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Chapter 11 Gas Laws
The Gas Phase• Gases have no distinct volume or shape.• Gases expand to fill the volume of their
container.
• Gas particles are miscible with each other.
• Evidence for gas particles being far apart : We can see through gases We can walk through gases Gases are compressible Gases have low densities
Composition of Earth’s AtmosphereCompound %(Volume) Mole Fractiona
Nitrogen 78.08 0.7808
Oxygen 20.95 0.2095
Argon 0.934 0.00934
Carbon dioxide 0.033 0.00033
Methane 2 x 10-4 2 x 10-6
Hydrogen 5 x 10-5 5 x 10-7
a. mole fraction = mol component/total mol in mixture.
The Air We Breathe
Kinetic Theory of GasesKinetic Theory Postulates:
• Gas particles are sizeless relative to the volume of the gas
• Gas particles are in constant rapid motion
• Gas particles have elastic collisions; means no kinetic energy is lost on impact.
• The absolute temperature is directly proportional to the kinetic energy of a gas.
• Gas particles have no attraction to each other; i.e. no inter particle froces.
Parameters Affecting Gases
• Pressure (P); atm, mmHg, torr, lbs/in2
• Volume (V); L, mL
• Temperature (T); K (only)
• Number of Moles (n)
PressurePressure is equal to force/unit area (P =F/A) lbs/in2
Force is a push which comes from gas particles striking a container wall
Pressure Units• SI units = Newton/meter2 = 1 Pascal (Pa)• 1 standard atmosphere (atm) = 101,325 Pa
1 atm =760 mm Hg 1 atm = 760 torr (torr is abbreviation of mmHg) 1 atm = 14.7 lbs/in2
1 atm = 1.013 barr Barr = 100 kPa
Measurement of Pressure
What is above mercury?
Measurement of Pressure
Elevation and Atmospheric Pressure
Units for Expressing Pressure
Unit Value
Atmosphere 1 atm
Pascal (Pa) 1 atm = 1.01325 x 105 Pa
Kilopascal (kPa) 1 atm = 101.325 kPa
mmHg 1 atm = 760 mmHg
Torr 1 atm = 760 torr
Bar 1 atm = 1.01325 bar
mbar 1 atm = 1013.25 mbar
psi 1 atm = 14.7 psi
Pressure MeasurementOpen Tube Manometer Is the atmosphere or the gas
in the canister pushing harder?
= 15 mm
Pressure MeasurementOpen Tube Manometer Is the atmosphere or the gas
in the canister pushing harder? Gas in the canister
If the atmospheric pressure is 766 mm, then what is the pressure of the canister?
= 15 mm
Pressure MeasurementOpen Tube Manometer Is the atmosphere or the gas in
the canister pushing harder? Gas in the canister
If the atmospheric pressure is 766 mm, then what is the pressure of the canister?P = 766 + 15 = 781 mm (torr)
= 15 mm
gas
Pressure MeasurementOpen Tube Manometer Is the atmosphere or the gas
in the canister pushing harder?
gas
Pressure MeasurementOpen Tube Manometer Is the atmosphere or the gas
in the canister pushing harder? The atmosphere
What is the pressure of the gas if the atmosphere is 766 mm?
= 13 mm
gas
Pressure MeasurementOpen Tube Manometer Is the atmosphere or the gas
in the canister pushing harder? The atmosphere
What is the pressure of the gas if the atmosphere is 766 mm? 753 mm
= 13 mm
gas
Pressure MeasurementOpen Tube Manometer
gas
Now what is pushing harder, the gas or the atomosphere?
Pressure MeasurementOpen Tube Manometer
gas
Now what is pushing harder, the gas or the atmosphere?Neither, both the same.
Pressure MeasurementOpen Tube Manometer
gas
Now what is pushing harder, the gas or the atmosphere?Neither, both the same.
Is the gas canister empty?
Pressure MeasurementOpen Tube Manometer
gas
Now what is pushing harder, the gas or the atmosphere?Neither, both the same.
Is the gas canister empty?No, completely full of gas!
Dalton’s Law of Partial Pressures
• For a mixture of gases in a container
• PTotal = P1 + P2 + P3 + . . .
