the sides of the container Pressure is defined as the force per unit area
Pressure is measured in units of atmosphere (atm) or mmHg or torr The SI unit of pressure is pascal (Pa) or kilopascal (kPa)
1 atm = 760 mmHg = 101325 kPa
1 mmHg = 1 torr
Tros Introductory Chemistry Chapter
9
Common Units of Pressure
Unit Average Air Pressure at Sea Level
pascal (Pa) 101325
kilopascal (kPa) 101325
atmosphere (atm) 1 (exactly)
millimeters of mercury (mmHg) 760 (exactly)
inches of mercury (inHg) 2992
torr (torr) 760 (exactly)
pounds per square inch (psi lbsin2) 147
10
PRESSURE ampITS MEASUREMENT
Atmospheric pressure can be measured with the use of a barometer
Mercury is used in a barometer due to its high density
At sea level the mercury stands at 760 mm above its base
Tros Introductory Chemistry Chapter
11
Atmospheric Pressure amp Altitude
bull the higher up in the atmosphere you go the lower the atmospheric pressure is around youat the surface the atmospheric pressure is
147 psi but at 10000 ft is is only 100 psibull rapid changes in atmospheric pressure
may cause your ears to ldquopoprdquo due to an imbalance in pressure on either side of your ear drum
Tros Introductory Chemistry Chapter
12
Pressure Imbalance in Ear
If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo
13
Example 1
The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm
1 mmHg = 1 torr
740 mmHg = 740 torr
740 mmHg atm
x mmHg760
1= 0974 atm
14
Example 2
The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr
1 mmHg = 1 torr
851 mmHg = 851 torr
112 atm mmHg
x atm1760
= 851 mmHg
15
PRESSURE amp MOLES OF A GAS
The pressure of a gas is directly proportional to the number of particles (moles) present
The greater the moles of the gas the greater the pressure
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
4
PROPERTIESOF GASES
Gases are characterized by four properties These are
1 Pressure (P)
2 Volume (V)
3 Temperature (T)
4 Amount (n)
5
KINETIC-MOLECULARTHEORY
Scientists use the kinetic-molecular theory (KMT) to describe the behavior of gases
The KMT consists of several postulates
1 Gases consist of small particles (atoms or molecules) that move randomly with rapid velocities
3 The distance between the particles is large compared to their size Therefore the volume occupied by gas molecules is small compared to the volume of the gas
2 Gas particles have little attraction for one another Therefore attractive forces between gas molecules can be ignored
6
KINETIC-MOLECULARTHEORY
5 The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin)
Animation
4 Gas particles move in straight lines and collide with each other and the container frequently The force of collision of the gas particles with the walls of the container causes pressure
Tros Introductory Chemistry Chapter
7
Kinetic Molecular Theory
8
PRESSURE ampITS MEASUREMENT
Pressure is the result of collision of gas particles with the sides of the container Pressure is defined as the force per unit area
Pressure is measured in units of atmosphere (atm) or mmHg or torr The SI unit of pressure is pascal (Pa) or kilopascal (kPa)
1 atm = 760 mmHg = 101325 kPa
1 mmHg = 1 torr
Tros Introductory Chemistry Chapter
9
Common Units of Pressure
Unit Average Air Pressure at Sea Level
pascal (Pa) 101325
kilopascal (kPa) 101325
atmosphere (atm) 1 (exactly)
millimeters of mercury (mmHg) 760 (exactly)
inches of mercury (inHg) 2992
torr (torr) 760 (exactly)
pounds per square inch (psi lbsin2) 147
10
PRESSURE ampITS MEASUREMENT
Atmospheric pressure can be measured with the use of a barometer
Mercury is used in a barometer due to its high density
At sea level the mercury stands at 760 mm above its base
Tros Introductory Chemistry Chapter
11
Atmospheric Pressure amp Altitude
bull the higher up in the atmosphere you go the lower the atmospheric pressure is around youat the surface the atmospheric pressure is
147 psi but at 10000 ft is is only 100 psibull rapid changes in atmospheric pressure
may cause your ears to ldquopoprdquo due to an imbalance in pressure on either side of your ear drum
Tros Introductory Chemistry Chapter
12
Pressure Imbalance in Ear
