Atkins CHAPTER 05 ( SOLUTIONS MANUAL )

Prévia do material em texto

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/V 2m 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 ofZ−1 over a range of pressures. So we expect no simple proportionality between φ andZ.
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) α = 1
V
×
(
∂V
∂T
)
p
κT = −
(
1
V
)
×
(
∂V
∂p
)
T(
∂S
∂p
)
T
= −
(
∂V
∂T
)
p
= −αV
E5.5(b) �G = nRT ln
(
pf
pi
)
at constant temperature,
pf
pi
= Vi
Vf
= nRT ln
(
Vi
Vf
)
= (2.5× 10−3 mol)× (8.314 J K−1 mol−1)× (298 K)× ln
(
72
100
)
= −2.035 = −2.0 J
76 INSTRUCTOR’S MANUAL
E5.6(b)
(
∂G
∂T
)
p
= −S
(
∂Gf
∂T
)
p
= −Sf and
(
∂Gi
∂T
)
p
= −Si
�S = Sf − Si = −
(
∂Gf
∂T
)
p
+
(
∂Gi
∂T
)
p
= −
{
∂(Gf −Gi)
∂T
}
p
= −
(
∂�G
∂T
)
p
= − ∂
∂T
{−73.1+ 42.8 T/K} J
= −42.8 J K−1
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
ρ = m
V
so V = m
ρ
= 25 g
0.791 g cm−3
= 31.61 cm3
�G = (31.61 cm3)×
(
1 m3
106 cm3
)
× (99.9× 106 Pa)
= +3.2 kJ
E5.8(b) (a) �S = nR ln
(
Vf
Vi
)
= nR ln
(
pi
pf
)
[Boyle’s Law]
Taking inverse logarithms
pf = pie−�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
(
252.0
92.0
)
= 2.71 kJ mol−1
E5.10(b) µ0 = µ −�− + RT ln
(
p
p −�−
)
µ = µ −�− + RT ln
(
f
p −�−
)
THE SECOND LAW: THE MACHINERY 77
µ− µ0 = RT ln
(
f
p
) [
f
p
≡ φ
]
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
E5.11(b) B ′ = B
RT
= −
(160.0 cm3 mol−1)×
(
1 m3
106 cm3
)
(8.314 J K−1 mol−1)× (100 K)
= −1.924× 10−7 Pa−1
φ = eB ′p+···
≈ e−
(
1.924×10−7 Pa−1)×(62×106 Pa)
≈ e−11.93
= 7× 10−6 or of the order of 10−6
E5.12(b) �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) V =
(
∂G
∂p
)
T
[5.10] = RT
p
+ B ′ + C′p +D′p2
which is the virial equation of state.
E5.15(b)
(
∂S
∂V
)
T
=
(
∂p
∂T
)
V
For a Dieterici gas
p = RT e
−a/RT Vm
Vm − b
(
∂p
∂T
)
Vm
=
R
(
1+ aRVmT
)
e−a/RVmT
Vm − b
dS =
(
∂S
∂Vm
)
T
dVm =
(
∂p
∂T
)
Vm
dVm
78 INSTRUCTOR’S MANUAL
�S =
∫ Vm,f
Vm,i
dS = R
(
1+ a
RVmT
)
e−a/RVmT
(
Vm,f − b
Vm,i − b
)
For a perfect gas �S = R ln
(
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.
