GASESGASES

Chapter 5

A GasA Gas

- Uniformly fills any container.

- Mixes completely with any other gas

- Exerts pressure on its surroundings.

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h = 760mm Hgfor standardatmosphere

Vacuum

Simple barometer invented by Evangelista Torricelli

BarometerBarometerThe pressure of the

atmosphere at sea level will hold a column of mercury 760 mm Hg.

1 atm = 760 mm Hg

1 atm Pressure

760 mm Hg

Vacuum

ManometerManometer

Gas

h

Column of mercury to measure pressure.

h is how much lower the pressure is than outside.

ManometerManometer

h is how much higher the gas pressure is than the atmosphere.h

Gas

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h h

Atmosphericpressure (Patm)

Atmosphericpressure (Patm)

Gaspressure (Pgas)less thanatmospheric pressure

Gaspressure (Pgas)greater thanatmospheric pressure

(Pgas) = (Patm) - h (Pgas) = (Patm) + h

(a) (b)

Simple Manometer

Units of pressureUnits of pressure1 atmosphere = 760 mm Hg

1 mm Hg = 1 torr

1 atm = 101,235 Pascals = 101.325 kPa

14.69 psi=1Atm

Can make conversion factors from these.

What is 724 mm Hg in atm

in torr?

in? kPa?

PressurePressure

- is equal to force/unit area

- SI units = Newton/meter2 = 1 Pascal (Pa)

- 1 standard atmosphere = 101,325 Pa

- 1 standard atmosphere = 1 atm =

760 mm Hg = 760 torr

Pressure Unit ConversionsPressure Unit ConversionsThe pressure of a tire is measured to be 28

psi. What would the pressure in atmospheres, torr, and pascals.

(28 psi)(1.000 atm/14.69 psi) = 1.9 atm

(28 psi)(1.000 atm/14.69 psi)(760.0 torr/1.000atm) = 1.4 x 103 torr

(28 psi)(1.000 atm/14.69 psi)(101,325 Pa/1.000 atm) = 1.9 x 105 Pa

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Pext

Pext

Volume is decreased

Volume of a gas decreases as pressure increases at constant temperature

Boyle’s LawBoyle’s Law**

P1V1 = P2V2 (T = constant)

(*Holds precisely only at very low pressures.)

Boyle’s LawPressure and volume are inversely related at constant temperature.

V

P (at constant T)

Boyle’s Law CalculationsBoyle’s Law CalculationsA 1.5-L sample of gaseous CCl2F2 has a pressure of

56 torr. If the pressure is changed to 150 torr, will the volume of the gas increase or decrease? What will the new volume be?

Decrease

P1 = 56 torr

P2 = 150 torr

V1 = 1.5 L

V2 = ?

V1P1 = V2P2

V2 = V1P1/P2

V2 = (1.5 L)(56 torr)/(150 torr)V2 = 0.56 L

Boyle’s Law CalculationsBoyle’s Law CalculationsIn an automobile engine the initial cylinder

volume is 0.725 L. After the piston moves up, the volume is 0.075 L. The mixture is 1.00 atm, what is the final pressure?

P1 = 1.00 atm

P2 = ?

V1 = 0.725 L

V2 = 0.075 L

V1P1 = V2P2

P2 = V1P1/V2

P2 = (0.725 L)(1.00 atm)/(0.075 L)P2 = 9.7 atm

Is this answer reasonable?

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Pext

Pext

Energy (heat) added

Volume of a gas increases as heat is addedwhen pressure is held constant.

Charles’s LawCharles’s Law

The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin.

Charles’s LawCharles’s Law

VT

VT

P1

1

2

2 ( constant)

ExamplesExamples

What would the final volume be if 247 mL of gas at 22ºC is heated to 98ºC , if the pressure is held constant?

Combined Gas LawCombined Gas Law

If the moles of gas remains constant, use this formula and cancel out the other things that don’t change.

