Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap05

5

The Second Law: the machinery

Solutions to exercises
Discussion questions
E5.1(b)

See the solution to Exercise 3.14(a) and Example 5.1, where it is demonstrated that πT = a/Vm2 for
a van der Waals gas. Therefore, there is no dependence on b for a van der Waals gas. The internal
pressure results from attractive interactions alone. For van der Waals gases and liquids with strong
attractive forces (large a) at small volumes, the internal pressure can be very large.

E5.2(b)

The relation (∂G/∂T )p = −S shows that the Gibbs function of a system decreases with T at
constant p in proportion to the magnitude of its entropy. This makes good sense when one considers
the definition of G, which is G = U + pV − T S. Hence, G is expected to decrease with T in
proportion to S when p is constant. Furthermore, an increase in temperature causes entropy to
increase according to
S=

f
i

dqrev /T

The corresponding increase in molecular disorder causes a decline in th Gibbs energy. (Entropy is
always positive.)
E5.3(b)

The fugacity coefficient, φ, can be expressed in terms of an integral involving the compression factor,
specifically an integral of Z − 1 (see eqn 5.20). Therefore, we expect that the variation with pressure
of the fugacity coefficient should be similar, in a very qualitative sense, to the variation with pressure
of the compression factor itself. Comparison of figures 1.27 and 5.8 of the text shows this to be roughly
the case, though the detailed shapes of the curves are necessarily different because φ is an integral
function of Z − 1 over a range of pressures. So we expect no simple proportionality between φ and Z.
But we find φ < 1 in pressure regions where attractive forces are expected to predominate and φ > 1
when repulsive forces predominate, which in behavior is similar to that of Z. See Section 5.5(b) for
a more complete discussion.

Numerical exercises
E5.4(b)

E5.5(b)

α=

1
×
V

∂V
∂T p

κT = −

1
V

×

∂V
∂p T

∂S
∂V
=−
= −αV
∂p T
∂T p
pf
pf
Vi
at constant temperature,
G = nRT ln
=
pi
pi
Vf
= nRT ln

Vi
Vf

= (2.5 × 10−3 mol) × (8.314 J K −1 mol−1 ) × (298 K) × ln
= −2.035 = −2.0 J

72
100

INSTRUCTOR’S MANUAL

76

∂G
= −S
∂T p

∂Gf
∂Gi
= −Sf and
= −Si
∂T p
∂T p
∂Gf
∂Gi
S = Sf − Si = −
+
∂T p
∂T p
∂(Gf − Gi )
∂ G
=−
=−
∂T
∂T p
p
∂
{−73.1 + 42.8 T /K} J

=−
∂T
= −42.8 J K−1

E5.6(b)

E5.7(b)

See the solution to Exercise 5.7(a). Without knowledge of the compressibility of methanol we can
only assume that V = V1 (1 − κT p) ≈ V1 . Then
G=V p
m
m
25 g
ρ=
= 31.61 cm3
so V =
=
V
ρ
0.791 g cm−3
1 m3
106 cm3

G = (31.61 cm3 ) ×

× (99.9 × 106 Pa)

= +3.2 kJ
E5.8(b)

(a)

pi
Vf
= nR ln
Vi
pf
Taking inverse logarithms
S = nR ln

[Boyle’s Law]

pf = pi e− S/nR = (150 kPa) exp −

−(−15.0 J K−1 )
(3.00 mol) × (8.314 J K −1 mol−1 )

= 274 kPa
(b)

G = nRT ln

pf
pi

= −T S

[ H = 0, constant temperature, perfect gas]

= −(230 K) × (−15.0 J K−1 )
= +3450 J = 3.45 kJ
E5.9(b)

µ = µf − µi = RT ln

pf
pi

= (8.314 J K−1 mol−1 ) × (323 K) × ln
= 2.71 kJ mol−1
E5.10(b)

µ0 = µ−− + RT ln
µ = µ−− + RT ln

p
p −−
f
p −−

252.0
92.0


THE SECOND LAW: THE MACHINERY

77

f

p

µ − µ0 = RT ln

f
≡φ
p

µ − µ0 = RT ln φ
= (8.314 J K−1 mol−1 ) × (290 K) × ln(0.68)
= −929.8 J mol−1
= −930 J mol−1

or

−0.93 kJ mol−1

(160.0 cm3 mol−1 ) × 1016mcm3
B
B =
=−
RT
(8.314 J K−1 mol−1 ) × (100 K)
3

E5.11(b)