Solutions to problems
Solutions to numerical problem
P5.2 For the reaction
N2(g)+ 3H2(g)→ 2NH3(g) �rG −�− = 2�fG −�− (NH3, g)
(a) �rG −�− (500 K) = τ�rG −�− (Tc)+ (1− τ)�rH −�− (Tc)
[
Problem 5.1, τ = T
Tc
]
=
(
500 K
298.15 K
)
× (2)× (−16.45 kJ mol−1)
+
(
1− 500 K
298.15 K
)
× (2)× (−46.11 kJ mol−1)
= −55.17+ 62.43 kJ mol−1 = +7 kJ mol−1
(b) �rG −�− (1000 K) =
(
1000 K
298.15 K
)
× (2)× (−16.45 kJ mol−1)
+
(
1− 1000 K
298.15 K
)
× (2)× (−46.11 kJ mol−1)
= (−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
∂S
)
V
dS +
(
∂U
∂V
)
S
dV
Comparing the two expressions, we see that(
∂U
∂S
)
V
= T and
(
∂U
∂V
)
S
= −p
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 = dqrev
T
= CV dT
T
Then
(
dU
dS
)
V
=
(
∂U
∂S
)
V
= CV dT(
CV dT
T
) = T
For a reversible adiabatic (constant-entropy) change in a perfect gas
dU = dw = −p dV
Therefore,
(
∂U
∂V
)
S
= −p
P5.8
(
∂p
∂S
)
V
= −
(
∂T
∂V
)
S
[Maxwell relation]
= 1(
∂S
∂T
)
V
(
∂V
∂S
)
T
[chain relation] =
(
∂S
∂V
)
T(
∂S
∂T
)
V
[inversion]
=
( ∂p
∂T
)
V(
∂S
∂U
)
V
(
∂U
∂T
)
V
[Maxwell relation] =
−( ∂p∂V )T ( ∂V∂T )
p(
∂S
∂U
)
V
(
∂U
∂T
)
V
[chain relation]
=
−
(
∂V
∂T
)
p
(
∂U
∂S
)
V(
∂V
∂p
)
T
(
∂U
∂T
)
V
[inversion twice] = αT
κT CV
[(
∂U
∂S
)
V
= T
]
P5.10
(
∂H
∂p
)
T
=
(
∂H
∂S
)
p
(
∂S
∂p
)
T
+
(
∂H
∂p
)
S
[Relation 1, Further information 1.7]
dH = T dS + V dp [Problem 5.6]
dH =
(
∂H
∂S
)
p
dS +
(
∂H
∂p
)
S
dp (H = H(p, S))

 compare
Thus,
(
∂H
∂S
)
p
= T ,
(
∂H
∂p
)
S
= V [dH exact]
Substitution yields,
(
∂H
∂p
)
T
= T
(
∂S
∂p
)
T
+ V = −T
(
∂V
∂T
)
p
+ V [Maxwell relation]
(a) For pV = nRT(
∂V
∂T
)
p
= nR
p
, hence
(
∂H
∂p
)
T
= −nRT
p
+ V = 0
(b) For p = nRT
V − nb −
an2
V 2
[Table 1.6]
T = p(V − nb)
nR
+ na(V − nb)
RV 2
80 INSTRUCTOR’S MANUAL
(
∂T
∂V
)
p
= p
nR
+ na
RV 2
− 2na(V − nb)
RV 3
Therefore,
(
∂H
∂p
)
T
= −T(
∂T
∂V
)
p
+ V [inversion] = −T
p
nR + naRV 2 −
2na(V−nb)
RV 3
+ V
which yields after algebraic manipulation
(
∂H
∂p
)
T
=
nb −
(
2na
RT
)
λ2
1−
(
2na
RT V
)
λ2
, λ = 1− nb
V
When
b
Vm
 1, λ ≈ 1 and
2na
RT V
= 2na
RT
× 1
V
≈ 2na
RT
× p
nRT
= 2pa
R2T 2
Therefore,
(
∂H
∂p
)
T
≈ nb−
( 2na
RT
)
1− ( 2pa
R2T 2
)
For argon, a = 1.337 L2 atm mol−2, b = 3.20× 10−2 L mol−1,
2na
RT
= (2)× (1.0 mol)× (1.337 L
2 atm mol−2)
(8.206× 10−2 L atm K−1 mol−1)× (298 K) = 0.11 L
2pa
R2T 2
= (2)× (10.0 atm)× (1.337 L
2 atm mol−2)[
(8.206× 10−2 L atm K−1 mol−1)× (298 K)]2 = 0.045
Hence,
(
∂H
∂p
)
T
≈ {(3.20× 10
−2)− (0.11)}L
1− 0.045 = −0.0817 L = −8.3 J atm
−1
�H ≈
(
∂H
∂p
)
T
�p ≈ (−8.3 J atm−1)× (1 atm) = −8 J
P5.12 πT = T
(
∂p
∂T
)
V
− p [5.8]
p = RT
Vm
+ BRT
V 2m
[The virial expansion, Table 1.6, truncated after the term in B]
(
∂p
∂T
)
V
= R
Vm
+ BR
V 2m
+ RT
V 2m
(
∂B
∂T
)
V
= p
T
+ RT
V 2m
(
∂B
∂T
)
V
Hence, πT = RT
2
V 2m
(
∂B
∂T
)
V
≈ RT
2�B
V 2m�T
Since πT represents a (usually) small deviation from perfect gas behaviour, we may approximate Vm.
