Instructor solution manual to accompany physical chemistry 7th ed by peter atkins chap03

3

The First Law: the machinery

Solutions to exercises
Discussion questions
E3.1(b)

The following list includes only those state functions that we have encountered in the first three
chapters. More will be encountered in later chapters.
Temperature, pressure, volume, amount, energy, enthalpy, heat capacity, expansion coefficient,
isothermal compressibility, and Joule–Thomson coefficient.

E3.2(b)

One can use the general expression for πT given in Justification 3.3 to derive its specific form for
a van der Waals gas as given in Exercise 3.14(a), that is, πT = a/Vm2 . (The derivation is carried
out in Example 5.1.) For an isothermal expansion in a van der Waals gas dUm = (a/Vm )2 . Hence
Um = −a(1/Vm,2 − 1/Vm,1 ). See this derivation in the solution to Exercise 3.14(a). This formula
corresponds to what one would expect for a real gas. As the molecules get closer and closer the molar
volume gets smaller and smaller and the energy of attraction gets larger and larger.

E3.3(b)

The solution to Problem 3.23 shows that the Joule–Thomson coefficient can be expressed in terms
of the parameters representing the attractive and repulsive interactions in a real gas. If the attractive
forces predominate then expanding the gas will reduce its energy and hence its temperature. This
reduction in temperature could continue until the temperature of the gas falls below its condensation
point. This is the principle underlying the liquefaction of gases with the Linde Refrigerator which
utilizes the Joule–Thomson effect. See Section 3.4 for a more complete discussion.

Numerical exercises
E3.4(b)

A function has an exact differential if its mixed partial derivatives are equal. That is, f (x, y) has an
exact differential if
∂f
∂y

∂
∂x
(a)

(b)

E3.5(b)
E3.6(b)

∂f
∂x
∂f
∂y
∂f
∂s
∂f
∂t

dz =
(a)
(b)

=

∂
∂y

∂f
∂x

∂
∂y
∂
= 2x 3 y and
∂x
∂
= tes + 1 and
∂t
∂
= 2t + es and
∂s
= 3x 2 y 2

and

∂f
= 6x 2 y
∂x
∂f
= 6x 2 y
∂y
∂f

= es
∂s
∂f
= es
∂t

Therefore, exact.

Therefore, exact.

∂z
∂z
dx
2x dy
dx +
dy =
−
∂x
∂y
(1 + y)2
(1 + y)3

dz =

∂z
∂z
dx +
dy = (3x 2 − 2y 2 ) dx − 4xy dy
∂x
∂y

∂
∂ 2z
=
(3x 2 − 2y 2 ) = −4y
∂y∂x
∂y
∂ 2z
∂
and
=
(−4xy) = −4y
∂x∂y
∂x


THE FIRST LAW: THE MACHINERY

E3.7(b)

dz =

45

∂z
∂z
dx +
dy = (2xy + y 2 ) dx + (x 2 + 2xy) dy
∂x
∂y

∂ 2z
∂
=
(2xy + y 2 ) = 2x + 2y
∂y∂x
∂y
and

∂ 2z
∂ 2
=
(x + 2xy) = 2x + 2y
∂x∂y
∂x
∂Cp
=
∂p T

E3.8(b)

∂
∂p

∂ 2H
∂H
=
∂T p
∂p∂T
T

∂
∂T

∂H
∂p T p

∂H
= 0 for a perfect gas, its temperature derivative also equals zero; thus
∂p T

Because

∂Cp
= 0.
∂p T
∂(U +pV )

∂V
(∂H /∂V )p
(∂U/∂V )p + p
∂H
p
p
=
=
=
= 1+
∂U p
(∂U/∂V )p

(∂U/∂V )p
(∂U/∂V )p
(∂U/∂V )p

E3.9(b)
E3.10(b)

∂p
dV +
∂V T

dp =

dp
1
=
p
p

d ln p =

We express

∂T
∂V p

so d ln p = −
E3.11(b)

