1000 Solved Problems in Modern Physics

276 4 Thermodynamics and Statistical Physics

=(V−b)+

2 a

RT V^3

(V−b)^3

T

(

∂V

∂T

)

P

−V=

2 a

RT

−b (∴bV)

Using this in the expression for Joule–Thompson effect (Problem 4.31),

ΔT=

1

Cp

(

2 a

RT

−b

)

ΔP

4.35 The equation of state for an imperfect gas is
(
p+

a

V^2

)

(V−b)=RT

It can be shown that

ΔT=

1

Cp

(

2 a

RT

−b

)

Δp

IfT< 2 a/bR,ΔT/Δpis positive and there will be cooling.

IfT> 2 a/bR,ΔT/Δpwill be negative and the gas is heated on undergo-

ing Joule–Kelvin expansion.

IfT= 2 a/bR,ΔT/Δp=0, there is neither heating nor cooling.

The temperature given byTi=bR^2 a is called the temperature of inversion

since on passing through this temperature the Joule–Kelvin effect changes its

sign. Figure 4.3 shows the required curve.

Fig. 4.3Joule-Thompson
effect

4.36 By definition

ET=−V

(

∂P

∂V

)

T

;ES=−V

(

∂P

∂V

)

S