Methane (CH4)

Ammonia (NH3)

Chlorine (Cl2)

Oxygen (O2)

Ethylene (C2H4)

natural deposits; domestic fuel

from N2+ H2; fertilizers, explosives

electrolysis of seawater; bleaching and

disinfecting

liquefied air; steelmaking

high-temperature decomposition of natural gas;

plastics

Name (Formula) Origin and Use

Atmosphere-Biosphere Redox Interconnections

Chapter 5

Gases and the Kinetic-Molecular Theory

Figure 5.1 The three states of matter.

An Overview of the Physical States of Gases

Note The Distinction of Gases from Liquids and Solids

1. Gas volume changes greatly with pressure.

2. Gas volume changes greatly with temperature.

3. Gases have relatively low viscosity.

4. Most gases have relatively low densities under normal conditions.

5. Gases are miscible.

6. Gases are compressible.

7. Gas particles have negligible attraction for each other (ideal

gases).

Figure 5.2 Effect of atmospheric pressure on objects

at the Earth’s surface.

Variations in pressure, temperature, and

composition of the Earth’s atmosphere.

Figure 5.3 A mercury barometer.

Figure 5.4

Two types of

manometer

closed-end

open-end

Table 5.2 Common Units of Pressure

Atmospheric PressureUnit Scientific Field

chemistryatmosphere(atm) 1 atm*

pascal(Pa);

kilopascal(kPa)

1.01325x105Pa;

101.325 kPa

SI unit; physics, chemistry

millimeters of

mercury(Hg)

760 mm Hg* chemistry, medicine, biology

torr 760 torr* chemistry

pounds per square

inch (psi or lb/in2)

14.7lb/in2 engineering

bar 1.01325 bar meteorology, chemistry,

physics

*This is an exact quantity; in calculations, we use as many significant figures as necessary.

A geochemist heats a limestone (CaCO3) sample and

collects the CO2 released in an evacuated flask attached

to a closed-end manometer. After the system comes to

room temperature, h = 291.4 mm Hg. Calculate the CO2

pressure in torrs, atmospheres, and kilopascals.

Exercise # 8: Do problems 1 & 2.

PROBLEM SOLVING

Boyle’s Law n and T are fixedV 1

P

Charles’s Law V T P and n are fixed

V

T= constant V = constant x T

Amontons’s Law P T V and n are fixed

P

T= constant P = constant x T

combined gas law V T

PV = constant x

T

P

PV

T= constant

V x P = constant V = constant / P

Basic Behavior of a Gas

Boyle’s Law: P1V1 = P2V2

Where T and n are held constant.

Figure 5.5 The relationship between the volume

and pressure of a gas.

Boyle’s Law

Problem Solving

Do #’s 3 & 4 of Exercise 8

Relationship of Pressure and

Temperature

Charles’ Law: V1 = V2

T1 T2

Where P and n are held constant.

**** Temperature must be in Kelvin****

Figure 5.6

The relationship between the

volume and temperature of a

gas.

Charles’s Law

Problem Solving

Do Problem #’s 5 & 6 of Exercise 8

Relationship Between Temperature

and Pressure

Amonton’s Law: P1 = P2

T1 T2

Where V and n are held constant.

**** Temperature must be in Kelvin****

Relationship Between Volume,

Temperature, and Pressure

The Combined Gas Law: P1V1 = P2V2

T1 T2

Where n is held constant.

Do problem #’s 7 & 8 of Exercise 8

The Relationship Between the Volume and Amount of a Gas.

Avogadro’s Law

Equal volumes of a gas will contain the

same number of particles if P & T are held

constant.

At standard temperature, 273.15 K and

pressure, 1.00 atm (STP)

1 mole of gas = 22.4 L

Figure 5.8 Standard molar volume.

Problem Solving

Do problems 9 & 10 of Exercise 8

Figure 5.9 The volume of 1 mol of an ideal gas compared with some

familiar objects.

THE IDEAL GAS LAW

PV = nRT

IDEAL GAS LAW

nRT

PPV = nRT or V =

Boyle’s Law

V =constant

P

R = PV

nT=

1atm x 22.414L

1mol x 273.15K=

0.0821atm*L

mol*K

V = V =

Charles’s Law

constant X T

Avogadro’s Law

constant X n

fixed n and T fixed n and P fixed P and T

Figure 5.10

R is the universal gas constant

3 significant figures

Problem Solving

Do problem #’s 11 & 12 of Exercise 8

Determination of Density of a Gas

Think about what units you will need. ? Grams

? Volume

Gas densities are usually expressed in g/L (m/V)

Consider: PV = nRT, then

Recall that n (moles) = m(mass)/MM(molecular mass)

So PV = m RT and rearranging we get m = d = MM x P

MM V RT

Do Problem #15 of Exercise 8

Determination of The Molar Mass of a Gas from its Density

n =mass

M=

PV

RT

M =

M = d RT

P

m RT

VPd =

m

V

Determination of the Molar Mass of

a Gas

Think about what units you will need to

determine.