U=

∂p
dT
∂T V

−1
∂p
∂p
1
∂V
=− V
so
=−
∂p T
∂V T
∂V T
κT V
1
∂p
in terms of κT and the expansion coefficient α =
∂T V
V

1
V

∂p
∂T V

1

∂p
dV +
∂V T
p

∂p
in terms of the isothermal compressibility κT
∂V T

We express
κT = −

∂p
dT
∂T V

∂V
= −1
∂p T

α
1
1
+
=
pV κT
pκT
pκT

3

nRT
2

so

(∂V /∂T )p
∂p
α
=−
=
∂T V
(∂V /∂p)T
κT

so

∂U
= 0
∂p T

α dT −

dV
V

by direct differentiation

H = U + pV = 23 nRT + nRT = 25 nRT ,
so
E3.12(b)

∂H
= 0 by direct differentiation
∂p T
1
∂V
nRT
α=
V =
V
∂T p
p
α=

1
V

×

V
T

=

1
T

∂V
V
nR

=
=
∂T p
p
T

∂V
∂T p


INSTRUCTOR’S MANUAL

46

E3.13(b)

κT = −

1
V

κT = −

1
V

× −

nRT
p2

=

1
p

The Joule–Thomson coefficient µ is the ratio of temperature change to pressure change under
conditions of isenthalpic expansion. So
µ=

E3.14(b)

∂V
nRT
=− 2
∂p T
p

∂V
∂p T

∂T
≈
∂p H

T
−10 K
= 0.48 K atm−1
=
(1.00 − 22) atm

p

Um = Um (T , Vm )

dUm =

dUm
dT +
∂T Vm

∂Um
∂Vm

dVm

dT = 0 in an isothermal process, so
dUm =

a
∂Um
dVm = 2 dVm
∂Vm T
Vm
Vm2

−1

−1

Vm2

22.1 L mol dV
a
a 22.1 L mol
m
Um =
dUm =
dV
=
a
=
−
m
2
2
Vm 1.00 L mol−1
Vm1
Vm1 Vm
1.00 L mol−1 Vm
a
a
21.1a
=−
+
=
= 0.95475a L−1 mol
−1
−1
22.1 L mol
1.00 L mol

22.1 L mol−1

a = 1.337 atm L2 mol−2
Um = (0.95475 mol L−1 ) × (1.337 atm L2 mol−2 )
= (1.2765 atm L mol−1 ) × (1.01325 × 105 Pa atm−1 ) ×

1 m3
103 L

= 129 Pa m3 mol−1 = 129 J mol−1
w=−
so w = −

pex dVm
RT
Vm − b

and p =
dVm +

RT
a
− 2 for a van der Waals gas
Vm − b V m
a
dVm = −q + Um
Vm2

Thus
q=+

22.1 L mol−1
1.00 L mol−1

RT
Vm − b

dVm = +RT ln(Vm − b)

22.1 L mol−1
1.00 L mol−1

= +(8.314 J K−1 mol−1 ) × (298 K)
× ln

22.1 − 3.20 × 10−2
1.00 − 3.20 × 10−2

= +7.7465 kJ mol−1
w = −7747 J mol−1 + 129 J mol−1 = −7618¯ J mol−1 = −7.62 kJ mol−1


THE FIRST LAW: THE MACHINERY

E3.15(b)

47

The expansion coefficient is
α=

=

1
V

V (3.7 × 10−4 K −1 + 2 × 1.52 × 10−6 T K−2 )
∂V
=
∂T p
V

V [3.7 × 10−4 + 2 × 1.52 × 10−6 (T /K)] K −1
V [0.77 + 3.7 × 10−4 (T /K) + 1.52 × 10−6 (T /K)2 ]

