Student Solutions
Manual to
Accompany Atkins’
Physical Chemistry
ELEVENTH EDITION

Peter Bolgar
Haydn Lloyd
Aimee North
Vladimiras Oleinikovas
Stephanie Smith
and
James Keeler
Department of Chemistry
University of Cambridge
UK

1


1
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Ninth edition 2010
Tenth edition 2014
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Table of contents
Preface

1

2


3

4

5

6

vii

The properties of gases

1

1A The perfect gas

1

1B The kinetic model

12

1C Real gases

24

Internal energy

41


2A Internal energy

41

2B Enthalpy

47

2C Thermochemistry

50

2D State functions and exact differentials

58

2E Adiabatic changes

66

The second and third laws

73

3A Entropy

73

3B Entropy changes accompanying specific processes


79

3C The measurement of entropy

91

3D Concentrating on the system

101

3E Combining the First and Second Laws

107

Physical transformations of pure substances

119

4A Phase diagrams of pure substances

119

4B Thermodynamic aspects of phase transitions

121

Simple mixtures

137


5A The thermodynamic description of mixtures

137

5B The properties of solutions

149

5C Phase diagrams of binary systems: liquids

165

5D Phase diagrams of binary systems: solids

173

5E Phase diagrams of ternary systems

179

5F

184

Activities

Chemical equilibrium

199


6A The equilibrium constant

199


iv

TABLE OF CONTENTS

7

8

9

6B The response of equilibria to the conditions

208

6C Electrochemical cells

221

6D Electrode potentials

228

Quantum theory


243

7A The origins of quantum mechanics

243

7B Wavefunctions

250

7C Operators and observables

254

7D Translational motion

263

7E Vibrational motion

277

7F

288

Rotational motion

Atomic structure and spectra


299

8A Hydrogenic Atoms

299

8B Many-electron atoms

308

8C Atomic spectra

311

Molecular Structure

321

9A Valence-bond theory

321

9B Molecular orbital theory: the hydrogen molecule-ion

324

9C Molecular orbital theory: homonuclear diatomic molecules

329


9D Molecular orbital theory: heteronuclear diatomic molecules

333

9E Molecular orbital theory: polyatomic molecules

339

10 Molecular symmetry

353

10A Shape and symmetry

353

10B Group theory

363

10C Applications of symmetry

374

11 Molecular Spectroscopy

385

11A General features of molecular spectroscopy


385

11B Rotational spectroscopy

394

11C Vibrational spectroscopy of diatomic molecules

408

11D Vibrational spectroscopy of polyatomic molecules

421

11E Symmetry analysis of vibrational spectroscopy

424

11F Electronic spectra

426

11G Decay of excited states

437


TABLE OF CONTENTS

12 Magnetic resonance


445

12A General principles

445

12B Features of NMR spectra

449

12C Pulse techniques in NMR

458

12D Electron paramagnetic resonance

467

13 Statistical thermodynamics

473

13A The Boltzmann distribution

473

13B Partition functions

477


13C Molecular energies

487

13D The canonical ensemble

496

13E The internal energy and entropy

497

13F Derived functions

511

14 Molecular Interactions

521

14A Electric properties of molecules

521

14B Interactions between molecules

533

14C Liquids


539

14D Macromolecules

542

14E Self-assembly

554

15 Solids

561

15A Crystal structure

561

15B Diffraction techniques

564

15C Bonding in solids

571

15D The mechanical properties of solids

576


15E The electrical properties of solids

578

15F The magnetic properties of solids

580

15G The optical properties of solids

583

16 Molecules in motion

589

16A Transport properties of a perfect gas

589

16B Motion in liquids

595

16C Diffusion

601

17 Chemical kinetics

17A The rates of chemical reactions

611
611

v


vi

TABLE OF CONTENTS

17B Integrated rate laws

617

17C Reactions approaching equilibrium

634

17D The Arrhenius equation

638

17E Reaction mechanisms

642

17F Examples of reaction mechanisms


648

17G Photochemistry

652

18 Reaction dynamics

671

18A Collision theory

671

18B Diffusion-controlled reactions

676

18C Transition-state theory

679

18D The dynamics of molecular collisions

691

18E Electron transfer in homogeneous systems

693


19 Processes at solid surfaces

699

19A An introduction to solid surfaces

699

19B Adsorption and desorption

704

19C Heterogeneous catalysis

717

19D Processes at electrodes

719


Preface
This manual provides detailed solutions to the (a) Exercises and the odd-numbered Discussion questions and Problems from the 11th edition of Atkins’ Physical Chemistry.

