1000 Solved Problems in Modern Physics

(Grace) #1

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
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