[3.7 × 10−4 + 2 × 1.52 × 10−6 (310)] K −1
= 1.27 × 10−3 K −1
0.77 + 3.7 × 10−4 (310) + 1.52 × 10−6 (310)2
Isothermal compressibility is
V
1 ∂V
V
so
p=−
≈−
κT = −
V ∂p T
V p
V κT
A density increase 0.08 per cent means V /V = −0.0008. So the additional pressure that must be
applied is

=

E3.16(b)

p=
E3.17(b)

0.0008
= 3.6 × 102 atm
2.21 × 10−6 atm−1

The isothermal Joule–Thomson coefficient is
∂H
= −µCp = −(1.11 K atm−1 ) × (37.11 J K −1 mol−1 ) = −41.2 J atm−1 mol−1
∂p T
If this coefficient is constant in an isothermal Joule–Thomson experiment, then the heat which must
be supplied to maintain constant temperature is H in the following relationship
H /n
= −41.2 J atm−1 mol−1
p

so

H = −(41.2 J atm−1 mol−1 )n p

H = −(41.2 J atm−1 mol−1 ) × (12.0 mol) × (−55 atm) = 27.2 × 103 J
E3.18(b)

The Joule–Thomson coefficient is
µ=

∂T
≈
∂p H

T
p

so

p=

T
−4.5 K
= −3.4 × 102 kPa
=
µ
13.3 × 10−3 K kPa−1

Solutions to problems
Assume that all gases are perfect and that all data refer to 298 K unless stated otherwise.

Solutions to numerical problems
P3.1

1 atm
1.013 × 105 Pa
For the change of volume with pressure, we use
κT = (2.21 × 10−6 atm−1 ) ×

= 2.18 × 10−11 Pa−1

∂V
1
dp[constant temperature] = −κT V dp
κT = −
∂p T
V
V = −κT V p [If change in V is small compared to V ]

dV =

∂V
∂p T

p = (1.03 × 103 kg m−3 ) × (9.81 m s−2 ) × (1000 m) = 1.010 × 107 Pa.


INSTRUCTOR’S MANUAL

48

Consequently, since V = 1000 cm3 = 1.0 × 10−3 m3 ,
V ≈ (−2.18 × 10−11 Pa−1 ) × (1.0 × 10−3 m3 ) × (1.010 × 107 Pa)
= −2.2 × 10−7 m3 ,

or

−0.220 cm3 .

For the change of volume with temperature, we use
1
∂V
dT [constant pressure] = αV dT
α=
∂T p
V
V = αV T [if change in V is small compared to V ]

dV =

∂V
∂T p

≈ (8.61 × 10−5 K −1 ) × (1.0 × 10−3 m3 ) × (−30 K)
≈ −2.6 × 10−6 m3 ,

or

− 2.6 cm3

V ≈ −2.8 cm3 V = 997.2 cm3

Overall,

Comment. A more exact calculation of the change of volume as a result of simultaneous pressure
and temperature changes would be based on the relationship
dV =

∂V

dp +
∂p T

∂V
dT = −κT V dp + αV dT
∂p p

This would require information not given in the problem statement.
P3.5

Use the formula derived in Problem 3.25.
Cp,m − CV ,m = λR
which gives γ =

1
(3Vr − 1)2
=1−
λ
4Vr3 Tr

Cp,m
CV ,m + λR
λR
=
=1+
CV ,m
CV ,m
CV ,m

In conjunction with CV ,m = 23 R for a monatomic, perfect gas, this gives

γ = 1 + 23 λ
Vm
Vm
T
27RbT
, Tr =
(Table 1.6) with a =
=
=
Vc
3b
Tc
8a
4.137 L2 atm mol−2 and b = 5.16 × 10−2 L mol−1 (Table 1.6). Hence, at 100◦ C and 1.00 atm,
RT
where Vm ≈
= 30.6 L mol−1
p