= −1.924 × 10−7 Pa−1
φ = eB p+···
−7

−1
6
≈ e− 1.924×10 Pa × 62×10 Pa

≈ e−11.93
= 7 × 10−6
E5.12(b)

or

of the order of 10−6

G = nVm p = V p
= (1.0 L) ×

1 m3
103 L

× (200 × 103 Pa)

= 200 Pa m3 = 200 J
E5.13(b)

Gm = RT ln

pf
pi

= (8.314 J K−1 mol−1 ) × (500 K) × ln

100.0 kPa
50.0 kPa

= +2.88 kJ mol−1
E5.14(b)

E5.15(b)

∂G
RT
+ B + C p + D p2
[5.10] =
∂p T
p
which is the virial equation of state.
∂S
∂p
=
∂V T
∂T V
V =

For a Dieterici gas
p=

RT e−a/RT Vm
Vm − b

R 1 + RVam T e−a/RVm T
∂p

=
Vm − b
∂T Vm
dS =

∂S
dVm =
∂Vm T

∂p
dVm
∂T Vm


INSTRUCTOR’S MANUAL

78

S=

Vm,f
Vm,i

dS = R 1 +

a
RVm T

e−a/RVm T

Vm,f − b
Vm,i − b

Vm,f
Vm,i
S for a Dieterici gas may be greater or lesser than S for a perfect gas depending on T and the
magnitudes of a and b. At very high T , S is greater. At very low T , S is less.

For a perfect gas

S = R ln

Solutions to problems
Solutions to numerical problem
P5.2

For the reaction
N2 (g) + 3H2 (g) → 2NH3 (g)
(a)

rG

−−

rG

−−

= 2 f G−− (NH3 , g)

(500 K) = τ r G−− (Tc ) + (1 − τ ) r H −− (Tc )
=

Problem 5.1, τ =

T
Tc

500 K
× (2) × (−16.45 kJ mol−1 )
298.15 K
500 K
+ 1−
× (2) × (−46.11 kJ mol−1 )
298.15 K

= −55.17 + 62.43 kJ mol−1 = +7 kJ mol−1
(b)

rG

−−

(1000 K) =

1000 K
× (2) × (−16.45 kJ mol−1 )
298.15 K
1000 K
+ 1−

× (2) × (−46.11 kJ mol−1 )
298.15 K

= (−110.35 + 217.09) kJ mol−1 = +107 kJ mol−1

Solutions to theoretical problems
P5.5

We start from the fundamental relation
dU = T dS − p dV [2]
But, since U = U (S, V ), we may also write
dU =

∂U
dS +
∂S V

∂U
dV
∂V S

Comparing the two expressions, we see that
∂U
=T
∂S V

and

∂U
= −p

∂V S

These relations are true in general and hence hold for the perfect gas. We can demonstrate this
more explicitly for the perfect gas as follows. For the perfect gas at constant volume
dU = CV dT


THE SECOND LAW: THE MACHINERY

79

and
dS =
Then

CV dT
dqrev
=
T
T
CV dT
∂U
=
CV dT
∂S V
T

dU
=
dS V

=T

For a reversible adiabatic (constant-entropy) change in a perfect gas
dU = dw = −p dV
∂U
= −p
∂V S
∂p
∂T
=−
[Maxwell relation]
∂S V
∂V S

Therefore,
P5.8

1

=

∂S
∂T

=

P5.10

V

∂S
∂U

−

=

∂p
∂T

V

V

∂U
∂T

∂V
∂T
∂V
∂p

∂H
=
∂p T

∂V
∂S

T

∂U
∂T

T

∂S
∂T

V

[Maxwell relation] =
V

∂U
∂S

p
T

[chain relation] =

∂S
∂V

V

[inversion twice] =

[inversion]

∂p
− ∂V
∂S
∂U

T

V

∂V
∂T
∂U
∂T

p

[chain relation]

V

αT
κT CV

V

∂U
=T
∂S V

∂S
+
∂p T

∂H
[Relation 1, Further information 1.7]
∂p S

dH = T dS + V dp [Problem 5.6]

∂H
∂H
dS +
dp (H = H (p, S)) compare
dH =
∂S p
∂p S
Thus,

∂H
∂S p

∂H
= T,
∂S p

Substitution yields,

∂H

= V [dH exact]
∂p S
∂H
=T
∂p T

∂S
+ V = −T
∂p T

∂V
+ V [Maxwell relation]
∂T p

(a) For pV = nRT
nR
∂V
,
=
∂T p
p

hence

∂H
−nRT
+V = 0
=
∂p T
p

(b) For p =

an2
nRT
− 2 [Table 1.6]
V − nb
V

T =

p(V − nb) na(V − nb)
+
nR
RV 2


INSTRUCTOR’S MANUAL

80

∂T
p
2na(V − nb)
na
=
−
+
∂V p
nR

RV 2
RV 3
−T
+ V [inversion] = p
+V
2na(V −nb)
na
3
nR + RV 2 −
RV
p
which yields after algebraic manipulation
Therefore,