Conventions used is presenting the solutions
We have included page-specific references to equations, sections, figures and other features
of the main text. Equation references are denoted [14B.3b–595], meaning eqn 14B.3b located
on page 595 (the page number is given in italics). Other features are referred to by name,
with a page number also given.
Generally speaking, the values of physical constants (from the first page of the main text)

are used to 5 significant figures except in a few cases where higher precision is required.
In line with the practice in the main text, intermediate results are simply truncated (not
rounded) to three figures, with such truncation indicated by an ellipsis, as in 0.123...; the
value is used in subsequent calculations to its full precision.
The final results of calculations, generally to be found in a box , are given to the precision
warranted by the data provided. We have been rigorous in including units for all quantities
so that the units of the final result can be tracked carefully. The relationships given on
the back of the front cover are useful in resolving the units of more complex expressions,
especially where electrical quantities are involved.
Some of the problems either require the use of mathematical software or are much easier
with the aid of such a tool. In such cases we have used Mathematica (Wolfram Research,
Inc.) in preparing these solutions, but there are no doubt other options available. Some of
the Discussion questions relate directly to specific section of the main text in which case we
have simply given a reference rather than repeating the material from the text.

Acknowledgements
In preparing this manual we have drawn on the equivalent volume prepared for the 10th edition of Atkins’ Physical Chemistry by Charles Trapp, Marshall Cady, and Carmen Giunta. In
particular, the solutions which use quantum chemical calculations or molecular modelling
software, and some of the solutions to the Discussion questions, have been quoted directly
from the solutions manual for the 10th edition, without significant modification. More
generally, we have benefited from the ability to refer to the earlier volume and acknowledge,
with thanks, the influence that its authors have had on the present work.
This manual has been prepared by the authors using the LATEX typesetting system, in
the implementation provided by MiKTEX (miktex.org); the vast majority of the figures
and graphs have been generated using PGFPlots. We are grateful to the community who
maintain and develop these outstanding resources.
Finally, we are grateful to the editorial team at OUP, Jonathan Crowe and Roseanna
Levermore, for their invaluable support in bringing this project to a conclusion.



viii

PREFACE

Errors and omissions
In such a complex undertaking some errors will no doubt have crept in, despite the authors’
best efforts. Readers who identify any errors or omissions are invited to pass them on to us
by email to


The properties of gases

1
1A

The perfect gas

Answers to discussion questions
D1A.1

An equation of state is an equation that relates the variables that define the
state of a system to each other. Boyle, Charles, and Avogadro established these
relations for gases at low pressures (perfect gases) by appropriate experiments.
Boyle determined how volume varies with pressure (V ∝ 1/p), Charles how
volume varies with temperature (V ∝ T), and Avogadro how volume varies
with amount of gas (V ∝ n). Combining all of these proportionalities into one
gives
nT
V∝
p

Inserting the constant of proportionality, R, yields the perfect gas equation
V =R

nT
p

or

pV = nRT

Solutions to exercises
E1A.1(a)

From the inside the front cover the conversion between pressure units is: 1 atm
≡ 101.325 kPa ≡ 760 Torr; 1 bar is 105 Pa exactly.
(i) A pressure of 108 kPa is converted to Torr as follows
108 kPa ×

E1A.2(a)

1 atm
760 Torr
×
= 810 Torr
101.325 kPa
1 atm

(ii) A pressure of 0.975 bar is 0.975 × 105 Pa, which is converted to atm as
follows
1 atm

0.975 × 105 Pa ×
= 0.962 atm
101.325 kPa

The perfect gas law [1A.4–8], pV = nRT, is rearranged to give the pressure,
p = nRT/V . The amount n is found by dividing the mass by the molar mass of
Xe, 131.29 g mol−1 .
n

(8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (298.15 K)
(131 g)
(131.29 g mol−1 )
1.0 dm3
= 24.4 atm

p=


2

1 THE PROPERTIES OF GASES

So no , the sample would not exert a pressure of 20 atm, but 24.4 atm if it were
a perfect gas.
E1A.3(a)

Because the temperature is constant (isothermal) Boyle’s law applies, pV =
const. Therefore the product pV is the same for the initial and final states
p f Vf = p i Vi


hence

p i = p f Vf /Vi

The initial volume is 2.20 dm3 greater than the final volume so Vi = 4.65+2.20 =
6.85 dm3 .
pi =

Vf
4.65 dm3
× pf =
× (5.04 bar) = 3.42 bar
Vi
6.85 dm3

(i) The initial pressure is 3.42 bar

(ii) Because a pressure of 1 atm is equivalent to 1.01325 bar, the initial pressure
expressed in atm is
1 atm
× 3.40 bar = 3.38 atm
1.01325 bar

E1A.4(a)

If the gas is assumed to be perfect, the equation of state is [1A.4–8], pV = nRT.
In this case the volume and amount (in moles) of the gas are constant, so it
follows that the pressure is proportional to the temperature: p ∝ T. The ratio
of the final and initial pressures is therefore equal to the ratio of the temperatures: p f /p i = Tf /Ti . The pressure indicated on the gauge is that in excess
of atmospheric pressure, thus the initial pressure is 24 + 14.7 = 38.7 lb in−2 .