For a van der Waals gas Vr =

Vr ≈

30.6 L mol−1
= 198
(3) × (5.16 × 10−2 L mol−1 )

Tr ≈

(27) × (8.206 × 10−2 L atm K−1 mol−1 ) × (5.16 × 10−2 L mol−1 ) × (373 K)

≈ 1.29
(8) × (4.317 L2 atm mol−2 )

Hence
[(3) × (198) − (1)]2
1
=1−
= 1 − 0.0088 = 0.9912,
λ
(4) × (198)3 × (1.29)
γ ≈ (1) + 23 × (1.009) = 1.67

λ = 1.009


THE FIRST LAW: THE MACHINERY

49

Comment. At 100◦ C and 1.00 atm xenon is expected to be close to perfect, so it is not surprising
that γ differs only slightly from the perfect gas value of 53 .
P3.7

See the solution to Problem 3.6. It does not matter whether the piston between chambers 2 and 3 is
diathermic or adiabatic as long as the piston between chambers 1 and 2 is adiabatic. The answers
are the same as for Problem 3.6. However, if both pistons are diathermic, the result is different. The
solution for both pistons being diathermic follows.
See Fig. 3.1.

Diathermic piston

Diathermic piston

Figure 3.1

Initial equilibrium state.
n = 1.00 mol diatomic gas in each section
pi = 1.00 bar
Ti = 298 K
For each section
nRTi
(1 mol) × (0.083 145 L bar K−1 mol−1 ) × (298 K)
Vi =
=
pi
1.00 bar
= 24.8 L
Vtotal = 3Vi = 74.3 L = constant
Final equilibrium state. The diathermic walls allow the passage of heat. Consequently, at equlibrium
all chambers will have the same temperature T1 = T2 = T3 = 348 K. The chambers will also be at
mechanical equlibrium so
p1 = p2 = p3 =

(n1 + n2 + n3 )RT1
Vtotal

(3 mol) × (0.083 145 L bar K−1 mol−1 ) × (348 K)
74.3 L
= 1.17 bar = p2 = p3

=

The chambers will have equal volume.


INSTRUCTOR’S MANUAL

50

Vtotal
= Vi = 24.8 L = V1 = V2 = V3
3

V1 =

U1 = n1 CV T1 = n1 25 R

T1

= (1 mol) × 25 × (8.314 51 J K −1 mol−1 ) × (348 K − 298 K)
U1 = 1.04 kJ =

U2 =

U3

Utotal = 3 U1 = 3.12 kJ =

Utotal

Solutions to theoretical problems
P3.11

dw =

∂w
dx +
∂x y,z

∂w
dy +
∂y x,z

∂w
dz
∂z x,y

dw = (y + z) dx + (x + z) dy + (x + y) dz
This is the total differential of the function w, and a total differential is necessarily exact, but here
we will demonstrate its exactness showing that its integral is independent of path.
Path a
dw = 2x dx + 2y dy + 2z dz = 6x dx
(1,1,1)
(0,0,0)

dw =

1
0

6x dx = 3

Path b
dw = 2x 2 dx + (y 1/2 + y) dy + (z1/2 + z) dz = (2x 2 + 2x + 2x 1/2 ) dx
(1,1,1)
(0,0,0)

dw =

1
0

(2x 2 + 2x + 2x 1/2 ) dx =

2
4
+1+ =3
3
3

Therefore, dw is exact.
P3.12

U = U (T , V )
∂U
∂U
dU =
dT +
dV = CV dT +
∂T V

∂V T
For U = constant, dU = 0, and
CV dT = −

∂U
dV
∂V T

or

CV = −

∂U
dV
∂V T

∂U
∂V T

dV
∂U
=−
dT U
∂V T

∂V
∂T U

This relationship is essentially the permuter [Relation 3, Further information 1.7].
P3.13

H = H (T , p)
∂H
dH =
dT +
∂T p

∂H
dp = Cp dT +
∂p T
For H = constant, dH = 0, and
∂H
dp = −Cp dT
∂p T
∂H
dT
= −Cp
= −Cp
∂p T
dp H

∂H
dp
∂p T

∂T
= −Cp µ = −µCp
∂p H

This relationship is essentially the permuter [Relation 3, Further information 1.7].