If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo
13
Example 1
The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm
1 mmHg = 1 torr
740 mmHg = 740 torr
740 mmHg atm
x mmHg760
1= 0974 atm
14
Example 2
The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr
1 mmHg = 1 torr
851 mmHg = 851 torr
112 atm mmHg
x atm1760
= 851 mmHg
15
PRESSURE amp MOLES OF A GAS
The pressure of a gas is directly proportional to the number of particles (moles) present
The greater the moles of the gas the greater the pressure
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
5
KINETIC-MOLECULARTHEORY
Scientists use the kinetic-molecular theory (KMT) to describe the behavior of gases
The KMT consists of several postulates
1 Gases consist of small particles (atoms or molecules) that move randomly with rapid velocities
3 The distance between the particles is large compared to their size Therefore the volume occupied by gas molecules is small compared to the volume of the gas
2 Gas particles have little attraction for one another Therefore attractive forces between gas molecules can be ignored
6
KINETIC-MOLECULARTHEORY
5 The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin)
Animation
4 Gas particles move in straight lines and collide with each other and the container frequently The force of collision of the gas particles with the walls of the container causes pressure
Tros Introductory Chemistry Chapter
7
Kinetic Molecular Theory
8
PRESSURE ampITS MEASUREMENT
Pressure is the result of collision of gas particles with the sides of the container Pressure is defined as the force per unit area
Pressure is measured in units of atmosphere (atm) or mmHg or torr The SI unit of pressure is pascal (Pa) or kilopascal (kPa)
1 atm = 760 mmHg = 101325 kPa
1 mmHg = 1 torr
Tros Introductory Chemistry Chapter
9
Common Units of Pressure
Unit Average Air Pressure at Sea Level
pascal (Pa) 101325
kilopascal (kPa) 101325
atmosphere (atm) 1 (exactly)
millimeters of mercury (mmHg) 760 (exactly)
inches of mercury (inHg) 2992
torr (torr) 760 (exactly)
pounds per square inch (psi lbsin2) 147
10
PRESSURE ampITS MEASUREMENT
Atmospheric pressure can be measured with the use of a barometer
Mercury is used in a barometer due to its high density
At sea level the mercury stands at 760 mm above its base
Tros Introductory Chemistry Chapter
11
Atmospheric Pressure amp Altitude
bull the higher up in the atmosphere you go the lower the atmospheric pressure is around youat the surface the atmospheric pressure is
147 psi but at 10000 ft is is only 100 psibull rapid changes in atmospheric pressure
may cause your ears to ldquopoprdquo due to an imbalance in pressure on either side of your ear drum
Tros Introductory Chemistry Chapter
12
Pressure Imbalance in Ear
If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo
13
Example 1
The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm
1 mmHg = 1 torr
740 mmHg = 740 torr
740 mmHg atm
x mmHg760
1= 0974 atm
14
Example 2
The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr
1 mmHg = 1 torr
851 mmHg = 851 torr
112 atm mmHg
x atm1760
= 851 mmHg
15
PRESSURE amp MOLES OF A GAS
The pressure of a gas is directly proportional to the number of particles (moles) present
The greater the moles of the gas the greater the pressure
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
6
KINETIC-MOLECULARTHEORY
5 The average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin)
Animation
4 Gas particles move in straight lines and collide with each other and the container frequently The force of collision of the gas particles with the walls of the container causes pressure
Tros Introductory Chemistry Chapter
7
Kinetic Molecular Theory
8
PRESSURE ampITS MEASUREMENT
Pressure is the result of collision of gas particles with the sides of the container Pressure is defined as the force per unit area
Pressure is measured in units of atmosphere (atm) or mmHg or torr The SI unit of pressure is pascal (Pa) or kilopascal (kPa)
1 atm = 760 mmHg = 101325 kPa
1 mmHg = 1 torr
Tros Introductory Chemistry Chapter
9
Common Units of Pressure
Unit Average Air Pressure at Sea Level
