4.3 Solutions
275
where H is the enthalpy
∴ T ΔS + V ΔP = 0
But by Problem 4.31
T ΔS = C P ΔT − T
∂ V
∂ T
P
ΔP
∴ C P ΔT +
V − T
∂ V
∂ T
P
ΔP = 0
or ΔT =
T
∂ V
∂ T
P
− V
ΔP
C P
4.34 (a) For perfect gas
PV = RT
P
∂ V
∂ T
P
= R
T
∂ V
∂ T
P
=
T R
P
= V
or T
∂ V
∂ T
P
− V = 0
∴ ΔT = 0 by Problem 4.31
(b) For imperfect gas
P +
a
V 2
(V − b) = RT
or PV = RT −
a
V
+ bP +
ab
V 2
P
∂ V
∂ T
P
= R +
a
V 2
∂ V
∂ T
P
−
2ab
V 3
∂ V
∂ T
P
Re-arranging
∂ V
∂ T
P
=
R
P −
a
V 2 +
2ab
V 3
=
R
RT
V −b
−
2a
V 2
1 −
b
V
Multiplying both numerator and denominator of RHS by (V − b)/R
T
∂ V
∂ T
P
= (V − b)
1 −
2a
RT V 3 (V − b)
2
−1
= (V − b)
1 +
2a
RT V 3 (V − b)
2