∂H
−T
=
∂T
∂p T
∂V

nb −
∂H
=
∂p T
1−
When

b
Vm

2na
RT

λ2

2na
RT V

λ2

,

λ=1−

nb
V

1, λ ≈ 1 and

2na
1
2na
p
2pa
2na
×
≈
×
= 2 2

=
RT
V
RT
nRT
RT V
R T
Therefore,

nb − 2na
∂H
RT
≈
∂p T
1 − R2pa
2T 2

For argon, a = 1.337 L2 atm mol−2 , b = 3.20 × 10−2 L mol−1 ,
2na
(2) × (1.0 mol) × (1.337 L2 atm mol−2 )
= 0.11 L
=
RT
(8.206 × 10−2 L atm K−1 mol−1 ) × (298 K)
2pa
(2) × (10.0 atm) × (1.337 L2 atm mol−2 )
=
= 0.045
2
R2 T 2

(8.206 × 10−2 L atm K−1 mol−1 ) × (298 K)
Hence,

∂H
{(3.20 × 10−2 ) − (0.11)} L
= −0.0817 L = −8.3 J atm−1
≈
∂p T
1 − 0.045

H ≈

P5.12

∂H
∂p T

p ≈ (−8.3 J atm−1 ) × (1 atm) = −8 J

∂p
− p [5.8]
∂T V
RT
BRT
p=
+
[The virial expansion, Table 1.6, truncated after the term in B]
Vm
Vm2

πT = T

∂p
R
=
+
∂T V
Vm
RT 2
Hence, πT = 2
Vm

BR
RT ∂B
p
RT
+ 2
= + 2
T
Vm2
Vm ∂T V
Vm
∂B
RT 2 B
≈
∂T V
Vm2 T

∂B
∂T V

Since πT represents a (usually) small deviation from perfect gas behaviour, we may approximate Vm .
Vm ≈

RT
p

πT ≈

p2
×
R

B
T


THE SECOND LAW: THE MACHINERY

From the data

81

B = ((−15.6) − (−28.0)) cm3 mol−1 = +12.4 cm3 mol−1

Hence,
(1.0 atm)2 × (12.4 × 10−3 L mol−1 )
= 3.0 × 10−3 atm
(8.206 × 10−2 L atm K−1 mol−1 ) × (50 K)

(a)

πT =

(b)

πT ∝ p 2 ;

so at p = 10.0 atm, πT = 0.30 atm

Comment. In (a) πT is 0.3 per cent of p; in (b) it is 3 per cent. Hence at these pressures the
approximation for Vm is justified. At 100 atm it would not be.
P5.13

Question. How would you obtain a reliable estimate of πT for argon at 100 atm?
∂H
∂U
and Cp =
CV =
∂T p
∂T V
(a)

∂ 2U
∂ 2U
∂CV
=
=
=
∂V ∂T

∂T ∂V
∂V T
∂CV
∂ 2U
∂ 2U
=
=
=
∂p∂T
∂T ∂p
∂p T
∂V
∂U
∂
=
∂T ∂V T ∂p
∂Cp
Since Cp = CV + R,
=
∂x T

∂
∂T
∂
∂T
T V

∂U
=0
∂V T V

∂U
∂p T V
=0

[πT = 0]

(πT = 0)

∂CV
for x = p or V
∂x T
d2 U dCV
dCV
CV and Cp may depend on temperature. Since
=
is nonzero if U depends on
,
dT
dT 2 dT
T through a nonlinear relation. See Chapter 20 for further discussion of this point. However, for
a perfect monatomic gas, U is a linear function of T ; hence CV is independent of T . A similar
argument applies to Cp .
(b) This equation of state is the same as that of Problem 5.12.
∂ 2U
∂CV
=
=
∂V T
∂T ∂V
=

=

=

P5.15

∂πT
[Part (a)]
∂T V
∂ RT 2
∂T Vm2
2RT
Vm2
RT
Vm2

∂B
[Problem 5.12]
∂T V
V

∂B
RT 2
+ 2
∂T V
Vm
∂ 2 (BT )
∂T 2

∂ 2B

∂T 2

V

V

∂p
− p [5.8]
∂T V
nRT
× e−an/RT V [Table 1.6]
p=
V − nb
nRT
nRT
na
nap
∂p
× e−an/RT V +
×
× e−an/RT V = p +
=
T
∂T V
V − nb
RT V
V − nb
RT V