pascal (Pa) 101325
kilopascal (kPa) 101325
atmosphere (atm) 1 (exactly)
millimeters of mercury (mmHg) 760 (exactly)
inches of mercury (inHg) 2992
torr (torr) 760 (exactly)
pounds per square inch (psi lbsin2) 147
10
PRESSURE ampITS MEASUREMENT
Atmospheric pressure can be measured with the use of a barometer
Mercury is used in a barometer due to its high density
At sea level the mercury stands at 760 mm above its base
Tros Introductory Chemistry Chapter
11
Atmospheric Pressure amp Altitude
bull the higher up in the atmosphere you go the lower the atmospheric pressure is around youat the surface the atmospheric pressure is
147 psi but at 10000 ft is is only 100 psibull rapid changes in atmospheric pressure
may cause your ears to ldquopoprdquo due to an imbalance in pressure on either side of your ear drum
Tros Introductory Chemistry Chapter
12
Pressure Imbalance in Ear
If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo
13
Example 1
The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm
1 mmHg = 1 torr
740 mmHg = 740 torr
740 mmHg atm
x mmHg760
1= 0974 atm
14
Example 2
The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr
1 mmHg = 1 torr
851 mmHg = 851 torr
112 atm mmHg
x atm1760
= 851 mmHg
15
PRESSURE amp MOLES OF A GAS
The pressure of a gas is directly proportional to the number of particles (moles) present
The greater the moles of the gas the greater the pressure
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
Tros Introductory Chemistry Chapter
7
Kinetic Molecular Theory
8
PRESSURE ampITS MEASUREMENT
Pressure is the result of collision of gas particles with the sides of the container Pressure is defined as the force per unit area
Pressure is measured in units of atmosphere (atm) or mmHg or torr The SI unit of pressure is pascal (Pa) or kilopascal (kPa)
1 atm = 760 mmHg = 101325 kPa
1 mmHg = 1 torr
Tros Introductory Chemistry Chapter
9
Common Units of Pressure
Unit Average Air Pressure at Sea Level
pascal (Pa) 101325
kilopascal (kPa) 101325
atmosphere (atm) 1 (exactly)
millimeters of mercury (mmHg) 760 (exactly)
inches of mercury (inHg) 2992
torr (torr) 760 (exactly)
pounds per square inch (psi lbsin2) 147
10
PRESSURE ampITS MEASUREMENT
Atmospheric pressure can be measured with the use of a barometer
Mercury is used in a barometer due to its high density
At sea level the mercury stands at 760 mm above its base
Tros Introductory Chemistry Chapter
11
Atmospheric Pressure amp Altitude
bull the higher up in the atmosphere you go the lower the atmospheric pressure is around youat the surface the atmospheric pressure is
147 psi but at 10000 ft is is only 100 psibull rapid changes in atmospheric pressure
may cause your ears to ldquopoprdquo due to an imbalance in pressure on either side of your ear drum
Tros Introductory Chemistry Chapter
12
Pressure Imbalance in Ear
If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo
13
Example 1
The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm
1 mmHg = 1 torr
740 mmHg = 740 torr
740 mmHg atm
x mmHg760
1= 0974 atm
14
Example 2
The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr
1 mmHg = 1 torr
851 mmHg = 851 torr
112 atm mmHg
x atm1760
= 851 mmHg
15
PRESSURE amp MOLES OF A GAS
The pressure of a gas is directly proportional to the number of particles (moles) present
The greater the moles of the gas the greater the pressure
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
8
PRESSURE ampITS MEASUREMENT
Pressure is the result of collision of gas particles with the sides of the container Pressure is defined as the force per unit area
Pressure is measured in units of atmosphere (atm) or mmHg or torr The SI unit of pressure is pascal (Pa) or kilopascal (kPa)
1 atm = 760 mmHg = 101325 kPa
1 mmHg = 1 torr
Tros Introductory Chemistry Chapter
9
Common Units of Pressure
Unit Average Air Pressure at Sea Level
pascal (Pa) 101325
kilopascal (kPa) 101325
atmosphere (atm) 1 (exactly)
millimeters of mercury (mmHg) 760 (exactly)
inches of mercury (inHg) 2992
torr (torr) 760 (exactly)
pounds per square inch (psi lbsin2) 147
10
PRESSURE ampITS MEASUREMENT
