# Deriving heat capacity using thermodynamics identity

1. Dec 14, 2006

### ultimateguy

1. The problem statement, all variables and given/known data
Use the thermodynamic identity to derive the heat capacity formula $$C_V=T\frac{\partial{S}}{\partial{T}}_V$$

2. Relevant equations
$$C_V=\frac{\partial{U}}{\partial{T}}$$
$$T=\frac{\partial{U}}{\partial{S}}$$
dU=TdS-PdV+$$\mu$$dN

3. The attempt at a solution

I used $$C_V=\frac{\partial{U}}{\partial{S}}\frac{\partial{S}}{\partial{T}}=T\frac{\partial{S}}{\partial{T}}_V=\frac{\partial{U}}{\partial{T}}=C_V$$.
This is the only solution I can think of, but I don't think I used the thermodynamics identity, did I?

Last edited: Dec 14, 2006
2. Dec 14, 2006

### Kurdt

Staff Emeritus
There are a distinct lack of thermal physicists on this website and with good reason as it is one of the most boring subjects ever. I think if you know that:

$$dS=(\frac{\partial{S}}{\partial{U}})_{PV}dU$$

Then you can substitute the following into the equation for dU,

$$dU=(\frac{\partial{U}}{\partial{T}})_VdT$$

and have a fiddle about, that should set you off on the right lines.

Last edited: Dec 15, 2006
3. Dec 14, 2006

### ultimateguy

This may be a dumb question, but what's the difference between E and U?

4. Dec 15, 2006

### Kurdt

Staff Emeritus
There is none. I apologise. I normally write E and I was trying to conform to your notation but obviously forgot half way through I will change it.

5. Dec 15, 2006

### dextercioby

The solution provided by the OP is correct.

Daniel.

6. Dec 15, 2006

### Kurdt

Staff Emeritus
I'd have thought if they wanted a derivation from the thermodynamic identity then they'd want you to start from that. However the OP proof might be accepted.