Boyles Law
Atm
●
● ●
●
Consider a gas in a closed system containing a movable plunger. If the plunger is not moving up or down, what can be said about the pressure of the gas relative to the atmospheric pressure?
Boyles Law
Atm
●
● ●
●
Suppose we add some red gas to the container, what would happen to the collisions of gas particles with container walls. Would they increase, decrease or stay the same?
●●
●
Boyles Law
Atm
●
● ●
●
Suppose we add some red gas to the container, what would happen to the collisions of gas particles with container walls. Would they increase, decrease or stay the same?
More particles, more collisions, and more pressure.
What happens to the plunger?
●●
●
Boyles Law
Atm
●
● ●
●
Suppose we add some red gas to the container, what would happen to the collisions of gas particles with container walls. Would they increase, decrease or stay the same?
More particles, more collisions, and more pressure.
What happens to the plunger?
●●
●
Boyles Law
●●
●
●●●
●●
●●
The number of particles remain the same, but the surface area they have to strike increases, thus the number of collisions per square inch decrease as the plunger goes up exposing more surface area causing a decrease in pressure.
Boyle’s Law
• P 1/V (T and n fixed)
• P V = Constant
• P1V1 = P2V2
Pressure and volume are inversely proportional.
Charles’s Law• The volume of a gas is
directly proportional to Kelvin temperature, and extrapolates to zero at zero Kelvin.
V T (P & n are constant)
V1 = V2
T1 T2
Combined Gas Law• Combining the gas laws the relationship
P T(n/V) can be obtained.
• If n (number of moles) is held constant, then PV/T = constant.
P1V1
T1
= P2V2
T2
Volume: L, mL, cm3, …Pressure: Atm, mmHg, Torr, PSI, KPaTemperature, K (only)
Example A balloon is filled with hydrogen to a A balloon is filled with hydrogen to a
pressure of 1.35 atm and has a volume of pressure of 1.35 atm and has a volume of 2.54 L. If the temperature remains constant, 2.54 L. If the temperature remains constant, what will the volume be when the pressure is what will the volume be when the pressure is increased to 2.50 atm?increased to 2.50 atm?
P1V1
T1
= P2V2
T2
(1.35 atm)(2.54 L)T1
= (2.50atm)V2T1
Constant Temp. means T1=T2
(1.35 atm)(2.54 L)(2.50atm)V2 =
V2 = 1.37 L
Example
A sample of oxygen gas is at 0.500 atm and occupies a volume of 11.2 L at 00C, what volume will the gas occupy at 6.00 atm at room temperature (250C)?
Ideal Gas Law
PV = nRT
R = universal gas constant = 0.08206 L atm K-1 mol-1
P = pressure in atm V = volume in liters n = moles T = temperature in Kelvin
STP• “STP” means standard temperature and standard
pressure P = 1 atmosphere T = 0C The molar volume of an ideal gas is 22.42 liters
at STP (put 1 mole, 1 atm, R, and 273 K in the ideal gas law and calculate V)
Example Calculate the pressure of a 1.2 mol sample of Calculate the pressure of a 1.2 mol sample of
methane gas in a 3.3 L container at 25°C.methane gas in a 3.3 L container at 25°C.
Example Calculate the pressure of a 1.2 mol sample of Calculate the pressure of a 1.2 mol sample of
methane gas in a 3.3 L container at 25°C.methane gas in a 3.3 L container at 25°C.
0.0821 L-atmMole-K
Example Calculate the pressure of a 1.2 mol sample of Calculate the pressure of a 1.2 mol sample of
methane gas in a 3.3 L container at 25°C.methane gas in a 3.3 L container at 25°C.
0.0821 L-atmMole-K 3.3 L
Example Calculate the pressure of a 1.2 mol sample of Calculate the pressure of a 1.2 mol sample of
methane gas in a 3.3 L container at 25°C.methane gas in a 3.3 L container at 25°C.
0.0821 L-atmMole-K 3.3 L
298 K
Example Calculate the pressure of a 1.2 mol sample of Calculate the pressure of a 1.2 mol sample of
methane gas in a 3.3 L container at 25°C.methane gas in a 3.3 L container at 25°C.