Atmospheric pressure can be measured with the use of a barometer
Mercury is used in a barometer due to its high density
At sea level the mercury stands at 760 mm above its base
Tros Introductory Chemistry Chapter
11
Atmospheric Pressure amp Altitude
bull the higher up in the atmosphere you go the lower the atmospheric pressure is around youat the surface the atmospheric pressure is
147 psi but at 10000 ft is is only 100 psibull rapid changes in atmospheric pressure
may cause your ears to ldquopoprdquo due to an imbalance in pressure on either side of your ear drum
Tros Introductory Chemistry Chapter
12
Pressure Imbalance in Ear
If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo
13
Example 1
The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm
1 mmHg = 1 torr
740 mmHg = 740 torr
740 mmHg atm
x mmHg760
1= 0974 atm
14
Example 2
The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr
1 mmHg = 1 torr
851 mmHg = 851 torr
112 atm mmHg
x atm1760
= 851 mmHg
15
PRESSURE amp MOLES OF A GAS
The pressure of a gas is directly proportional to the number of particles (moles) present
The greater the moles of the gas the greater the pressure
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
Tros Introductory Chemistry Chapter
9
Common Units of Pressure
Unit Average Air Pressure at Sea Level
pascal (Pa) 101325
kilopascal (kPa) 101325
atmosphere (atm) 1 (exactly)
millimeters of mercury (mmHg) 760 (exactly)
inches of mercury (inHg) 2992
torr (torr) 760 (exactly)
pounds per square inch (psi lbsin2) 147
10
PRESSURE ampITS MEASUREMENT
Atmospheric pressure can be measured with the use of a barometer
Mercury is used in a barometer due to its high density
At sea level the mercury stands at 760 mm above its base
Tros Introductory Chemistry Chapter
11
Atmospheric Pressure amp Altitude
bull the higher up in the atmosphere you go the lower the atmospheric pressure is around youat the surface the atmospheric pressure is
147 psi but at 10000 ft is is only 100 psibull rapid changes in atmospheric pressure
may cause your ears to ldquopoprdquo due to an imbalance in pressure on either side of your ear drum
Tros Introductory Chemistry Chapter
12
Pressure Imbalance in Ear
If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo
13
Example 1
The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm
1 mmHg = 1 torr
740 mmHg = 740 torr
740 mmHg atm
x mmHg760
1= 0974 atm
14
Example 2
The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr
1 mmHg = 1 torr
851 mmHg = 851 torr
112 atm mmHg
x atm1760
= 851 mmHg
15
PRESSURE amp MOLES OF A GAS
The pressure of a gas is directly proportional to the number of particles (moles) present
The greater the moles of the gas the greater the pressure
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
10
PRESSURE ampITS MEASUREMENT
Atmospheric pressure can be measured with the use of a barometer
Mercury is used in a barometer due to its high density
At sea level the mercury stands at 760 mm above its base
Tros Introductory Chemistry Chapter
11
Atmospheric Pressure amp Altitude
bull the higher up in the atmosphere you go the lower the atmospheric pressure is around youat the surface the atmospheric pressure is
147 psi but at 10000 ft is is only 100 psibull rapid changes in atmospheric pressure
may cause your ears to ldquopoprdquo due to an imbalance in pressure on either side of your ear drum
Tros Introductory Chemistry Chapter
12
Pressure Imbalance in Ear
If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo
13
Example 1
The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm
1 mmHg = 1 torr
740 mmHg = 740 torr
740 mmHg atm
x mmHg760
1= 0974 atm
14
Example 2
The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr
1 mmHg = 1 torr
851 mmHg = 851 torr
112 atm mmHg
x atm1760
= 851 mmHg
15
PRESSURE amp MOLES OF A GAS
The pressure of a gas is directly proportional to the number of particles (moles) present
The greater the moles of the gas the greater the pressure
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
Tros Introductory Chemistry Chapter
11
Atmospheric Pressure amp Altitude