THE FIRST LAW: THE MACHINERY

P3.16

51

The reasoning here is that an exact differential is always exact. If the differential of heat can be shown
to be inexact in one instance, then its differential is in general inexact, and heat is not a state function.
Consider the cycle shown in Fig. 3.2.

1

4
A
Isotherm at
B
2

3

Isotherm at

Isotherm at

Figure 3.2
The following perfect gas relations apply at points labelled 1, 2, 3 and 4 in Fig. 3.2.
(1) p1 V1 = p2 V2 = nRT ,
Define

T =T −T ,

(2) p2 V1 = nRT ,

T =T −T

Subtract (2) from (1)
−nRT + nRT = −p2 V1 + p1 V1
V1 (p1 − p2 )
RT
Subtracting (1) from (3) we obtain

giving

T =

T =

V2 (p1 − p2 )
RT

Since V1 = V2 ,

T =

T

qA = Cp T − CV T = (Cp − CV ) T
qB = −CV T + Cp T = (Cp − CV ) T

giving qA = qB and q(cycle) = qA − qB = 0.
dq = 0 and dq is not exact.
a
RT
−
p = p(T , V ) =
Vm − b Vm2

Therefore
P3.18

dp =

∂p
dT +
∂T V

∂p
dV
∂V T

(3) p1 V2 = nRT


INSTRUCTOR’S MANUAL

52

In what follows adopt the notation Vm = V .
∂p

R
;
=
∂T V
V −b
then, dp =

∂p
RT
2a
=−
+ 3
∂V T
(V − b)2
V

R
V −b

dT +

RT
2a
−
V3
(V − b)2

dV
∂V
is more readily evaluated with the use

∂T p

Because the van der Waals equation is a cubic in V ,
of the permuter.
∂p

R
∂T V
∂V
V −b
=−
=−
∂p
∂T p
+ V2a3
− (VRT
∂V T
−b)2

RV 3 (V − b)
RT V 3 − 2a(V − b)2

=

For path a
T2 ,V2
T1 ,V1

dp =
=

T2
T1

V2
R
RT2
2a
−
+ 3
dT +
2
V1 − b
(V
−
b)
V
V1

dV

R
RT2
RT2
(T2 − T1 ) +
−
−a
V1 − b
(V2 − b) (V1 − b)

=−

RT1
RT2
+
−a
V1 − b V 2 − b

1
V22

−

1
V22

−

1
V12

1
V12

For path b
T2 ,V2
T1 ,V1

dp =
=

V2
V1

−

T2
RT1
R
2a
dV
+
+
dT
2
3
(V − b)
V
T1 V2 − b

RT1
RT1
−
−a
V 2 − b V1 − b

=−

RT1
RT2

+
−a
V1 − b V 2 − b

1
V22

1

−

1
V22

V12
−

+

R
(T2 − T1 )
V2 − b

1
V12

Thus, they are the same and dp satisfies the condition of an exact differential, namely, that its integral
between limits is independent of path.
P3.20

p = p(V , T )
Therefore,
dp =

∂p
dV +
∂V T

∂p
dT
∂T V

with p =

∂p
−nRT
2n2 a
−p
+
=
+
=
2
3
∂V T
V
− nb
(V − nb)
V
∂p

p
n2 a
nR
= +
=
∂T V
V − nb
T
TV2

n2 a
nRT
− 2 [Table 1.6]
V − nb
V
n2 a
V3

×

V − 2nb
V − nb


THE FIRST LAW: THE MACHINERY

53

Therefore, upon substitution
dp =

=

−p dV
V − nb

n2 a
V3

+

× (V − 2nb) ×

(n2 a) × (V − nb)/V 3 − p
V − nb

=

a(Vm − b)/Vm3 − p
Vm − b

dV +

dV
V − nb

p + n2 a/V 2
T

p + a/Vm2
T

dVm +

+

p dT
T

+

n2 a
V2

×

dT
T

dT

dT

Comment. This result may be compared to the expression for dp obtained in Problem 3.18.
P3.21

p=

nRT
n2 a
− 2 (Table 1.6)