Solving for the final pressure p f (remember to use absolute temperatures) gives
Tf
× pi
Ti
(35 + 273.15) K
=
× (38.7 lb in−2 ) = 44.4... lb in−2
(−5 + 273.15) K

pf =

The pressure indicated on the gauge is this final pressure, minus atmospheric
pressure: 44.4... − 14.7 = 30 lb in−2 . This assumes that (i) the gas is behaving
perfectly and (ii) that the tyre is rigid.
E1A.5(a)

The perfect gas law pV = nRT is rearranged to give the pressure
p=
=

nRT
V

n

255 × 10−3 g (8.3145 × 10−2 dm3 bar K−1 mol−1 ) × (122 K)
×
20.18 g mol−1
3.00 dm3


= 0.0427 bar


SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY

Note the choice of R to match the units of the problem. An alternative is to
use R = 8.3154 J K−1 mol−1 and adjust the other units accordingly, to give a
pressure in Pa.
[(255 × 10−3 g)/(20.18 g mol−1 )] × (8.3145 J K−1 mol−1 ) × (122 K)
3.00 × 10−3 m3
5
= 4.27 × 10 Pa

p=

where 1 dm3 = 10−3 m3 has been used along with 1 J = 1 kg m2 s−2 and 1 Pa =
1 kg m−1 s−2 .
E1A.6(a)

The vapour is assumed to be a perfect gas, so the gas law pV = nRT applies. The
task is to use this expression to relate the measured mass density to the molar
mass.
First, the amount n is expressed as the mass m divided by the molar mass M to
give pV = (m/M)RT; division of both sides by V gives p = (m/V )(RT/M).
The quantity (m/V ) is the mass density ρ, so p = ρRT/M, which rearranges
to M = ρRT/p; this is the required relationship between M and the density.
M=

ρRT (3.710 kg m−3 ) × (8.3145 J K−1 mol−1 ) × ([500 + 273.15] K)
=

p
93.2 × 103 Pa

= 0.255... kg mol−1

where 1 J = 1 kg m2 s−2 and 1 Pa = 1 kg m−1 s−2 have been used. The molar mass
of S is 32.06 g mol−1 , so the number of S atoms in the molecules comprising
the vapour is (0.255... × 103 g mol−1 )/(32.06 g mol−1 ) = 7.98. The result is
expected to be an integer, so the formula is likely to be S8 .
E1A.7(a)

The vapour is assumed to be a perfect gas, so the gas law pV = nRT applies; the
task is to use this expression to relate the measured data to the mass m. This
is done by expressing the amount n as m/M, where M is the the molar mass.
With this substitution it follows that m = MPV /RT.

The partial pressure of water vapour is 0.60 times the saturated vapour pressure
M pV
RT
(18.0158 g mol−1 ) × (0.60 × 0.0356 × 105 Pa) × (400 m3 )
=
(8.3145 J K−1 mol−1 ) × ([27 + 273.15] K)

m=

= 6.2 × 103 g = 6.2 kg

E1A.8(a)

Consider 1 m3 of air: the mass of gas is therefore 1.146 kg. If perfect gas behaviour is assumed, the amount in moles is given by n = pV /RT

n=

pV
(0.987 × 105 Pa) × (1 m3 )
=
= 39.5... mol
RT (8.3145 J K−1 mol−1 ) × ([27 + 273.15] K)

3


4

1 THE PROPERTIES OF GASES

(i) The total amount in moles is n = n O2 + n N2 . The total mass m is computed
from the amounts in moles and the molar masses M as
m = n O2 × M O2 + n N2 × M N2

These two equations are solved simultaneously for n O2 to give the following expression, which is then evaluated using the data given
n O2 =
=

m − M N2 n
M O2 − M N2

(1146 g) − (28.02 g mol−1 ) × (39.5... mol)
= 9.50... mol
(32.00 g mol−1 ) − (28.02 g mol−1 )


The mole fractions are therefore
x O2 =

n O2 9.50... mol
=
= 0.240
n
39.5... mol

x N2 = 1 − x O2 = 0.760

The partial pressures are given by p i = x i p tot

p O2 = x O2 p tot = 0.240(0.987 bar) = 0.237 bar

p N2 = x N2 p tot = 0.760(0.987 bar) = 0.750 bar

(ii) The simultaneous equations to be solved are now
n = n O2 + n N2 + n Ar

m = n O2 M O2 + n N2 M N2 + n Ar M Ar

Because it is given that x Ar = 0.01, it follows that n Ar = n/100. The two
unknowns, n O2 and n N2 , are found by solving these equations simultaneously to give
n N2 =

100m − n(M Ar + 99M O2 )
100(M N2 − M O2 )

100×(1146 g)−(39.5... mol)×[(39.95 g mol−1 )+99×(32.00 g mol−1 )]

100 × [(28.02 g mol−1 ) − (32.00 g mol−1 )]
= 30.8... mol
=

From n = n O2 + n N2 + n Ar it follows that

n O2 = n − n Ar − n N2
= (39.5... mol) − 0.01 × (39.5... mol) − (30.8... mol) = 8.31... mol

The mole fractions are
x N2 =

n N2 30.8... mol
=
= 0.780
n
39.5... mol

The partial pressures are

x O2 =

n O2 8.31... mol
=
= 0.210
n
39.5... mol

p N2 = x N2 p tot = 0.780 × (0.987 bar) = 0.770 bar


p O2 = x O2 p tot = 0.210 × (0.987 bar) = 0.207 bar


SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY

Note: the final values are quite sensitive to the precision with which the intermediate results are carried forward.
E1A.9(a)

The vapour is assumed to be a perfect gas, so the gas law pV = nRT applies. The
task is to use this expression to relate the measured mass density to the molar
mass.
First, the amount n is expressed as the mass m divided by the molar mass M to
give pV = (m/M)RT; division of both sides by V gives p = (m/V )(RT/M).
The quantity (m/V ) is the mass density ρ, so p = ρRT/M, which rearranges
to M = ρRT/p; this is the required relationship between M and the density.
M=
=

ρRT
p

(1.23 kg m−3 ) × (8.3145 J K−1 mol−1 ) × (330 K)
20.0 × 103 Pa

= 0.169 kg mol−1

E1A.10(a)

The relationships 1 J = 1 kg m2 s−2 and 1 Pa = 1 kg m−1 s−2 have been used.