bull the higher up in the atmosphere you go the lower the atmospheric pressure is around youat the surface the atmospheric pressure is
147 psi but at 10000 ft is is only 100 psibull rapid changes in atmospheric pressure
may cause your ears to ldquopoprdquo due to an imbalance in pressure on either side of your ear drum
Tros Introductory Chemistry Chapter
12
Pressure Imbalance in Ear
If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo
13
Example 1
The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm
1 mmHg = 1 torr
740 mmHg = 740 torr
740 mmHg atm
x mmHg760
1= 0974 atm
14
Example 2
The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr
1 mmHg = 1 torr
851 mmHg = 851 torr
112 atm mmHg
x atm1760
= 851 mmHg
15
PRESSURE amp MOLES OF A GAS
The pressure of a gas is directly proportional to the number of particles (moles) present
The greater the moles of the gas the greater the pressure
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
Tros Introductory Chemistry Chapter
12
Pressure Imbalance in Ear
If there is a differencein pressure acrossthe eardrum membranethe membrane will bepushed out ndash what we commonly call a ldquopopped eardrumrdquo
13
Example 1
The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm
1 mmHg = 1 torr
740 mmHg = 740 torr
740 mmHg atm
x mmHg760
1= 0974 atm
14
Example 2
The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr
1 mmHg = 1 torr
851 mmHg = 851 torr
112 atm mmHg
x atm1760
= 851 mmHg
15
PRESSURE amp MOLES OF A GAS
The pressure of a gas is directly proportional to the number of particles (moles) present
The greater the moles of the gas the greater the pressure
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
13
Example 1
The atmospheric pressure at Walnut CA is 740 mmHg Calculate this pressure in torr and atm
1 mmHg = 1 torr
740 mmHg = 740 torr
740 mmHg atm
x mmHg760
1= 0974 atm
14
Example 2
The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr
1 mmHg = 1 torr
851 mmHg = 851 torr
112 atm mmHg
x atm1760
= 851 mmHg
15
PRESSURE amp MOLES OF A GAS
The pressure of a gas is directly proportional to the number of particles (moles) present
The greater the moles of the gas the greater the pressure
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
14
Example 2
The barometer at a location reads 112 atm Calculate the pressure in mmHg and torr
1 mmHg = 1 torr
851 mmHg = 851 torr
112 atm mmHg
x atm1760
= 851 mmHg
15
PRESSURE amp MOLES OF A GAS
The pressure of a gas is directly proportional to the number of particles (moles) present
The greater the moles of the gas the greater the pressure
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
15
PRESSURE amp MOLES OF A GAS
The pressure of a gas is directly proportional to the number of particles (moles) present
The greater the moles of the gas the greater the pressure
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
16
BOYLErsquoS LAW
The relationship of pressure and volume in gases is called Boylersquos Law
At constant temperature the volume of a fixed amount of gas is inversely proportional to its pressure
As pressure increases the volume of the gas decreases
Pressure and volume of a
gas are inversely
proportional
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
17
BOYLErsquoS LAW
Boylersquos Law can be mathematically expressed as
P1 x V1 = P2 x V2
P and V under initial
condition
P and V under final
condition
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
18
Example 1
A sample of gas has a volume of 12-L and a pressure of 4500 mmHg What is the volume of the gas when the pressure is reduced to 750 mmHg
1 12
2
P VV =
P
V2 = 72 L
P1= 4500 mmHg
P2= 750 mmHg
V1 = 12 L
V2 =
P1 x V1 = P2 x V2
(4500 mmHg)(12 L)=
(750 mmHg)
Pressure decreases Volume
increases
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
19
Example 2
A sample of hydrogen gas occupies 40 L at 650 mmHg What volume would it occupy at 20 atm
1 12
2
P VV =
P
V2 = 17 L
P1= 650 mmHg
P2= 20 atm
V1 = 40 L
V2 =
P1 x V1 = P2 x V2
(650 mmHg)(40 L)=
(1520 mmHg)Pressures must be in the same
unit
P2 = 20 atm = 1520 mmHgVolume decreases
Pressure increases
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
20
CHARLESrsquoS LAW
