270 4 Thermodynamics and Statistical Physics
δQ=Ldm (4)
Ifν 1 andν 2 are the specific volumes (volumes per unit mass) of the liquid and
vapor respectively
δν=(ν 2 −ν 1 )dm (5)
Using (4) and (5) in (3)
L
ν 2 −ν 1
=T
(
∂P
∂T
)
V
(6)
Here, various thermodynamic quantities refer to a mixture of the liquid and
vapor in equilibrium. In this case
(
∂P
∂T
)
V
=
(
∂V
∂T
)
sat
since the pressure is due to the saturated vapor and is therefore independent of
V, being only a function ofT. Thus (6) can be written as
(
∂P
∂T
)
sat
=
L
T(ν 2 −ν 1 )
(Clapeyron’s equation) (7)
4.25 L=T(ν 2 −ν 1 )
dP
dT
= 373 .2(1, 674 −1)×
(
2. 71
76
)
× 1. 013 × 106
= 2. 255 × 1010 erg g−^1
= 2 .255 J/g
=
2. 255
4. 18
= 539 .5 cal/g
4.26
(
∂S
∂V
)
T
=
(
∂P
∂T
)
V
(1)
Substitute
dS=
dU+PdV
T
(2)
in (1)
(
∂U
∂V
)
T
=T
(
∂P
∂T
)
V
−P (3)
Ifuis the energy density andPthe total pressure,
(∂U
∂V
)
=uand the total
pressureP=u/3, since the radiation is diffuse. Hence (3) reduces to