P1 V1 = P2 V2

. T1 T2

ExamplesExamples

A deodorant can has a volume of 175 mL and a pressure of 3.8 atm at 22ºC. What would the pressure be if the can was heated to 100.ºC?

Avogadro’s LawAvogadro’s Law

For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures).

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Increase volume to return to originalpressure

Gas cylinder

Moles of gasincreases

Pext Pext

Pext

At constant temperature and pressure, increasing the moles of a gas increases its volume.

AVOGADRO’S LAWAVOGADRO’S LAW

V1/n1 = V2/n2

Avogadro simpleAvogadro simple

A 5.20L sample at 18.0 C and 2.00 atm pressure contains 0.436 moles of gas . If we add an additional 1.27 moles of the gas at the same temperature and pressure, what will the total volume occupied by the gas be ?

AVOGADRO’S LAW difAVOGADRO’S LAW dif

A 12.2 L sample containing 0.50 mol of oxygen gas, O2, at a pressure of 1.00 atm and a temperature of 25 oC is converted to ozone, O3, at the same temperature and pressure, what will be the volume of the ozone? 3 O2(g) ---> 2 O3(g)

(0.50 mol O2)(2 mol O3/3 mol O2) = 0.33 mol O3

V1 = 12.2 L

V2 = ?

n1 = 0.50 mol

n2 = 0.33 mol

V2 = (12.2 L)(0.33 mol)/(0.50 mol)V2 = 8.1 L

Ideal Gas LawIdeal Gas Law

- An equation of state for a gas.

- “state” is the condition of the gas at a given time.

PV = nRT

IDEAL GASIDEAL GAS

1. Molecules are infinitely far apart.

2. Zero attractive forces exist between the molecules.

3. Molecules are infinitely small--zero molecular volume.

REAL GASREAL GAS

1. Molecules are relatively far apart compared to their size.

2. Very small attractive forces exist between molecules.

3. The volume of the molecule is small compared to the distance between molecules.

Ideal Gas LawIdeal Gas Law

PV = nRT

R = proportionality constant

= 0.0821 L atm mol

P = pressure in atm

V = volume in liters

n = moles

T = temperature in Kelvins

Holds closely at P < 1 atm

Ideal Gas Law CalculationsIdeal Gas Law CalculationsA 1.5 mol sample of radon gas has a volume of 21.0 L at

33 oC. What is the pressure of the gas?

p = ?

V = 21.0 L

n = 1.5 mol

T = 33 oC + 273

T = 306 K

R = 0.0821 Latm/molK

pV = nRTp = nRT/Vp = (1.5mol)(0.0821Latm/molK)(306K)

(21.0L)p = 1.8 atm

Ideal Gas Law CalculationsIdeal Gas Law CalculationsA sample of hydrogen gas, H2, has a volume of 8.56 L at a

temperature of O oC and a pressure of 1.5 atm. Calculate the number of moles of hydrogen present.

p = 1.5 atm

V = 8.56 L

R = 0.0821 Latm/molK

n = ?

T = O oC + 273

T = 273K

pV = nRTn = pV/RTn = (1.5 atm)(8.56L) (0.08206 Latm/molK)(273K)n = 0.57 mol

Standard Temperature Standard Temperature and Pressureand Pressure

“STP”

P = 1 atmosphere

T = C or 273K

The molar volume of an ideal gas is 22.42 liters at STP

Molar VolumeMolar Volume

pV = nRT

V = nRT/p

V = (1.00 mol)(0.08206 Latm/molK)(273K)

(1.00 atm)

V = 22.4 L

Gases at STPGases at STP

A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N2 are present?

(1.75L N2)(1.000 mol/22.4 L) = 7.81 x 10-2 mol N2

MOLAR MASS OF A GASMOLAR MASS OF A GAS

n = m/M

n = number of moles

m = mass

M = molar mass

MOLAR MASS OF A GASMOLAR MASS OF A GAS

P = mRT/VM

or

P = DRT/M

therefore:

M = DRT/P

A gas at 34.0C and 1.75 atm has a density of 3.40 g/L. Calculate the molar mass.

