1000 Solved Problems in Modern Physics

(Romina) #1

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
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