The relationship of temperature and volume in gases is called Charlesrsquo Law
At constant pressure the volume of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the volume of the gas increases
Temperature and volume of a gas
are directly proportional
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
21
CHARLESrsquoS LAW
Charlesrsquo Law can be mathematically expressed as
V and T under initial
condition
1 2
1 2
V V=
T T
V and T under final condition
T must be in units of K
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
22
Example 1
A 20-L sample of a gas is cooled from 298K to 278 K at constant pressure What is the new volume of the gas
22 1
1
TV = V x
T
V2 = 19 L
V1 = 20 L
V2 =
T1 = 298 K
T2 = 278 K
278 K= 20 L x
298 K
Volume decreases1 2
1 2
V V=
T T
Temperature decreases
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
23
Example 2
If 200 L of oxygen is cooled from 100ordmC to 0ordmC what is the new volume
22 1
1
TV = V x
T
V2 = 146 L
V1 = 200 L
V2 =
T1 = 100ordmC
T2 = 0ordmC
273 K= 200 L x
373 K
T1= 100ordmC + 273 = 373 K
T2= 0ordmC + 273 = 273 K
Temperatures must be in
KVolume
decreasesTemperature
decreases
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
24
GAY-LUSSACrsquoSLAW
The relationship of temperature and pressure in gases is called Gay-Lussacrsquos Law
At constant volume the pressure of a fixed amount of gas is directly proportional to its absolute temperature
As temperature increases the
pressure of the gas increases
Temperature and pressure of a gas
are directly proportional
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
25
GAY-LUSSACrsquoSLAW
Gay-Lussacrsquos Law can be mathematically expressed as
P and T under initial
condition
1 2
1 2
P P=
T T
P and T under final conditionT must be in
units of K
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
26
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
P1 = 40 atm
P2 =
T1 = 25ordmC
T2 = 400ordmC
T1= 25ordmC + 273 = 298 K
T2= 400ordmC + 273 = 673 K
Temperatures must be in K
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
27
Example 1
An aerosol spray can has a pressure of 40 atm at 25C What pressure will the can have if it is placed in a fire and reaches temperature of 400C
22 1
1
TP = P x
T
P2 = 90 atm
P1 = 40 atm
P2 =
T1 = 298 K
T2 = 673 K
673 K= 40 L x
298 K
Pressure increases
1 2
1 2
P P=
T T
Temperature increases
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
28
Example 2
A cylinder of gas with a volume of 150-L and a pressure of 965 mmHg is stored at a temperature of 55C To what temperature must the cylinder be cooled to reach a pressure of 850 mmHg
22 1
1
PT = T x
P
T2 = 289 K
P1 = 965 mmHg
P2 = 850 mmHg
T1 = 55ordmC
T2 =
850 mmHg= 328 K x
965 mmHg
T1= 55ordmC + 273 = 328 K
T2 = 16ordmC
Temperature decreases
Pressure decreases
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
29
VAPOR PRESSUREamp BOILING POINT
In an open container liquid molecules at the surface that possess sufficient energy can break away from the surface and become gas particles or vapor
In a closed container these gas particles can accumulate and create pressure called vapor pressure
Vapor pressure is defined as the pressure above a liquid at a given temperature Vapor pressure varies with each liquid and increases with temperature
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
30
VAPOR PRESSUREamp BOILING POINT
Listed below is the vapor pressure of water at various temperatures
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
31
VAPOR PRESSUREamp BOILING POINT
A liquid reaches its boiling point when its vapor pressure becomes equal to the external pressure (atmospheric pressure)
For example at sea level water reaches its boiling point at 100C since its vapor pressure is 760 mmHg at this temperature
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
32
VAPOR PRESSUREamp BOILING POINT
At higher altitudes where atmospheric pressure is lower water reaches boiling point at temperatures lower than 100C
For example in Denver where atmospheric pressure is 630 mmHg water boils at 95C since its vapor pressure is 630 mmHg at this temperature
