PHYSICAL CHEMISTRY IN BRIEF

CHAP. 3: FUNDAMENTALS OF THERMODYNAMICS [CONTENTS] 87

Proof

For an ideal gas we have

T

(

∂p

∂T

)

V

−p=T

nR

V

−p= 0.

It follows from equation (3.55) that

(

∂U

∂V

)

T

= 0. Hence the internal energy of an ideal gas does

not depend on volume. In order to prove the independence of the internal energy of an ideal

gas on pressure, we first write its total differential as a function ofTandp. By combining the

equation

dV =

(

∂V

∂T

)

p

dT+

(

∂V

∂p

)

T

dp

with equation (3.51) we get

dU=





CV+

[

T

(

∂p

∂T

)

V

−p

](

∂V

∂T

)

p





dT+

[

T

(

∂p

∂T

)

V

−p

](

∂V

∂p

)

T

dV.

From this expression it follows that

(

∂U

∂p

)

T

=

[

T

(

∂p

∂T

)

V

−p

](

∂V

∂p

)

T

As was shown above, the expression in brackets for an ideal gas is zero. Hence the internal energy

of an ideal gas is independent of pressure.

3.4.6 Conditions of thermodynamic equilibrium

The state of thermodynamic equilibrium is defined in1.4.1. From the second law of thermody-
namics [see equations (3.6) and (3.7)] it follows that during irreversible processes the entropy
of an isolated system increases

dS > 0 , [isolated system, irreversible process] (3.58)