? Grams

? Moles

Do Problem #’s 13 and 17 of Exercise 8

Figure 5.11

Determining the molar

mass of an unknown

volatile liquid.

based on the method of

J.B.A. Dumas (1800-1884)

Problem Solving

An organic chemist isolates a colorless liquid from a

petroleum sample. She uses the Dumas method and

obtains the following data:

Volume of flask =

213 mL

Mass of flask + gas =

78.416 g

T = 100.00C

Mass of flask =

77.834 g

P = 754 torr

Calculate the molar mass of the liquid.

Dalton’s Law of Partial Pressures

Ptotal = P1 + P2 + P3 + ...

P1= 1 x Ptotal where 1 is the mole fraction

1 = n1

n1 + n2 + n3 +...=

n1

ntotal

Mixtures of Gases

•Gases mix homogeneously in any proportions.

•Each gas in a mixture behaves as if it were the only gas present.

Problem Solving

Do problem # 21 of Exercise 8

Collecting a water-insoluble gaseous reaction product and

determining its pressure (Week 8 Lab).

Table 5.3 Vapor Pressure of Water (P ) at Different TH2O

T(0C) P (torr) T(0C) P (torr)

0

5

10

11

12

13

14

15

16

18

20

22

24

26

28

30

35

40

45

50

55

60

65

70

75

80

85

90

95

100

31.8

42.2

55.3

71.9

92.5

118.0

149.4

187.5

233.7

289.1

355.1

433.6

525.8

633.9

760.0

4.6

6.5

9.2

9.8

10.5

11.2

12.0

12.8

13.6

15.5

17.5

19.8

22.4

25.2

28.3

Patm

Ptotal

Water-insoluble

gaseous product

bubbles through

water into

collection vessel

1

Pgas adds to vapor pressure of

water (PH2O) to give Ptotal-

As shown Ptotal < Patm

2

Ptotal

Patm

Water-insoluble

gaseous product

bubbles through

water into

collection vessel

1

Pgas adds to vapor pressure of

water (PH2O) to give Ptotal-

As shown Ptotal < Patm

2

Patm

Ptotal is made equal to

Patm by adjusting height

of vessel until water level

equals that in beaker

3

PtotalPatm

Ptotal

Water-insoluble

gaseous product

bubbles through

water into

collection vessel

1

=

Pgas adds to vapor pressure of

water (PH2O) to give Ptotal-

As shown Ptotal < Patm

2

Patm

Ptotal is made equal to

Patm by adjusting height

of vessel until water level

equals that in beaker

3

PtotalPatm

Ptotal

PH2O

Pgas

Ptotal

Water-insoluble

gaseous product

bubbles through

water into

collection vessel

1

Pgas adds to vapor pressure of

water (PH2O) to give Ptotal-

As shown Ptotal < Patm

2

PH2O

Pgas

Patm =

Ptotal is made equal to

Patm by adjusting height

of vessel until water level

equals that in beaker

3

PtotalPatm

Ptotal

4 Ptotal equals Pgas plus

PH2O at temperature of

experiment. Therefore,

Pgas = Ptotal – PH2O

Ptotal

Water-insoluble

gaseous product

bubbles through

water into

collection vessel

1

Draw Diagram of Lab Apparatus

PT = PH2O + Pgas

PT = barometric pressure

PH2O = vapor pressure of H2O at

a particular temperature

Problem Solving

Do Problem # 24 of Exercise 8

P,V,T

of

reactant

A

amount

(mol)

of reactant A

amount

(mol)

of gas B

P,V,T

of gas B

ideal

gas

law

ideal

gas

law

molar ratio from

balanced equation

Figure 15.13

Summary of the stoichiometric relationships among the

amount (mol,n) of a reactant or product and the gas variables

of pressure (P), volume (V), and temperature (T).

Gas Stoichiometry Problem Solving

Do problem #’s 18,19, 20, and 25 of

Exercise 8

Postulates of the Kinetic-Molecular Theory

Because the volume of an individual gas particle is so

small compared to the volume of its container, the gas

particles are considered to have mass, but no volume.

Gas particles are in constant, random, straight-line

motion except when they collide with each other or with

the container walls.

Collisions are elastic therefore the total kinetic

energy(Ek) of the particles is constant.

Postulate 1: Particle Volume

Postulate 2: Particle Motion

Postulate 3: Particle Collisions

Figure 5.14 Distribution of molecular speeds at three temperatures.

A molecular description of Boyle’s Law.

A molecular description of Dalton’s law of partial pressures.

A molecular description of Charles’s Law.