V − nb
V

Hence T =

p
× (V − nb) +
nR

na
RV 2

∂T
V − nb
Vm − b
=
=
=
nR
∂p V
R

× (V − nb)

1
∂p
∂T V

For Euler’s chain relation, we need to show that
Hence, in addition to

∂T
and
∂p V

which can be found from

∂p
∂V T

∂T
∂p V

∂V
= −1
∂T p

∂p
[Problem 3.20] we need
∂V T

∂T
p
=
+
∂V p
nR
=

T

V − nb

na
RV 2
−

−

2na
RV 3

2na
RV 3

∂V
=
∂T p

1
∂T
∂V p

× (V − nb)

× (V − nb)

Therefore,
∂T
∂p V

∂p
∂V T

∂p
∂T
∂p V ∂V T
∂T
∂V p

∂V
=
∂T p

=

V −nb
nR

a
× (V−nRT
+ 2n
−nb)2
V3

T
V −nb

2na
− RV
× (V − nb)

3

2

=

−T
V −nb

2na
+ RV
× (V − nb)
3

T
V −nb

2na
− RV
× (V − nb)
3

= −1
P3.23

µCp = T

∂V
−V =
∂T p

T
∂T
∂V p

− V [Relation 2, Further information 1.7]

∂T
2na
T
−
=
(V − nb) [Problem 3.21]
∂V p
V − nb RV 3


INSTRUCTOR’S MANUAL

54

Introduction of this expression followed by rearrangement leads to
(2na) × (V − nb)2 − nbRT V 2
×V
RT V 3 − 2na(V − nb)2

µCp =

Then, introducing ζ =

µCp =

1 − nbζ
V
ζ −1

RT V 3
to simplify the appearance of the expression
2na(V − nb)2
V =

1 − Vbζm
ζ −1

V

For xenon, Vm = 24.6 L mol−1 , T = 298 K, a = 4.137 L2 atm mol−2 , b = 5.16 × 10−2 L mol−1 ,
b
5.16 × 10−2 L mol−1
nb
=
=
= 2.09 × 10−3
V
Vm
24.6 L mol−1
ζ =

(8.206 × 10−2 L atm K−1 mol−1 ) × (298 K) × (24.6 L mol−1 )3
= 73.0

(2) × (4.137 L2 atm mol−2 ) × (24.6 L mol−1 − 5.16 × 10−2 L mol−1 )2

Therefore, µCp =

1 − (73.0) × (2.09 × 10−3 )
× (24.6 L mol−1 ) = 0.290 L mol−1
72.0

Cp = 20.79 J K−1 mol−1 [Table 2.6], so
µ=

0.290 × 10−3 m3 mol−1
0.290 L mol−1
=
−1
−1
20.79 J K mol
20.79 J K−1 mol−1

= 1.393 × 10−5 K m3 J−1 = 1.393 × 10−5 K Pa−1
= (1.393 × 10−5 ) × (1.013 × 105 K atm−1 ) = 1.41 K atm−1
The value of µ changes at T = T1 and when the sign of the numerator 1 −
is positive). Hence
bζ
= 1 at T = T1
Vm
that is, T1 =

For xenon,

2a
Rb

or

× 1−

RT1 bV 3
=1
2na(V − nb)2 Vm

nbζ
changes sign (ζ − 1
V

implying that T1 =

2a(Vm − b)2
RbVm2

b 2
b 2
27
Tc 1 −
=
Vm
4
Vm

2a

(2) × (4.137 L2 atm mol−2 )
=
= 1954 K
Rb
(8.206 × 10−2 L atm K−1 mol−1 ) × (5.16 × 10−2 L mol−1 )

5.16 × 10−2
and so T1 = (1954 K) × 1 −
24.6

2

= 1946 K

Question. An approximate relationship for µ of a van der Waals gas was obtained in Problem 3.17.
Use it to obtain an expression for the inversion temperature, calculate it for xenon, and compare to
the result above.