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
33
COMBINEDGAS LAW
All pressure-volume-temperature relationships can be combined into a single relationship called the combined gas law
1 1 2 2
1 2
P V P V=
T T
Initial condition
This law is useful for studying the effect of changes in two variables
Final condition
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
34
COMBINEDGAS LAW
The individual gas laws studied previously are embodied in the combined gas law
1 1 2 2
1 2
P V P V=
T T
Boylersquos LawCharlesrsquos
LawGay-Lussacrsquos
Law
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
35
Example 1
A 250-mL sample of gas has a pressure of 400 atm at a temperature of 10C What is the volume of the gas at a pressure of 100 atm and a temperature of 18C
1 22 1
2 1
P TV = V x x
P T
V2 = 103 mL
P1= 400 atm V1= 250 mL T1= 283 K P2= 100 atm V2= T2 = 291 K
2
400 atm 291 KV = 250 mL x x
100 atm 283 K
1 1 2 2
1 2
P V P V=
T TBoth pressure
and temp change
Must use combined
gas law
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
36
AVOGADROrsquoSLAW
The relationship of moles and volume in gases is called Avogadrorsquos Law
At constant temperature and pressure the volume of a fixed amount of gas is directly proportional to the number of moles
As number of moles increases
the volume of the gas
increases
Number of moles and volume of a gas are directly
proportional
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
Tros Introductory Chemistry Chapter
37
Avogadrorsquos Law
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
38
AVOGADROrsquoSLAW
As a result of Avogadrorsquos Law equal volumes of different gases at the same temp and pressure contain equal number of moles (molecules)
2 Tanks of gas of equal volume at the same T amp P contain the same number of molecules
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
39
AVOGADROrsquoSLAW
For example
2 H2 (g) + 1 O2 (g) 2 H2O (g)
2 molecules 1 molecule 2 molecules
2 moles 1 mole 2 moles
2 Liters 1 Liter 2 Liters
Avogadrorsquos Law also allows chemists to relate volumes and moles of a gas in a chemical reaction
This relationship is only valid for
gaseous substances
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
40
Example 1
A sample of helium gas with a mass of 180 g occupies 16 Liters at a particular temperature and pressure What mass of oxygen would occupy16 L at the same temperature and pressure
1 mol= 180 g x
400 gmol of He = 450 mol
mol of O2 = mol of He = 450 mol
Same Temp amp Pressure
mass of O2
320 g= 450 mol x
1 mol= 144 g
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
41
Example 2
How many Liters of NH3 can be produced from reaction of 18 L of H2 with excess N2 as shown below
18 L H23
2
L NHx
L H= 12 L NH3
N2 (g) + 3 H2 (g) 2 NH3 (g)
32
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
42
STP amp MOLAR VOLUME
To better understand the factors that affect gas behavior a set of standard conditions have been chosen for use and are referred to as Standard Temperature and Pressure (STP)
STP =
1 atm (760 mmHg)
0ordmC (273 K)
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
43
STP amp MOLAR VOLUME
At STP conditions one mole of any gas is observed to occupy a volume of 224 L
V = 224 L
Molar Volume at
STP
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
44
Example 1
If 200 L of a gas at STP has a mass of 323 g what is the molar mass of the gas
mol200 L x
Lmol of gas =
2241
= 00893 mol
g
molMolar mass =
323 g=
00893 mol= 362 gmol Molar
volume at STP
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
45
Example 2
A sample of gas has a volume of 250 L at 730 mmHg and 20C What is the volume of this gas at STP
1 22 1
2 1
P TV = V x x
P T
V2 = 224 L
P1= 730 mmHg V1= 250 L T1= 293 K P2= 760 mmHg V2= T2 = 273 K
2
730 mmHg 273 KV = 250 L x x
760 mmHg 293 K
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
46
IDEAL GAS LAW
Combining all the laws that describe the behavior of gases one can obtain a useful relationship that relates the volume of a gas to the temperature pressure and number of moles
n R TV =
P
Universal gas constant
R = 00821 L atmmol K
