4.3 Solutions 273
Use (2) and (3) in (4)
CP−CV=−T
(
∂P
∂V
)
T
(
∂V
∂T
) 2
P
(5)
CP−CV=−T
(
∂V
∂P
)
T
(
∂P
∂T
) 2
V
(6)
Equation (5) can be written in terms of the bulk modulusEat constant tem-
perature and the coefficient of volume expansion∝.
E=−
(
∂P
∂V/V
)
; α=
1
V
(
∂V
∂T
)
(7)
Cp−Cν=TEα^2 V (8)
4.30 TakingTandVas independent variables
S=f(T,V)
dS=
(
∂S
∂T
)
V
dT+T
(
∂S
∂V
)
T
dV
Multiplying byT,
TdS=T
(
∂S
∂T
)
V
dT+T
(
∂S
∂V
)
T
dV
=CVdT+T
(
∂S
∂V
)
T
dV
But
(
∂S
∂V
)
T
=
(
∂P
∂T
)
ν
∴TdS=CVdT+T
(
∂P
∂T
)
V
dV
Also,
(
∂P
∂T
)
V
=−
(
∂P
∂V
)(
∂V
∂T
)
P
∴TdS=CVdT−T
(
∂P
∂V
)(
∂V
∂T
)
P
dV
Introducing relationsα=V^1 (∂V/∂T)PandET=−V(∂P/∂V)Tfor volume
coefficient of expansion and isothermal elasticity
TdS=CVdT+TαETdV