Vm ≈ RT
p
πT ≈ p
2
R
× �B
�T
THE SECOND LAW: THE MACHINERY 81
From the data �B = ((−15.6)− (−28.0)) cm3 mol−1 = +12.4 cm3 mol−1
Hence,
(a) πT = (1.0 atm)
2 × (12.4× 10−3 L mol−1)
(8.206× 10−2 L atm K−1 mol−1)× (50 K) = 3.0× 10
−3 atm
(b) πT ∝ p2; 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.
Question. How would you obtain a reliable estimate of πT for argon at 100 atm?
P5.13 CV =
(
∂U
∂T
)
V
and Cp =
(
∂H
∂T
)
p
(a)
(
∂CV
∂V
)
T
= ∂
2U
∂V ∂T
= ∂
2U
∂T ∂V
=
(
∂
∂T
(
∂U
∂V
)
T
)
V
= 0 [πT = 0](
∂CV
∂p
)
T
= ∂
2U
∂p∂T
= ∂
2U
∂T ∂p
=
(
∂
∂T
(
∂U
∂p
)
T
)
V
=
(
∂
∂T
(
∂U
∂V
)
T
(
∂V
∂p
)
T
)
V
= 0 (πT = 0)
Since Cp = CV + R,
(
∂Cp
∂x
)
T
=
(
∂CV
∂x
)
T
for x = p or V
CV and Cp may depend on temperature. Since
dCV
dT
= d
2U
dT 2
,
dCV
dT
is nonzero if U depends on
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.
(
∂CV
∂V
)
T
= ∂
2U
∂T ∂V
=
(
∂πT
∂T
)
V
[Part (a)]
=
(
∂
∂T
RT 2
V 2m
(
∂B
∂T
)
V
)
V
[Problem 5.12]
= 2RT
V 2m
(
∂B
∂T
)
V
+ RT
2
V 2m
(
∂2B
∂T 2
)
V
= RT
V 2m
(
∂2(BT )
∂T 2
)
V
P5.15 πT = T
(
∂p
∂T
)
V
− p [5.8]
p = nRT
V − nb × e
−an/RT V [Table 1.6]
T
(
∂p
∂T
)
V
= nRT
V − nb × e
−an/RT V + na
RT V
× nRT
V − nb × e
−an/RT V = p + nap
RT V
82 INSTRUCTOR’S MANUAL
Hence, πT = nap
RT V
πT → 0 as p → 0, V → ∞, a → 0, and T → ∞. The fact that πT > 0 (because a > 0)
is consistent with a representing attractive contributions, since it implies that
(
∂U
∂V
)
T
> 0 and the
internal energy rises as the gas expands (so decreasing the average attractive interactions).