What is the density of 2L of O2 gas at STP?

GAS STOICHIOMETRYGAS STOICHIOMETRY

1. Mass-Volume

2. Volume-Volume

Gas StoichiometryGas Stoichiometryat STPat STP

Quicklime, CaO, is produced by heating calcium carbonate, CaCO3. Calculate the volume of CO2 produced at STP from the decomposition of 152 g of CaCO3. CaCO3(s) ---> CaO(s) + CO2(g)

(152g CaCO3)(1 mol/100.1g)(1mol CO2/1mol CaCO3) (22.4L/1mol) = 34.1L CO2

Note: This method only works when the gas is at STP!!!!!

Volume-VolumeVolume-Volume

If 25.0 L of hydrogen reacts with an excess of nitrogen gas, how much ammonia gas will be produced? All gases are measured at the standard temperature and pressure.

2N2(g) + 3H2(g) ----> 2NH3(g)

16.7 L NH3

Gas StoichiometryGas StoichiometryNot at STP (Continued)Not at STP (Continued)

p = 1.00 atm

V = ?n = 1.28 x 10-1 mol

R = 0.08206 Latm/molK

T = 25 oC + 273 = 298 K

pV = nRTV = nRT/pV = (1.28 x 10-1mol)(0.08206Latm/molK)(298K)

(1.00 atm)V = 3.13 L O2

Dalton’s Law of Dalton’s Law of Partial PressuresPartial Pressures

For a mixture of gases in a container,

PTotal = P1 + P2 + P3 + . . .

Dalton’s Law of Partial Dalton’s Law of Partial Pressures CalculationsPressures Calculations

A mixture of nitrogen gas at a pressure of 1.25 atm, oxygen at 2.55 atm, and carbon dioxide at .33 atm would have what total pressure?

PTotal = P1 + P2 + P3

PTotal = 1.25 atm + 2.55 atm + .33 atm

Ptotal = 4.13 atm

Water Vapor PressureWater Vapor Pressure2KClO3(s) ----> 2KCl(s) + 3O2(g)

When a sample of potassium chlorate is decomposed and the oxygen produced is collected by water displacement, the oxygen has a volume of 0.650 L at a temperature of 22 oC. The combined pressure of the oxygen and water vapor is 754 torr (water vapor pressure at 22 oC is 21 torr). How many moles of oxygen are produced?

Pox = Ptotal - PHOH

Pox = 754 torr - 21 torr

pox = 733 torr

Dalton’s Law 

Mixtures of helium and oxygen can be used in scuba diving tanks to help prevent “the bends.” For a particular dive, 46L He at 25ºC and 1.0 atm and 12 L O2 at 25ºC and 1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure in the tank at 25ºC.

A Volume of 2.0 L of He at 46° C and 1.2 ATM pressure was added to a vessel that contained 4.5L of N2 at STP. What is the total pressure of each gas at STP after the He is added .

RB P 128

MOLE FRACTIONMOLE FRACTION

-- the ratio of the number of moles of a given component in a mixture to the total number of moles of the mixture.

1 = n1/ ntotal

1 = V1/ Vtotal

1 = P1 / Ptotal (volume & temperature constant)

The partial pressure of Oxygen gas was observed to be 156 torr in the air with a total atmosphere pressure of 743 torr . Calculate the mole fraction of O2 present.

Mole FractionMole Fraction

The mole Fraction of nitrogen in the air is .7808. Calculate the partial pressure of N2 in the air when the atmospheric pressure is 760.

Root mean square velociyyRoot mean square velociyy

Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.)

M

RTrms

3

Calculate the root mean Calculate the root mean square velocity for the square velocity for the atoms in a sample of atoms in a sample of helium gas at 25helium gas at 25°°C.C.

and100 and100°°C. C.

