GASESGASES

Chapter 5

A GasA Gas

- Uniformly fills any container.

- Mixes completely with any other gas

- Exerts pressure on its surroundings.

05_47

h = 760mm Hgfor standardatmosphere

Vacuum

Simple barometer invented by Evangelista Torricelli

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

05_48

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

05_1541

Pext

Pext

Volume is decreased

Volume of a gas decreases as pressure increases at constant temperature

05_50

P (

in H

g)

0

V(in3)

20 40 60

50

100

PP2

V

2V

(a)

00

1/P (in Hg)

0.01 0.02 0.03

20

40slope = k

(b)

V(in3 )

BOYLE’S LAW DATA

P vs V V vs 1/P

Boyle’s LawBoyle’s Law**

(Pressure)( Volume) = Constant (T = constant)

P1V1 = P2V2 (T = constant)

V 1/P (T = constant)

(*Holds precisely only at very low pressures.)

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?

A gas that A gas that strictly obeys strictly obeys Boyle’s Law is called an Boyle’s Law is called an

ideal gasideal gas..

05_51

PV

(L

• at

m)

022.25

P (atm)

1.000.750.500.25

22.30

22.35

22.40

22.45Ne

O2

CO2

Ideal

Plot of PV vs. P for several gases at pressuresbelow 1 atm.

05_53

V(L

)

-300

T(ºC)

-200 -100 0 100 200

1

2

3

6

4

5

300

-273.2 ºC

N2O

H2

H2O

CH4

He

Plot of V vs. T(oC) for several gases

05_1543

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.

V = bT (P = constant)

b = a proportionality constant

Charles’s LawCharles’s Law

VT

VT

P1

1

2

2 ( constant)

Charles’s Law CalculationsCharles’s Law CalculationsConsider a gas with a a volume of 0.675 L at 35 oC and

1 atm pressure. What is the temperature (in Co) of the gas when its volume is 0.535 L at 1 atm pressure?

V1 = 0.675 L

V2 = 0.535 L

T1 = 35 oC + 273 = 308 K

T2 = ?

V1/V2 = T1/T2

T2 = T1 V2/V1

T2 = (308 K)(0.535 L)/(0.675 L)T2 = 244 K -273 T2 = - 29 oC

05_1544

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

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

V = an

a = proportionality constant

V = volume of the gas

n = number of moles of gas

AVOGADRO’S LAWAVOGADRO’S LAW

V1/V2 = n1/n2

AVOGADRO’S LAWAVOGADRO’S LAW

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

V1/V2 = n1/n2

V2 = V1 n2/n1

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

COMBINED GAS LAWCOMBINED GAS LAW

V1/ V2 = P2 T1/ P1 T2

COMBINED GAS LAWCOMBINED GAS LAW

V1/ V2 = P2 T1/ P1 T2

V2 = V1P1T2/P2T1

V2 = (309 K)(0.454 atm)(3.48 L)

(258 K)(0.616 atm)

V2 = 3.07 L

What will be the new volume of a gas under the following conditions?

V1 = 3.48 LV2 = ?P1 = 0.454 atmP2 = 0.616 atmT1 = - 15 oC + 273 = 258 KT2 = 36 oC + 273T2= 309 K

05_1542

Pext

Pext

Temperature is increased

Pressure exerted by a gas increases as temperature increases provided volume remains constant.

If the volume of a gas is held constant, then V1 / V2 = 1.Therefore:

P1 / P2 = T1 / T2

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.

What is an example of an ideal gas?

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.

What is an example of a real gas?

Ideal Gas LawIdeal Gas Law

PV = nRT

R = proportionality constant

= 0.08206 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 = 0.08206 Latm/molK

pV = nRTp = nRT/Vp = (1.5mol)(0.08206Latm/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 = 0.08206 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

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

GAS STOICHIOMETRYGAS STOICHIOMETRY

1. Mass-Volume

2. Volume-Volume

Molar VolumeMolar Volume

pV = nRT

V = nRT/p

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

(1.00 atm)

V = 22.4 L

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

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

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-VolumeIf 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 same temperature and pressure.

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

(25.0 L H2)(2 mol NH3/3 mol H2) = 16.7 L NH3

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

Molar Mass CalculationsMolar Mass Calculations

M = ? d = 1.95 g/L

T = 27 oC + 273 p = 1.50 atm

T = 300. K R = 0.08206 Latm/mol K

M = dRT/pM = (1.95g/L)(0.08206Latm/molK)(300.K)

(1.5atm)M = 32.0 g/mol

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 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

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)

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.

05_58

Rela

tive

nu

mbe

r o

f O

2 m

ole

cule

sw

ith g

ive

n v

elo

city

0

Molecular velocity (m/s)

4 x 102 8 x102

Plot of relative number of oxygen molecules with a given velocity at STP (Boltzmann Distribution).

05_59

Rel

ativ

e nu

mbe

r of

N2

mol

ecul

esw

ith g

iven

vel

ocity

0Velocity (m/s)1000 2000 3000

273 K

1273 K

2273 K

Plot of relative number of nitrogen molecules witha given velocity at three different temperatures.

The Meaning of The Meaning of TemperatureTemperature

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

(K E )3

2av g R T

M

RTrms

3

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.

05_60

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:

Real GasesReal Gases

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

05_62

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

05_68

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!!