P5.17 dG =
(
∂G
∂p
)
T
dp = V dp
G(pf )−G(pi) =
∫ pf
pi
V dp
In order to complete the integration, V as a function of p is required.(
∂V
∂p
)
T
= −κT V (given), so d lnV = −κ dp
Hence, the volume varies with pressure as
∫ V
V0
d lnV = −κT
∫ p
pi
dp
or V = V0e−κT (p−pi) (V = V0 when p = pi)
Hence,
∫ pf
pi
dG =
∫
V dp = V0
∫ pf
pi
e−κT (p−pi) dp
G(pf ) = G(pi)+ (V0)×
(
1− e−κT (pf−pi)
κT
)
= G(pi)+ (V0)×
(
1− e−κT �p
κT
)
If κT �p 
 1, 1− e−κT �p ≈ 1−
(
1− κT �p + 12κ2T �p2
) = κT �p − 12κ2T �p2
Hence, G′ = G+ V0�p
(
1− 1
2
κT �p
)
For the compression of copper, the change in molar Gibbs function is
�Gm = Vm�p
(
1− 1
2
κT �p
)
=
(
M�p
ρ
)
×
(
1− 1
2
κT �p
)
=
(
63.54 g mol−1
8.93× 106 g m−3
)
× (500)× (1.013× 105 Pa)×
(
1− 1
2
κT �p
)
= (360.4 J)×
(
1− 12κT �p
)
If we take κT = 0 (incompressible), �Gm = +360 J. For its actual value
1
2κT �p =
( 1
2
)× (0.8× 10−6 atm−1)× (500 atm) = 2× 10−4
1− 12κT �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 83
P5.19 κS = −
(
1
V
)
×
(
∂V
∂p
)
S
= − 1
V
(
∂p
∂V
)
S
The only constant-entropy changes of state for a perfect gas are reversible adiabatic changes, for
which
pV γ = const
Then,
(
∂p
∂V
)
S
=
(
∂
∂V
const
V γ
)
S
= −γ ×
(
const
V γ+1
)
= −γp
V
Therefore, κS = −1
V
(−γp
V
) = +1
γp
Hence, pγ κS = +1
P5.21 S = S(T , p)
dS =
(
∂S
∂T
)
p
dT +
(
∂S
∂p
)
T
dp
T dS = T
(
∂S
∂T
)
p
dT + T
(
∂S
∂p
)
T
dp
Use
(
∂S
∂T
)
p
=
(
∂S
∂H
)
p
(
∂H
∂T
)
p
= 1
T
× Cp
[(
∂H
∂S
)
p
= T , Problem 5.6
]
(
∂S
∂p
)
T
= −
(
∂V
∂T
)
p
[Maxwell relation]
Hence, T dS = Cp dT − T
(
∂V
∂T
)
p
dp = Cp dT − αT V dp
For reversible, isothermal compression, T dS = dqrev, dT = 0; hence
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
Vm
− a
V 2m
and hence
Z = 1− a
RT Vm
Then substituting into eqn 5.20 we get
ln φ =
∫ p
o
(
Z − 1
p
)
dp =
∫ p
o
−a
pRT Vm
dp
84 INSTRUCTOR’S MANUAL
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
V 2m −
RT
p
Vm + a
p
= 0
The solution is
Vm = 12
(
RT/p ± 1
p
√
(RT )2 − 4ap
)
applying the approximation (RT )2 � 4ap we obtain
Vm = 12
(
RT
p
± RT
p
)
Choosing the + sign we get
Vm = RT
p
which is the perfect volume
Then
ln φ =
∫ p
0
− a
RT 2
dp = − ap
(RT )2
For ammonia a = 4.169 atm L2 mol−2
ln φ = − 4.169 atm L
2 mol−2 × 10.00 atm
(0.08206 L atm K−1mol−1 × 298.15 K)2
= −0.06965
φ = 0.9237 = f
p
f = φp = 0.9237× 10.00 atm = 9.237 atm