Charles’ law [1A.3b–7] states that V ∝ T at constant n and p, and p ∝ T at
constant n and V . For a fixed amount the density ρ is proportional to 1/V , so it
follows that 1/ρ ∝ T. At absolute zero the volume goes to zero, so the density
goes to infinity and hence 1/ρ goes to zero. The approach is therefore to plot
1/ρ against the temperature (in ○ C) and then by extrapolating the straight line
find the temperature at which 1/ρ = 0. The plot is shown in Fig 1.1.
θ/○ C
−85
0
100

ρ/(g dm−3 )
1.877
1.294
0.946

(1/ρ)/(g−1 dm3 )
0.532 8
0.772 8
1.057 1

The data are a good fit to a straight line, the equation of which is
(1/ρ)/(g−1 dm3 ) = 2.835 × 10−3 × (θ/○ C) + 0.7734

The intercept with 1/ρ = 0 is found by solving

0 = 2.835 × 10−3 × (θ/○ C) + 0.7734

E1A.11(a)


This gives θ = −273 ○ C as the estimate of absolute zero.
(i) The mole fractions are
x H2 =

n H2
2.0 mol
=
=
n H2 + n N2 2.0 mol + 1.0 mol

2
3

x N2 = 1 − x H2 =

1
3

5


1 THE PROPERTIES OF GASES

1.0
(1/ρ)/(g−1 dm3 )

6

0.5


0.0
−300

Figure 1.1

−200

−100
θ/○ C

0

100

(ii) The partial pressures are given by p i = x i p tot . The total pressure is given
by the perfect gas law: p tot = n tot RT/V
2 (3.0 mol) × (8.3145 J K−1 mol−1 ) × (273.15 K)
×
3
22.4 × 10−3 m3
5
= 2.0 × 10 Pa

p H2 = x H2 p tot =

1 (3.0 mol) × (8.3145 J K−1 mol−1 ) × (273.15 K)
×
3
22.4 × 10−3 m3
= 1.0 × 105 Pa


p N2 = x N2 p tot =

Expressed in atmospheres these are 2.0 atm and 1.0 atm, respectively.
(iii) The total pressure is
(3.0 mol) × (8.3145 J K−1 mol−1 ) × (273.15 K)
= 3.0 × 105 Pa
22.4 × 10−3 m3
or 3.00 atm.

Alternatively, note that 1 mol at STP occupies a volume of 22.4 dm3 , which is
the stated volume. As there are a total of 3.0 mol present the (total) pressure
must therefore be 3.0 atm.

Solutions to problems
P1A.1

(a) The expression ρgh gives the pressure in Pa if all the quantities are in
SI units, so it is helpful to work in Pa throughout. From the front cover,
760 Torr is exactly 1 atm, which is 1.01325×105 Pa. The density of 13.55 g cm−3
is equivalent to 13.55 × 103 kg m−3 .
p = p ex + ρgh

= 1.01325 × 105 Pa + (13.55 × 103 kg m−3 ) × (9.806 m s−2 )
× (10.0 × 10−2 m) = 1.15 × 105 Pa


SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY

(b) The calculation of the pressure inside the apparatus proceeds as in (a)

p = 1.01325 × 105 Pa + (0.9971 × 103 kg m−3 ) × (9.806 m s−2 )
× (183.2 × 10−2 m) = 1.192... × 105 Pa

The value of R is found by rearranging the perfect gas law to R = pV /nT
R=

(1.192... × 105 Pa) × (20.000 × 10−3 m3 )
pV
=
nT [(1.485 g)/(4.003 g mol−1 )] × ([500 + 273.15] K)

= 8.315 J K−1 mol−1

The perfect gas law pV = nRT implies that pVm = RT, where Vm is the molar
volume (the volume when n = 1). It follows that p = RT/Vm , so a plot of p
against T/Vm should be a straight line with slope R.

