Problem 3.1
Ê ∂Cv ˆ
= 0 for an ideal gas, etc.
Show that
Ë ∂V ¯T
Since Cv =
È ∂ Ê ∂U ˆ ˘
Ê ∂Cv ˆ
Ê ∂U ˆ
, then: Ë ∂V ¯ = ÍV Ë ∂T ¯ ˙
Ë ∂T ¯V
T Î
V ˚T
Since U is a state function ( dU is an exact differential), the mixed partial derivatives are
equal, hence:
È ∂ Ê ∂U ˆ ˘
È ∂ Ê ∂U ˆ ˘
=
ÍÎV Ë ∂T ¯ V ˙˚
Í T Ë ∂V ¯T ˙˚
T Î
V
For an ideal gas,
Ê ∂U ˆ
= 0 (U = U(T) only, remember the Joule experiment), so that
Ë ∂V ¯T
Ê ∂Cv ˆ
= 0 for an ideal gas.
Ë ∂V ¯T
For other equations of state, we need a relation for this derivative in terms of state
variables, i.e.,
Ê ∂U ˆ
Ê ∂P ˆ
=T
- P , which will be derived later in Chapter 4. See the
Ë ∂V ¯T
Ë ∂T ¯V
next few pages for parts 3.1(a)-(c) using this relation.