0.0821 L-atmMole-K 3.3 L
298 K 1.2 mole
Example Calculate the pressure of a 1.2 mol sample of Calculate the pressure of a 1.2 mol sample of
methane gas in a 3.3 L container at 25°C.methane gas in a 3.3 L container at 25°C.
0.0821 L-atmMole-K 3.3 L
298 K 1.2 mole = 8.9 atm
Example
An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.
Example
An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.
mole-K0.0821 L-atm
Example
An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.
mole-K0.0821 L-atm 0.495 g
Example
An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.
mole-K0.0821 L-atm 0.495 g
10-3 LmL
Example
An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.
mole-K0.0821 L-atm 0.495 g
10-3 LmL
127 mL
Example
An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.
mole-K0.0821 L-atm 0.495 g
10-3 LmL
127 mL760 torratm
Example
An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.
mole-K0.0821 L-atm 0.495 g
10-3 LmL
127 mL760 torratm 754 torr
Example
An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.
mole-K0.0821 L-atm 0.495 g
10-3 LmL
127 mL760 torratm 754 torr
371 K
Example
An experiment shows that a 0.495 g sample An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at of an unknown gas occupies 127 mL at 98°C and 754 torr pressure. Calculate the 98°C and 754 torr pressure. Calculate the molar mass of the gas.molar mass of the gas.
mole-K0.0821 L-atm 0.495 g
10-3 LmL
127 mL760 torratm 754 torr
371 K
= 120 g/mole
Collecting a Gas Over Water
A sample of KClO3 is heated and decomposes to produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?
Hint: The gas collected is a mixture so use Hint: The gas collected is a mixture so use Dalton’s Law to calculate the pressure of oxygen Dalton’s Law to calculate the pressure of oxygen then the ideal gas law to find the number of then the ideal gas law to find the number of moles oxygen.moles oxygen.
PT = P + P
Practice
O2 H2O
Vapor Pressure of Water
A sample of KClO3 is heated and decomposes to produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?
Hint: The gas collected is a mixture so use Hint: The gas collected is a mixture so use Dalton’s Law to calculate the pressure of oxygen Dalton’s Law to calculate the pressure of oxygen then the ideal gas law to find the number of then the ideal gas law to find the number of moles oxygen.moles oxygen.
PT = P + P
Practice
O2 H2O
755 torr = PO2+ 23.8 torr
P = 755 – 23.8 = 731 torr O2
Practice• A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?
mole-K0.0821 L-atm
Practice• A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?
mole-K0.0821 L-atm
atm760 torr
Practice• A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?
mole-K0.0821 L-atm
atm760 torr
731 torr
Practice• A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?
mole-K0.0821 L-atm
atm760 torr
731 torr298 K
Practice• A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?
mole-K0.0821 L-atm
atm760 torr
731 torr298 K mL
10-3 L
Practice• A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water displacement at 25°C. The total volume of the collected gas is 229 mL at a pressure of 755 torr. How many moles of oxygen formed?
mole-K0.0821 L-atm
atm760 torr
731 torr298 K mL
10-3 L 229 mL
= 9.00 X 10-3 mole
Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)
13.4 g Zn
Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)
13.4 g Zn
65.39 g Zn
mole Zn
Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)
13.4 g Zn
65.39 g Zn
mole Znmole Znmole H2
Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)
13.4 g Zn
65.39 g Zn
mole Znmole Znmole H2 0.08206 L-atm
mole H2-K
Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)
13.4 g Zn
65.39 g Zn
mole Znmole Znmole H2 0.08206 L-atm
mole H2-K298.15 K
Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)
13.4 g Zn
65.39 g Zn
mole Znmole Znmole H2 0.08206 L-atm
mole H2-K298.15 K 760 torr
Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)
13.4 g Zn
65.39 g Zn
mole Znmole Znmole H2 0.08206 L-atm
mole H2-K298.15 K 760 torr
atm 766 torr
Stoichiometry and GasesCalculate the volume of hydrogen gas at 25°C and 766 torr from 13.4 g of zinc and an excess of hydrochloric acid.
Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)
13.4 g Zn
65.39 g Zn
mole Znmole Znmole H2 0.08206 L-atm
mole H2-K298.15 K 760 torr
atm 766 torr= 4.97 L
The End