However, real gases only become ideal in the limit of zero pressure, so what is
needed is a method of extrapolating the data to zero pressure. One approach is
to rearrange the perfect gas law into the form pVm /T = R and then to realise
that this implies that for a real gas the quantity pVm /T will tend to R in the limit
of zero pressure. Therefore, the intercept at p = 0 of a plot of pVm /T against p
is an estimate of R. For the extrapolation of the line back to p = 0 to be reliable,
the data points must fall on a reasonable straight line. The plot is shown in
Fig 1.2.
p/atm
0.750 000
0.500 000
0.250 000


(pVm /T)/(atm dm3 mol−1 K−1 )

P1A.3

Vm /(dm3 mol−1 ) (pVm /T)/(atm dm3 mol−1 K−1 )
29.8649
0.082 001 4
44.8090
0.082 022 7
89.6384
0.082 041 4

0.08206
0.08204
0.08202
0.08200

Figure 1.2

0.0

0.2

0.4
p/atm

0.6

0.8


7


1 THE PROPERTIES OF GASES

The data fall on a reasonable straight line, the equation of which is
(pVm /T)/(atm dm3 mol−1 K−1 ) = −7.995 × 10−5 × (p/atm) + 0.082062

The estimate for R is therefore the intercept, 0.082062 atm dm3 mol−1 K−1 .
The data are given to 6 figures, but they do not fall on a very good straight line
so the value for R has been quoted to one fewer significant figure.
P1A.5

For a perfect gas pV = nRT which can be rearranged to give p = nRT/V . The
amount in moles is n = m/M, where M is the molar mass and m is the mass of
the gas. Therefore p = (m/M)(RT/V ). The quantity m/V is the mass density
ρ, and hence
p = ρRT/M

It follows that for a perfect gas p/ρ should be a constant at a given temperature.
Real gases are expected to approach this as the pressure goes to zero, so a
suitable plot is of p/ρ against p; the intercept when p = 0 gives the best estimate
of RT/M. The plot is shown in Fig. 1.3.
ρ/(kg m−3 )
0.225
0.456
0.664
1.062
1.468
1.734


p/kPa
12.22
25.20
36.97
60.37
85.23
101.30

(p/ρ)/(kPa kg−1 m3 )
54.32
55.26
55.68
56.85
58.06
58.42

58

(p/ρ)/(kPa kg−1 m3 )

8

56

54
0

20


40

60
p/kPa

80

100

Figure 1.3

The data fall on a reasonable straight line, the equation of which is
(p/ρ)/(kPa kg−1 m3 ) = 0.04610 × (p/kPa) + 53.96


SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY

The intercept is (p/ρ)lim p→0 , which is equal to RT/M.
M=

(8.3145 J K−1 mol−1 ) × (298.15 K)
RT
=
= 4.594×10−2 kg mol−1
(p/ρ)lim p→0
53.96 × 103 Pa kg−1 m3

The estimate of the molar mass is therefore 45.94 g mol−1 .
P1A.7


(a) For a perfect gas pV = nRT so it follows that for a sample at constant
volume and temperature, p 1 /T1 = p 2 /T2 . If the pressure increases by
∆p for an increase in temperature of ∆T, then with p 2 = p 1 + ∆p and
T2 = T1 + ∆T is follows that
p 1 p 1 + ∆p
=
T1 T1 + ∆T

hence

∆p =

For an increase by 1.00 K, ∆T = 1.00 K and hence
∆p =

p 1 ∆T
T1

p 1 ∆T (6.69 × 103 Pa) × (1.00 K)
=
= 24.5 Pa
T1
273.16 K

Another way of looking at this is to write the rate of change of pressure
with temperature as
∆p p 1 6.69 × 103 Pa
=
=
= 24.5... Pa K−1

∆T T1
273.16 K

(b) A temperature of 100.00 ○ C is equivalent to an increase in temperature
from the triple point by 100.00 + 273.15 − 273.16 = 99.99 K
∆p′ = ∆T ′ × (

∆p
6.69 × 103 Pa
) = (99.99 K) ×
= 2.44... × 103 Pa
∆T
273.16 K

The final pressure is therefore 6.69 + 2.44... = 9.14 kPa .

(c) For a perfect gas ∆p/∆T is independent of the temperature so at 100.0 ○ C
a 1.00 K rise in temperature gives a pressure rise of 24.5 Pa , just as in (a).
P1A.9

The molar mass of SO2 is 32.06+2×16.00 = 64.06 g mol−1 . If the gas is assumed
to be perfect the volume is calculated from pV = nRT
n

V=

nRT
200 × 106 g (8.3145 J K−1 mol−1 ) × ([800 + 273.15] K)
)
=(

p
1.01325 × 105 Pa
64.06 g mol−1

= 2.7 × 105 m3

Note the conversion of the mass in t to mass in g; repeating the calculation for
300 t gives a volume of 4.1 × 105 m3 .

The volume of gas is therefore between 0.27 km3 and 0.41 km3 .

9


10

1 THE PROPERTIES OF GASES

P1A.11

Imagine a column of the atmosphere with cross sectional area A. The pressure
at any height is equal to the force acting down on that area; this force arises
from the gravitational attraction on the gas in the column above this height –
that is, the ‘weight’ of the gas.
Suppose that the height h is increased by dh. The force on the area A is reduced
because less of the atmosphere is now bearing down on this area. Specifically,
the force is reduced by that due to the gravitational attraction on the gas contained in a cylinder of cross-sectional area A and height dh. If the density of
the gas is ρ, the mass of the gas in the cylinder is ρ × A dh and the force due to
gravity on this mass is ρgA dh, where g is the acceleration due to free fall. The
change in pressure dp on increasing the height by dh is this force divided by

the area, so it follows that
dp = −ρgdh

The minus sign is needed because the pressure decreases as the height increases.
The density is related to the pressure by starting from the perfect gas equation,
pV = nRT. If the mass of gas is m and the molar mass is M, it follows that
n = m/M and hence pV = (m/M)RT. Taking the volume to the right gives
p = (m/MV )RT. The quantity m/V is the mass density ρ, so p = (ρ/M)RT;
this is rearranged to give an expression for the density: ρ = M p/RT.