P5.27 The equation of state
pVm
RT
= 1+ qT
Vm
is solved for Vm =
(
RT
2p
)[
1+
(
1+ 4pq
R
)1/2]
so
Z − 1
p
=
pVm
RT − 1
p
= qT
pVm
=
2q
R
1+
(
1+ 4pqR
)1/2
ln φ =
∫ p
0
(
Z − 1
p
)
dp[24] = 2q
R
∫ p
0
dp
1+
(
1+ 4pqR
)1/2
Defining, a ≡ 1+
(
1+ 4pq
R
)1/2
, dp = R(a − 1)
2q
da, gives
ln φ =
∫ a
2
(
a − 1
a
)
da [a = 2, when p = 0]
= a − 2− ln 1
2
a =
(
1+ 4pq
R
)1/2
− 1− ln
{
1
2
(
1+ 4pq
R
)1/2
+ 1
2
}
Hence, φ = 2e
{(1+4pq/R)1/2−1}
1+
(
1+ 4pqR
)1/2
THE SECOND LAW: THE MACHINERY 85
This function is plotted in Fig. 5.1(a) when 4pq
R
 1, and using the approximations
1.0
0.8
0.10.01
1.0
1.2
–0.01–0.1–1.0
Figure 5.1(a)
ex ≈ 1+ x, (1+ x)1/2 ≈ 1+ 1
2
x, and (1+ x)−1 ≈ 1− x [x 
 1]
φ ≈ 1+ pq
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 [4.38]
�rG
−�− (37◦C) = τ�rG −�− (Tc)+ (1− τ)�rH −�− (Tc)
[
Problem 5.1, τ = T
Tc
]
=
(
310 K
298.15 K
)
× (−6333 kJ mol−1)+
(
1− 310 K
298.15 K
)
× (−5797 kJ mol−1)
= −6354 kJ mol−1
The difference is �rG −�− (37◦C)−�rG −�− (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.
86 INSTRUCTOR’S MANUAL
Comment. As shown by Problem 5.1, increasing the temperature does not necessarily increase the
maximum non-expansion work. The relative magnitude of �rG −�− and �rH −�− is the determining
factor.
P5.31 The Gibbs–Helmholtz equation is
∂
∂T
(
�G
T
)
= −�H
T 2
so for a small temperature change
�
(
�rG
−�−
T
)
= �rH
−�−
T 2
�T and
�rG
−�−
2
T2
= �rG
−�−
1
T1
− �rH
−�−
T 2�T
so
∫
d
�rG
−�−
T
= −
∫
�rH
−�− dT
T 2
and
�rG
−�−
190
T190
= �rG
−�−
220
T220
+�rH −�−
(
1
T190
− 1
T220
)
�rG
−�−
190 = �rG −�−220
T190
T220
+�rH −�−
(
1− T190
T220
)
For the monohydrate
�rG
−�−
190 = (46.2 kJ mol−1)×
(
190 K
220 K
)
+ (127 kJ mol−1)×
(
1− 190 K
220 K
)
,
�rG
−�−
190 = 57.2 kJ mol−1
For the dihydrate
�rG
−�−
190 = (69.4 kJ mol−1)×
(
190 K
220 K
)
+ (188 kJ mol−1)×
(
1− 190 K
220 K
)
,
�rG
−�−
190 = 85.6 kJ mol−1
For the monohydrate
�rG
−�−
190 = (93.2 kJ mol−1)×
(
190 K
220 K
)
+ (237 kJ mol−1)×
(
1− 190 K
220 K
)
,
�rG
−�−
190 = 112.8 kJ mol−1
P5.32 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
f = −
(
∂A
∂l
)
T
= −
(
∂U
∂l
)
T
+ T
(
∂S
∂l
)
T
assuming that (∂U/∂l)T = 0 (valid for rubbers)
f = T
(
∂S
∂l
)
T
= T
(
∂
∂l
)
T
{
−3kBl
2
2Na2
+ C
}
= T
{
−3kBl
Na2
}
= −
(
3kBT
Na2
)
l
This tensile force has the Hooke’s law form f = −kHl with kH = 3kBT/Na2.