πT = T

INSTRUCTOR’S MANUAL

82

Hence, πT =

P5.17

nap
RT V

πT → 0 as p → 0, V → ∞, a → 0, and T → ∞. The fact that πT > 0 (because a > 0)
∂U
is consistent with a representing attractive contributions, since it implies that
> 0 and the
∂V T
internal energy rises as the gas expands (so decreasing the average attractive interactions).
∂G
dG =
dp = V dp
∂p T
G(pf ) − G(pi ) =

pf

pi

V dp

In order to complete the integration, V as a function of p is required.
∂V
= −κT V (given),
∂p T

d ln V = −κ dp

so

Hence, the volume varies with pressure as
V
V0

d ln V = −κT

p
pi

dp

or V = V0 e−κT (p−pi ) (V = V0 when p = pi )
pf

Hence,

pi

dG =

V dp = V0

G(pf ) = G(pi ) + (V0 ) ×

pf

pi

e−κT (p−pi ) dp

1 − e−κT (pf −pi )
κT

= G(pi ) + (V0 ) ×

1 − e−κT p
κT

1, 1 − e−κT p ≈ 1 − 1 − κT p + 21 κT2 p 2 = κT p − 21 κT2 p 2

If κT p

1
Hence, G = G + V0 p 1 − κT p
2
For the compression of copper, the change in molar Gibbs function is
1
G m = V m p 1 − κT p =
2
=

63.54 g mol−1
8.93 × 106 g m−3

M p
ρ

1
× 1 − κT p
2

1
× (500) × (1.013 × 105 Pa) × 1 − κT p
2

= (360.4 J) × 1 − 21 κT p
If we take κT = 0 (incompressible),
1
2 κT

Gm = +360 J. For its actual value

p = 21 × (0.8 × 10−6 atm−1 ) × (500 atm) = 2 × 10−4

1 − 21 κT p = 0.9998
Hence,

Gm differs from the simpler version by only 2 parts in 104 (0.02 per cent)

THE SECOND LAW: THE MACHINERY

P5.19

1
V

κS = −

×

83

1
∂V
=−
∂p
∂p S
V ∂V

S

The only constant-entropy changes of state for a perfect gas are reversible adiabatic changes, for
which
pV γ = const
∂p
=
∂V S

∂ const

= −γ ×
∂V V γ S
+1
−1
Therefore, κS =
=
−γp
γp
V V

Then,

const
V γ +1

=

−γp
V

Hence, pγ κS = +1
P5.21

S = S(T , p)
∂S
dS =
dT +
∂T p

∂S

dp
∂p T
∂S
∂S
dT + T
dp
∂T p
∂p T

T dS = T
Use

∂S
=
∂T p

∂S
∂H p

∂H
1
= × Cp
∂T p
T

∂H
= T , Problem 5.6
∂S p

∂S

∂V
=−
[Maxwell relation]
∂p T
∂T p
∂V
dp = Cp dT − αT V dp
∂T p
For reversible, isothermal compression, T dS = dqrev , dT = 0; hence
Hence, T dS = Cp dT − T

dqrev = −αT V dp
qrev =

pf
pi

−αT V dp = −αT V p

[α and V assumed constant]

For mercury
qrev = (−1.82 × 10−4 K −1 ) × (273 K) × (1.00 × 10−4 m−3 ) × (1.0 × 108 Pa) = −0.50 kJ
P5.25

When we neglect b in the van der Waals equation we have
p=

RT
a

− 2
Vm
Vm

and hence
Z =1−

a
RT Vm

Then substituting into eqn 5.20 we get
ln φ =

p
o

p −a
Z−1
dp =
dp
RT Vm
p
o p


INSTRUCTOR’S MANUAL

84

In order to perform this integration we must eliminate the variable Vm by solving for it in terms of p.