This expression for ρ is substituted into dp = −ρgdh to give dp = −(M p/RT)gdh.
Division by p results in separation of the variables (1/p) dp = −(M/RT)gdh.
The left-hand side is integrated between p0 , the pressure at h = 0 and p, the
pressure at h. The right-hand side is integrated between h = 0 and h
∫

h Mg
1
dp = ∫ −
dh
RT
p0 p
0
Mg
p
h
[ln p] p 0 = −
[h]0
RT
p

M gh
ln
=−
p0
RT
p

The exponential of each side is taken to give
p = p 0 e−h/H

with

It is assumed that g and T do not vary with h.

H=

RT
Mg

(a) The pressure decrease across such a small distance will be very small because h/H ≪ 1. It is therefore admissible to expand the exponential and
retain just the first two terms: ex ≈ 1 + x
p = p 0 (1 − h/H)

This is rearranged to give an expression for the pressure decrease, p − p 0
p − p 0 = −p 0 h/H


SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY

If it is assumed that p 0 is one atmosphere and that H = 8 km,

p − p 0 = −p 0 h/H = −

(1.01325 × 105 Pa) × (15 × 10−2 m)
= −2 Pa
8 × 103 m

(b) The pressure at 11 km is calculated using the full expression

p = p 0 e−h/H = (1 atm) × e−(11 km)/(8 km) = 0.25 atm
P1A.13

Imagine a volume V of the atmosphere, at temperature T and pressure p tot .
If the concentration of a trace gas is expressed as X parts per trillion (ppt), it
means that if that gas were confined to a volume X × 10−12 × V at temperature
T is would exert a pressure p tot . From the perfect gas law it follows that n =
pV /RT, which in this case gives
n trace =

p tot (X × 10−12 × V )
RT

Taking the volume V to the left gives the molar concentration, c trace
c trace =

n trace X × 10−12 × p tot
=
V
RT

An alternative way of looking at this is to note that, at a given temperature and

pressure, the volume occupied by a gas is proportional to the amount in moles.
Saying that a gas is present at X ppt implies that the volume occupied by the
gas is X × 10−12 of the whole, and therefore that the amount in moles of the gas
is X × 10−12 of the total amount in moles
n trace = (X × 10−12 ) × n tot

This is rearranged to give an expression for the mole fraction x trace
x trace =

n trace
= X × 10−12
n tot

The partial pressure of the trace gas is therefore

p trace = x trace p tot = (X × 10−12 ) × p tot

The concentration is n trace /V = p trace /RT, so
c trace =

n trace X × 10−12 × p tot
=
V
RT

11


12


1 THE PROPERTIES OF GASES

(a) At 10 ○ C and 1.0 atm

X CCl3 F × 10−12 × p tot
RT
261 × 10−12 × (1.0 atm)
=
(8.2057 × 10−2 dm3 atm K−1 mol−1 ) × ([10 + 273.15] K)

c CCl3 F =

= 1.1 × 10−11 mol dm−3

X CCl2 F2 × 10−12 × p tot
RT
509 × 10−12 × (1.0 atm)
=
−2
(8.2057 × 10 dm3 atm K−1 mol−1 ) × ([10 + 273.15] K)

c CCl2 F2 =

= 2.2 × 10−11 mol dm−3

(b) At 200 K and 0.050 atm

X CCl3 F × 10−12 × p tot
RT
261 × 10−12 × (0.050 atm)

=
(8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (200 K)

c CCl3 F =

= 8.0 × 10−13 mol dm−3

X CCl2 F2 × 10−12 × p tot
RT
509 × 10−12 × (0.050 atm)
=
(8.2057 × 10−2 dm3 atm K−1 mol−1 ) × (200 K)

c CCl2 F2 =

= 1.6 × 10−12 mol dm−3

1B The kinetic model
Answer to discussion questions
D1B.1

The three assumptions on which the kinetic model is based are given in Section 1B.1 on page 11.
1. The gas consists of molecules in ceaseless random motion obeying the
laws of classical mechanics.
2. The size of the molecules is negligible, in the sense that their diameters
are much smaller than the average distance travelled between collisions;
they are ‘point-like’.
3. The molecules interact only through brief elastic collisions.
An elastic collision is a collision in which the total translational kinetic energy
of the molecules is conserved.



SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY

None of these assumptions is strictly true; however, many of them are good approximations under a wide range of conditions including conditions of ambient
temperature and pressure. In particular,
(a) Molecules are subject to laws of quantum mechanics; however, for all
but the lightest gases at low temperatures, non-classical effects are not
important.
(b) With increasing pressure, the average distance between molecules will decrease, eventually becoming comparable to the dimensions of the molecules
themselves.
(c) Intermolecular interactions, such as hydrogen bonding, and the interactions of dipole moments, operate when molecules are separated by small
distances. Therefore, as assumption (2) breaks down, so does assumption
(3), because the molecules are often close enough together to interact even
when not colliding.
D1B.3

For an object (be it a space craft or a molecule) to escape the gravitational
field of the Earth it must acquire kinetic energy equal in magnitude to the
gravitational potential energy the object experiences at the surface of the Earth.
The gravitational potential between two objects with masses m 1 and m 2 when
separated by a distance r is
V =−

Gm 1 m 2
r

where G is the (universal) gravitational constant. In the case of an object of
mass m at the surface of the Earth, it turns out that the gravitational potential
is given by

GmM
V =−
R
where M is the mass of the Earth and R its radius. This expression implies that
the potential at the surface is the same as if the mass of the Earth were localized
at a distance equal to its radius.
As a mass moves away from the surface of the Earth the potential energy increases (becomes less negative) and tends to zero at large distances. This change
in potential energy must all be converted into kinetic energy if the mass is to
escape. A mass m moving at speed υ has kinetic energy 12 mυ 2 ; this speed will
be the escape velocity υ e when
√
GmM
2GM
2
1
mυ e =
hence
υe =
2
R
R

The quantity in the square root is related to the acceleration due to free fall,
g, in the following way. A mass m at the surface of the Earth experiences
a gravitational force given GMm/R 2 (note that the force goes as R−2 ). This
force accelerates the mass towards the Earth, and can be written mg. The two
expressions for the force are equated to give
GMm
= mg
R2


hence

GM
= gR
R

13


14

1 THE PROPERTIES OF GASES

This expression for GM/R is substituted into the above expression for υ e to give
υe =

√

2GM √
= 2Rg
R

The escape velocity is therefore a function of the radius of the Earth and the
acceleration due to free fall.

The radius of the Earth is 6.37×106 m and g = 9.81 m s−2 so the escape velocity
is 1.11×104 m s−1 . For comparison, the mean speed of He at 298 K is 1300 m s−1
and for N2 the mean speed is 475 m s−1 . For He, only atoms with a speed in
excess of eight times the mean speed will be able to escape, whereas for N2 the

speed will need to be more than twenty times the mean speed. The fraction of
molecules with speeds many times the mean speed is small, and because this
2
fraction goes as e−υ it falls off rapidly as the multiple increases. A tiny fraction
of He atoms will be able to escape, but the fraction of heavier molecules with
sufficient speed to escape will be utterly negligible.

Solutions to exercises
E1B.1(a)

(i) √
The mean speed is given by [1B.9–16], υ mean = (8RT/πM)1/2 , so υ mean ∝
1/M. The ratio of the mean speeds therefore depends on the ratio of
the molar masses
M Hg
υ mean,H2
=(
)
υ mean,Hg
M H2

1/2

=(

200.59 g mol−1
)
2 × 1.0079 g mol−1

1/2


= 9.975

(ii) The mean translational kinetic energy ⟨E k ⟩ is given by 12 m⟨υ 2 ⟩, where
⟨υ 2 ⟩ is the mean square speed, which is given by [1B.7–15], ⟨υ 2 ⟩ = 3RT/M.
The mean translational kinetic energy is therefore
1
1
3RT
⟨E k ⟩ = m⟨υ 2 ⟩ = m (
)
2
2
M

The molar mass M is related to the mass m of one molecule by M = mN A ,
where N A is Avogadro’s constant, and the gas constant can be written R =
kN A , hence
1
3
3RT
1
3kN A T
⟨E k ⟩ = m (
) = kT
) = m(
2
M
2
mN A

2

The mean translational kinetic energy is therefore independent of the
identity of the gas, and only depends on the temperature: it is the same
for H2 and Hg.
This result is related to the principle of equipartition of energy: a molecule
has three translational degrees of freedom (x, y, and z) each of which
contributes 12 kT to the average energy.


SOLUTIONS MANUAL TO ACCOMPANY ATKINS’ PHYSICAL CHEMISTRY

E1B.2(a)

The rms speed is given by [1B.8–15], υ rms = (3RT/M)1/2 .
υ rms,H2 = (

3RT
)
M H2

1/2

=(

= 1.90 km s−1

3 × (8.3145 J K−1 mol−1 ) × (293.15 K)
)
2 × 1.0079 × 10−3 kg mol−1


1/2

where 1 J = 1 kg m2 s−2 has been used. Note that the molar mass is in kg mol−1 .
υ rms,O2 = (

E1B.3(a)

3 × (8.3145 J K−1 mol−1 ) × (293.15 K)
)
2 × 16.00 × 10−3 kg mol−1

1/2

= 478 m s−1

The Maxwell–Boltzmann distribution of speeds, f (υ), is given by [1B.4–14].
The fraction of molecules with speeds between υ 1 and υ 2 is given by the integral
∫