THE FIRST LAW: THE MACHINERY

P3.25

Cp,m − CV ,m =

55

α2 T V
[3.21] = αT V
κT

∂p
[Justification 3.3]
∂T V

∂p
nR
=
[Problem 3.20]
∂T V
V − nb
∂V
1
αV =
=
∂T
∂T p
∂V p

Substituting,
T

Cp,m − CV ,m =

∂p
∂T V
∂T
∂V p

2na

∂T
T
−
=
(V − nb) [Problem 3.21]
∂V p
V − nb RV 3

so

Substituting,
nRT
(V −nb)

Cp,m − CV ,m =

T
(V −nb)

−

2na
RV 3

× (V − nb)

= nλR

with λ =

1
1−

2na
RT V 3

× (V − nb)2

For molar quantities,
Cp,m − CV ,m = λR

with

2a(Vm − b)2
1
=1−
λ
RT Vm3

Now introduce the reduced variables and use Tc =

8a
, Vc = 3b.
27Rb

After rearrangement,
1
(3Vr − 1)2
=1−
λ

4Tr Vr3
For xenon, Vc = 118.1 cm3 mol−1 , Tc = 289.8 K. The perfect gas value for Vm may be used as any
1
error introduced by this approximation occurs only in the correction term for .
λ
Hence, Vm ≈ 2.45 L mol−1 , Vc = 118.8 cm3 mol−1 , Tc = 289.8 K, and Vr = 20.6 and Tr = 1.03;
therefore
1
(61.8 − 1)2
= 0.90,
=1−
λ
(4) × (1.03) × (20.6)3
and

giving λ ≈ 1.1

Cp,m − CV ,m ≈ 1.1R = 9.2 J K−1 mol−1
P3.27

(a)

µ=−

1
Cp

∂H
1
=

∂p T
Cp

RT
+ aT 2
p
R
∂Vm
= + 2aT
∂T p
p

Vm =

T

∂Vm
− Vm
∂T p

[Justification 3.1 and Problem 3.24]


INSTRUCTOR’S MANUAL

56

µ=
µ=
(b)

1
Cp

RT
RT
+ 2aT 2 −
− aT 2
p
p

aT 2
Cp
∂p
∂T V
∂p
∂Vm
∂T p ∂T V

CV = Cp − αT Vm
= Cp − T

RT
Vm − aT 2
∂p
R
RT (−2aT )
=
−
∂T V

(Vm − aT 2 )2
Vm − aT 2
R
2aRT 2
=
+
(RT /p) (RT /p)2
2ap 2
p
= +
T
R
Therefore
R
p
2ap 2
CV = Cp − T
+ 2aT ×
+
p
T
R

But, p =

= Cp −

RT
p

1+

CV = Cp − R 1 +

2apT
R

× 1+

2apT
R

×

p
T

2apT 2
R

Solutions to additional problems
P3.29

(a) The Joule–Thomson coefficient is related to the given data by
µ = −(1/Cp )(∂H /∂p)T = −(−3.29 × 103 J mol−1 MPa−1 )/(110.0 J K−1 mol−1 )
= 29.9 K MPa−1
(b) The Joule–Thomson coefficient is defined as
µ = (∂T /∂p)H ≈ ( T / p)H
Assuming that the expansion is a Joule–Thomson constant-enthalpy process, we have
T = µ p = (29.9 K MPa−1 ) × [(0.5 − 1.5) × 10−1 MPa] = −2.99 K