Example Example

Calculate the root mean square velocity of carbon dioxide at 25ºC.

Calculate the root mean square velocity of chlorine at 25ºC.

Effusion: describes the passage of gas into an evacuated chamber.

Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.

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Gas

Vacuum

Pinhole

Effusion of a gas into an evacuated chamber

Rate of effusion for gas 1Rate of effusion for gas 2

2

1

MM

Distance traveled by gas 1Distance traveled by gas 2

2

1

MM

Effusion:Effusion:

Diffusion:Diffusion:

Grams Law Grams Law

Calculate the ratio of effusion rates of hydrogen gas and uranium hexafluoride.

A compound effuses through a porous cylinder 3.20 time faster than helium. What is it’s molar mass?

Kinetic Molecular TheoryKinetic Molecular Theory

1. Volume of individual particles is zero.

2. Collisions of particles with container walls cause pressure exerted by gas.

3. Particles exert no forces on each other.

4. Average kinetic energy Kelvin temperature of a gas.

Real GasesReal Gases

Real molecules do take up space and they do interact with each other (especially polar molecules).

Need to add correction factors to the ideal gas law to account for these.

Volume CorrectionVolume Correction

The actual volume free to move in is less because of particle size.

More molecules will have more effect.

Corrected volume V’ = V - nb

b is a constant that differs for each gas.

P’ = nRT

(V-nb)

Pressure correctionPressure correction

Because the molecules are attracted to each other, the pressure on the container will be less than ideal

depends on the number of molecules per liter.

since two molecules interact, the effect must be squared.

Pressure correctionPressure correction Because the molecules are attracted Because the molecules are attracted

to each other, the pressure on the to each other, the pressure on the container will be less than idealcontainer will be less than ideal

depends on the number of molecules depends on the number of molecules per liter.per liter.

since two molecules interact, the since two molecules interact, the effect must be squared.effect must be squared.

Pobserved = P’ - a

2

( )Vn

AltogetherAltogetherPobs= nRT - a n 2

V-nb V

Called the Van der Wall’s equation if

rearranged

Corrected Corrected Pressure Volume

( )

P + an

V x V - nb nRTobs

2

Where does it come fromWhere does it come from

a and b are determined by experiment.

Different for each gas.

Bigger molecules have larger b.

a depends on both size and polarity.

once given, plug and chug.

ExampleExample

Calculate the pressure exerted by 0.5000 mol Cl2 in a 1.000 L container at 25.0ºC

Using the ideal gas law.

Van der Waal’s equation– a = 6.49 atm L2 /mol2 – b = 0.0562 L/mol

Real GasesReal Gases

Must correct ideal gas behavior when at high pressure (smaller volume) and low temperature (attractive forces become important).

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00

P(atm)

200 400 600 800

1.0

2.0

PV

nRTCO2

H2

CH4N2

Ideal

gas

1000

Plots of PV/nRT vs. P for several gases at 200 K.Note the significant deviation from ideal behavior.

Real GasesReal Gases

[ ]P a V nb nRTobs2( / ) n V

corrected pressurecorrected pressure corrected volumecorrected volume

PPidealideal VVidealideal

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Con

cen

tra

tion (

pp

m)

4:0

00

Time of day

0.1

0.2

0.3

0.4

0.5

6:0

0

8:0

0

10

:00

Noo

n

2:0

0

4:0

0

6:0

0

NO

NO2

O3

Molecules of unburnedfuel (petroleum)

Otherpollutants

Concentration for some smog componentsvs. time of day

NO2(g) NO(g) + O(g)

O(g) + O2(g) O3(g)

NO(g) + 1/2 O2(g) NO2(g)__________________________

3/2 O2(g) O3(g)

What substances represent intermediates?

Which substance represents the catalyst?

Chemistry is a gas!!Chemistry is a gas!!