υ2

υ1

f (υ) dυ

If the range υ 2 − υ 1 = δυ is small, the integral is well-approximated by
f (υ mid ) δυ

where υ mid is the mid-point of the velocity range: υ mid = 12 (υ 2 + υ 1 ). In this

exercise υ mid = 205 m s−1 and δυ = 10 m s−1 .
fraction = f (υ mid ) δυ = 4π × (

−Mυ 2mid
M 3/2 2
) υ mid exp (
) δυ
2πRT
2RT

2 × 14.01 × 10−3 kg mol−1
)
= 4π × (
2π × (8.3145 J K−1 mol−1 ) × (400 K)
× exp (

3/2

× (205 m s−1 )2

−(2 × 14.01 × 10−3 kg mol−1 ) × (205 m s−1 )2
) × (10 m s−1 )
2 × (8.3145 J K−1 mol−1 ) × (400 K)

= 6.87 × 10−3

where 1 J = 1 kg m2 s−2 has been used. Thus, 0.687% of molecules have velocities in this range.
E1B.4(a)

The mean relative speed is given by [1B.11b–16], υ rel = (8kT/πµ)1/2 , where

µ = m A m B /(m A + m A ) is the effective mass. Multiplying top and bottom of
the expression for υ rel by N A and using N A k = R gives υ rel = (8RT/πN A µ)1/2
in which N A µ is the molar effective mass. For the relative motion of N2 and H2
this effective mass is
NA µ =

M N2 M H2
(2 × 14.01 g mol−1 ) × (2 × 1.0079 g mol−1 )
=
= 1.88... g mol−1
M N2 + M H2 (2 × 14.01 g mol−1 ) + (2 × 1.0079 g mol−1 )

8 × (8.3145 J K−1 mol−1 ) × (298.15 K)
) = 1832 m s−1
π × (1.88... × 10−3 kg mol−1 )
The value of the effective mass µ is dominated by the mass of the lighter molecule,
in this case H2 .
υ rel = (

8RT
)
πN A µ

1/2

=(

1/2

15



16

1 THE PROPERTIES OF GASES

E1B.5(a)

The most probable speed is given by [1B.10–16], υ mp = (2RT/M)1/2 , the mean
speed is given by [1B.9–16], υ mean = (8RT/πM)1/2 , and the mean relative
speed
between two molecules of the same mass is given by [1B.11a–16], υ rel =
√
2υ mean .
M CO2 = 12.01 + 2 × 16.00 = 44.01 g mol−1 .
υ mp = (

2RT 1/2
2 × (8.3145 J K−1 mol−1 ) × (293.15 K)
)
) =(
M
44.01 × 10−3 kg mol−1

υ mean = (
E1B.6(a)

1/2

8RT 1/2

8 × (8.3145 J K−1 mol−1 ) × (293.15 K)
) =(
)
πM
π × (44.01 × 10−3 kg mol−1 )
√
√
υ rel = 2υ mean = 2 × (376 m s−1 ) = 531 m s−1

= 333 m s−1

1/2

= 376 m s−1

The collision frequency is given by [1B.12b–17], z = σ υ rel p/kT, with the
√ relative
speed for two molecules of the same type given by [1B.11a–16], υ rel = 2υ mean .
The mean speed is given by [1B.9–16], υ mean = (8RT/πM)1/2 . From the Resource section the collision cross-section σ is 0.27 nm2 .
z=

8RT 1/2
σ υ rel p σ p √
=
× 2×(
)
kT
kT
πM
(0.27 × 10−18 m2 ) × (1.01325 × 105 Pa) √

=
× 2
(1.3806 × 10−23 J K−1 ) × (298.15 K)

8 × (8.3145 J K−1 mol−1 ) × (298.15 K)
×(
)
π × (2 × 1.0079 × 10−3 kg mol−1 )

1/2

= 1.7 × 1010 s−1

E1B.7(a)

where 1 J = 1 kg m2 s−2 and 1 Pa = 1 kg m−1 s−2 have been used. Note the
conversion of the collision cross-section σ to m2 : 1 nm2 = (1 × 10−9 )2 m2 =
1 × 10−18 m2 .

The mean speed is given by [1B.9–16], υ mean = (8RT/πM)1/2 . The collision
frequency is given by [1B.12b–17], z = σ υ rel p/kT, with the
√ relative speed for
two molecules of the same type given by [1B.11a–16], υ rel = 2υ mean . The mean
free path is given by [1B.14–18], λ = kT/σ p
(i) The mean speed is calculated as
υ mean

8RT 1/2
8 × (8.3145 J K−1 mol−1 ) × (298.15 K)
=(

) =(
)
πM
π × (2 × 14.01 × 10−3 kg mol−1 )

1/2

= 475 m s−1

(ii) The collision cross-section σ is calculated from the collision diameter d
as σ = πd 2 = π × (395 × 10−9 m)2 = 4.90... × 10−19 m2 . With this value
the mean free path is calculated as
λ=

kT
(1.3806 × 10−23 J K−1 ) × (298.15 K)
=
= 82.9×10−9 m = 82.9 nm
σ p (4.90... × 10−19 m2 ) × (1.01325 × 105 Pa)

where 1 J = 1 kg m2 s−2 and 1 Pa = 1 kg m−1 s−2 have been used.