Figure 5.18 A molecular description of Avogadro’s Law.

Summary: A Molecular View of the

Gas Laws

Particles exert a force when they collide with the walls of a container. The force is proportional to the pressure.

More particles more collisions more pressure

Greater T particles move faster more collisionsmore pressure.

Smaller volume more collisions more pressure.

Graham’s Law of Effusion

Graham’s Law of Effusion

The rate of effusion of a gas is inversely related to the square root of its molar mass.

rate of effusion 1

√M

Recall that KE = ½ mv2

If at some temperature a

heavy particle’s KE = a light particle’s KE

Then the heavy gas must be moving slower than the lighter gas.

Then KEA = KEB

(1/2 mv2)A = (1/2 mv2)B

or

(mA/mB)1/2 = vB/vA

Where m = molar mass

T

H2 (2)

Molecular speed at a given T

Rela

tive n

um

ber

of

mo

lecu

les

wit

h a

giv

en

sp

eed

Relationship Between Molar Mass and Molecular Speed

He (4)

H2 (2)

Molecular speed at a given T

Rela

tive n

um

ber

of

mo

lecu

les

wit

h a

giv

en

sp

eed

H2O (18)

He (4)

H2 (2)

Molecular speed at a given T

Rela

tive n

um

ber

of

mo

lecu

les

wit

h a

giv

en

sp

eed

H2O (18)

He (4)

H2 (2)

Molecular speed at a given T

Rela

tive n

um

ber

of

mo

lecu

les

wit

h a

giv

en

sp

eed

N2 (28)

O2 (32)

H2O (18)

He (4)

H2 (2)

Molecular speed at a given T

Rela

tive n

um

ber

of

mo

lecu

les

wit

h a

giv

en

sp

eed

N2 (28)

Effusion: Gaseous particles passing through a tiny

hole into an evacuated space.

Rateeffusion is proportional to 1/(MW)1/2

Thus . . . , RateA = (MWB)1/2

RateB = (MWA)1/2

Do Problem # 28 of Exercise 8.

Do Problem # 29 of Exercise 8.

Diffusion: Gaseous particles passing

through other gaseous particles.

Ratediffusion is proportional to 1/(MW)1/2

Thus . . . , RateA = (MWB)1/2

RateB = (MWA)1/2

Figure 5.20 Diffusion of a gas particle through a

space filled with other particles.

distribution of molecular speeds

mean free path

collision frequency

Real Gases

• At high temperatures and low pressures

most gases behave “ideally”.

• Consider gases at high pressure.

Figure 5.23 The effect of molecular volume on measured gas volume.

Figure 5.22 The effect of intermolecular attractions on

measured gas pressure.

Real (Non-Ideal) Gases

At high pressures we can no longer say the attractions between gases are negligible.

We can no longer say the volume of the gaseous particles themselves is negligible.

Thus our ideal gas law equation, PV = nRT has to have some correction factors, a and b.

( P + n2/ V)( V + nb ) = nRT

1.0

0.5

0.0

1.5

2.0

PV

RT

Pext (atm)

200 400 600 800 10000

The behavior of several real gases with increasing

external pressure.

1.0

0.5

0.0

1.5

2.0

PV

RT

Pext (atm)

200 400 600 800 10000

Pext (atm)

1.0PV

RT

0 10 20

Ideal gas

Ideal gas

1.0

0.5

0.0

1.5

2.0

PV

RT

Pext (atm)

200 400 600 800 10000

Pext (atm)

1.0PV

RT

0 10 20

Ideal gas

Ideal gas

H2

He

H2

He

1.0

0.5

0.0

1.5

2.0

PV

RT

Pext (atm)

200 400 600 800 10000

Pext (atm)

1.0PV

RT

0 10 20

PV/RT > 1Effect of molecular

volume predominates

H2

Ideal gas

CH4

CO2

He

H2

CH4

CO2

He

Ideal gas

1.0

0.5

0.0

1.5

2.0

PV

RT

Pext (atm)

200 400 600 800 10000

Pext (atm)

1.0PV

RT

0 10 20

PV/RT > 1Effect of molecular

volume predominates

PV/RT < 1Effect of intermolecular

attractions predominates

Table 5.5 Van der Waals Constants for Some Common Gases

0.034

0.211

1.35

2.32

4.19

0.244

1.39

1.36

6.49

3.59

2.25

4.17

5.46

He

Ne

Ar

Kr

Xe

H2

N2

O2

Cl2CO2

CH4

NH3

H2O

0.0237

0.0171

0.0322

0.0398

0.0511

0.0266

0.0391

0.0318

0.0562

0.0427

0.0428

0.0371

0.0305

Gasa

atm*L2

mol2b

L

mol

(Pn2a

V 2)(V nb) nRTVan der Waals

equation for n

moles of a real gas adjusts P up adjusts V down