4.3 Solutions 269
(b) Let the temperature and pressure be independent variables. Putx=Tand
y=Pin (20).
(
∂T
∂x)
y=
(
∂P
∂y)
x=1;
(
∂T
∂y)
x=
(
∂P
∂x)
y= 0
(
∂S
∂P
)
T=−
(
∂V
∂T
)
P(22)
4.22 In Problem 4.21 let the entropy and volume be independent variables. Put
x=sandy=Vin Eq. (19)
(
∂S
∂x
)
y=
(
∂V
∂y)
x=1;
(
∂S
∂y)
x=
(
∂V
∂x)
y= 0
(
∂T
∂V
)
T=−
(
∂P
∂S
)
V
(23)4.23 Let the entropy and pressure be independent variables. Putx=sandy=p
in Eq. (19) of Problem 4.21.
(
∂S
∂x
)
y=
(
∂P
∂y)
x= 1
Therefore,
(
∂T
∂p)
s=
(
∂V
∂S
)
p(24)
4.24 Consider Maxwell’s relation (21) of Problem 4.21
(
∂S
∂V
)
T=
(
∂P
∂T
)
V(1)
Multiply both sides by T,T
(
∂S
∂V
)
T=T
(
∂P
∂T
)
V(2)
or(
∂Q
∂V
)
T=T
(
∂P
∂T
)
V(3)
which means that the latent heat of isothermal expansion is equal to the prod-
uct of the absolute temperature and the rate of increase of pressure with tem-
perature at constant volume. Apply (3) to the phase transition of a substance.
Consider a vessel containing a liquid in equilibrium with its vapor. The pres-
sure is due to the saturated vapor pressure which is a function of temperature
only and is independent of the volume of liquid and vapor present. If the vessel
is allowed to expand at constant temperature the vapor pressure would remain
constant. However, some liquid of massδmwould evaporate to fill the extra
space with vapor. IfLis the latent heat absorbed per unit mass,