Physical Chemistry , 1st ed.

and can now evaluate ( V/ T)p:



V

T



p



R

p



Substituting:

JT

C

1

p

T

R

p

V

C

1

p



R

p

T

V

But RT/pequals V, according to the ideal gas law. Substituting:

JT

C

1

p

(VV) 

C

1

p

(0)  0

which shows once again that the Joule-Thomson coefficient for an ideal gas

is exactly zero.

Example 4.10

Starting with the natural variable equation for dU, derive an expression for

the isothermal volume dependence of the internal energy, ( U/ V)T,in terms

of measurable properties (T,V,or p) and and/or .Hint:you will have to

invoke the cyclic rule of partial derivatives (see Chapter 1).

Solution

The natural variable equation for dUis (from equation 4.14)

dUT dSp dV

In order to get ( U/ V)T, we hold the temperature constant and divide both

sides by dV.We get



U

V



T

T

V

S



T

p

Now we use a Maxwell relationship and substitute for ( S/ V)T, which ac-

cording to Maxwell’s relationships equals ( p/ T)V. Therefore,



U

V



T

T

T

p



V

p

Now we invoke the hint. The definitions for ,, and the partial derivative

( p/ T)Vall use p,T, and V. The cyclic rule for partial derivatives relates the

three possible independent partial derivatives of any three variables A,B,C:



A

B



C



C

B



A



C

A



B

 1

For the variables p,V, and T, this means that



V

T



p



T

p



V



V

p



T

 1

V

V

(^1) 

(^1) 
where we are showing how the coefficients and relate to the derivatives
in this cyclic-rule equation. The middle partial derivative involves pand Tat
104 CHAPTER 4 Free Energy and Chemical Potential
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