Rewriting the expression for p in the form of a quadratic we have
RT
a
Vm2 −
Vm + = 0
p
p
The solution is
1
1
Vm =
(RT )2 − 4ap
RT /p ±
2
p
applying the approximation (RT )2
Vm =

4ap we obtain

RT
RT
±
p
p

1
2

Choosing the + sign we get

RT
Vm =
which is the perfect volume
p
Then
p
a
ap
ln φ =
−
dp = −
2
RT
(RT )2
0
For ammonia a = 4.169 atm L2 mol−2
4.169 atm L2 mol−2 × 10.00 atm

ln φ = −

(0.08206 L atm K−1 mol−1 × 298.15 K)2
= −0.06965
f
φ = 0.9237 =
p

f = φp = 0.9237 × 10.00 atm = 9.237 atm
P5.27

The equation of state

qT
pVm
=1+
is solved for Vm =
RT
Vm

pVm
2q
−1
Z−1
qT
R
= RT
=
=
p
p
pVm
1 + 1 + 4pq
R

ln φ =

p
0

Z−1
p

Defining, a ≡ 1 + 1 +
ln φ =

a
2

a−1
a

dp[24] =

RT
2p

1+ 1+

1/2

dp
2q p
R 0 1 + 1 + 4pq
R

1/2

4pq 1/2
R(a − 1)
da, gives
, dp =

R
2q
da

[a = 2, when p = 0]

1
4pq 1/2
4pq 1/2 1
1
= a − 2 − ln a = 1 +
1+
− 1 − ln
+
2
R
2
R
2
Hence, φ =

1/2
2e{(1+4pq/R) −1}

1 + 1 + 4pq
R

1/2

4pq 1/2

so
R


THE SECOND LAW: THE MACHINERY

85

This function is plotted in Fig. 5.1(a) when

4pq
R

1, and using the approximations

1.2
–1.0

–0.1

–0.01

1.0
0.01

0.1

1.0

0.8

Figure 5.1(a)
1
ex ≈ 1 + x, (1 + x)1/2 ≈ 1 + x, and (1 + x)−1 ≈ 1 − x [x
1]
2
pq
φ ≈1+
R
When φ is plotted against x = 4pq/R on a linear rather than exponential scale, the apparent curvature
seen in Fig. 5.1(a) is diminished and the curve seems almost linear. See Fig. 5.1(b).
2

2

–2

Figure 5.1(b)

Solution to applications
P5.28

wadd,max =
rG

−−

rG

[4.38]

(37◦ C) = τ r G−− (Tc ) + (1 − τ ) r H −− (Tc )
=

310 K
298.15 K

Problem 5.1, τ =

× (−6333 kJ mol−1 ) + 1 −

310 K
298.15 K

T
Tc

× (−5797 kJ mol−1 )

= −6354 kJ mol−1
The difference is r G−− (37◦ C) − r G−− (Tc ) = {−6354 − (−6333)} kJ mol−1 = −21 kJ mol−1
Therefore, an additional 21 kJ mol−1 of non-expansion work may be done at the higher temperature.


INSTRUCTOR’S MANUAL

86

Comment. As shown by Problem 5.1, increasing the temperature does not necessarily increase the
maximum non-expansion work. The relative magnitude of r G−− and r H −− is the determining

factor.
P5.31

The Gibbs–Helmholtz equation is
∂
∂T

G
T

=−

H
T2

so for a small temperature change
rG

−−

=

T
so

d

r

G−−

T

−−
r G190

=

rH
T2

−−

r

=−

T

H −− dT
T2

−− T190
r G220
T220

+

−−
r G2

and

T2

−−
r G190

and
−−

rH

T190
1−

=
=

−−
r G1

T1

−−
r G220

T220

−

−−
rH
T2 T

+

rH

−−

1
T190

−

1
T220

T190
T220

For the monohydrate
−−
r G190

= (46.2 kJ mol−1 ) ×

−−
r G190

= 57.2 kJ mol−1

190 K
220 K

+ (127 kJ mol−1 ) × 1 −

190 K
,
220 K

190 K
220 K

+ (188 kJ mol−1 ) × 1 −

190 K
,
220 K

190 K
220 K

+ (237 kJ mol−1 ) × 1 −

190 K
,
220 K

For the dihydrate

−−
r G190

= (69.4 kJ mol−1 ) ×

−−
r G190

= 85.6 kJ mol−1

For the monohydrate

P5.32

−−
r G190

= (93.2 kJ mol−1 ) ×

−−
r G190

= 112.8 kJ mol−1

The change in the Helmholtz energy equals the maximum work associated with stretching the polymer.
Then
dwmax = dA = −f dl
For stretching at constant T
∂A
∂U

∂S
=−
+T
∂l T
∂l T
∂l T
assuming that (∂U/∂l)T = 0 (valid for rubbers)
f =−

f = T

∂S
=T
∂l T

=T −

3kB l
N a2

∂
∂l T

=−

−

3kB T
N a2

3kB l 2
+C
2N a 2
l

This tensile force has the Hooke’s law form f = −kH l with kH = 3kB T /N a 2 .