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
47
IDEAL GAS LAW
This relationship is called the Ideal Gas Law and commonly written as
P V = n R T
Pressure in atm Volume
in Liters
Number of moles
Temp in K
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
48
Example 1
A sample of H2 gas has a volume of 856 L at a temp of 0C and pressure of 15 atm Calculate the moles of gas present
PVn =
RT
= 057 mol
P = 15 atm V = 856 L T = 273 K n = R = 00821
PV = nRT
(15 atm)(856 L)=
L atm(00821 )(273 K)
mol K
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
49
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
P =750 mmHg V = T = 283 K n = not given m = 400 g R = 00821
mol of N2 =1 mol
400 g x280 g
= 143 mol
P = 750 mmHg1 atm
x760 mmHg
Moles must be calculated
from mass
Pressure must be
converted to atm
P = 0987 atm
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
50
Example 2
What volume does 400 g of N2 gas occupy at 10C and 750 mmHg
nRTV =
P
V = 337 L
P =0987 atm V = T = 283 K n = 143 R = 00821
L atm(143 mol)(00821 )(283 K)
mol KV = 0987 atm
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
Tros Introductory Chemistry Chapter
51
Ideal vs Real Gases
bull Real gases often do not behave like ideal gases at high pressure or low temperature
bull Ideal gas laws assume1) no attractions between gas molecules2) gas molecules do not take up space based on the Kinetic-Molecular Theory
bull at low temperatures and high pressures these assumptions are not valid
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
Tros Introductory Chemistry Chapter
52
Ideal vs Real Gases
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
53
PARTIAL PRESSURES
Many gas samples are mixture of gases For example the air we breathe is a mixture of mostly oxygen and nitrogen gases
Since gas particles have no attractions towards one another each gas in a mixture behaves as if it is present by itself and is not affected by the other gases present in the mixture
In a mixture each gas exerts a pressure as if it was the only gas present in the container This pressure is called partial pressure of the gas
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
54
DALTONrsquoSLAW
In a mixture the sum of all the partial pressures of gases in the mixture is equal to the total pressure of the gas mixture
This is called Daltonrsquos law of partial pressures
Total pressure of a gas mixture
Ptotal = P1 + P2 + P3 +
Sum of the partial pressures of the gases in the mixture
=
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
55
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = PHe + PAr
Total pressure of a gas mixture is the sum of partial pressures of each gas in the mixture
= 60 atm
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
56
PARTIAL PRESSURES
The partial pressure of each gas in a mixture is proportional to the amount (mol) of gas present in the mixture
For example in a mixture of gases consisting of 1 mole of nitrogen and 1 mol of hydrogen gas the partial pressure of each gas is one-half of the total pressure in the container
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
57
PARTIAL PRESSURES
PHe = 20 atm
+
PAr = 40 atm PTotal = 60 atm
The partial pressure of each gas is proportional to the amount (mol) of the gas present in the mixture
Twice as many moles of Ar
compared to He
PAr = 2 PHe
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
58
Example 1A scuba tank contains a mixture of oxygen and helium gases with total pressure of 700 atm If the partial pressure of oxygen in the tank is 1140 mmHg what is the partial pressure of helium in the tank
Ptotal = Poxygen + Phelium
Poxygen = 1140 mmHg x
1 atm
760 mmHg= 150 atm
Phelium
= PTotal - Poxygen = 700 atm ndash 150 atm = 550 atm
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
59
Example 2
A mixture of gases contains 20 mol of O2 gas and 40 mol of N2 gas with total pressure of 30 atm What is the partial pressure of each gas in the mixture
Ptotal = Poxygen + Pnitrogen
Pnitrogen = 2 Poxygen
Twice as many moles of N2
compared to O2
Pnitrogen = 23 (Ptotal ) = 20 atm
Poxygen = 13 (Ptotal ) = 10 atm
60
THE END
60
THE END