# Thermodynamic Potentials

1. Apr 8, 2015

### Persefonh

1. The problem statement, all variables and given/known data
My question is how can one derive U(S(V,T),V) from the relation U(S,V,N).This is the beginning of a given solution which is not explained in more detail.

2. Relevant equations

3. The attempt at a solution
I can understand U(S,V,N) as dU=TdS-pdV+μdN but not the transition: U(S,V,N)-->U(S(V,T),V).
A full solution of this exercise is given at http://www.thphys.uni-heidelberg.de/~amendola/statphys/problems-students.pdf [Broken] on page 3.

Last edited by a moderator: May 7, 2017
2. Apr 8, 2015

### jitu16

Are you referring to eqn 13?? If that's the case here's explanation:

$\frac{\partial U(S,V)}{\partial V}= \frac{\partial U}{\partial S} \frac{\partial S}{\partial V}+\frac{\partial U}{\partial V} \frac{\partial V}{\partial V} =\frac{\partial U}{\partial S} \frac{\partial S}{\partial V}+\frac{\partial U}{\partial V}$

This is a simple chain rule. If you don't know chain rule, review your calculus course, I don't think you'll go anywhere without that.

3. Apr 8, 2015

### Persefonh

Thank you for your reply.I am not referring to equation 13, but before that.
The exercise starts with U(S(V,T),V). My question is how one can derive that from U(S,V,N)?

4. Apr 8, 2015

### jitu16

Entropy is a function of T,V or T,P.
If you still can't recall,
$S=-\frac{\partial G(T,V)}{\partial T}$ and,
$S=-\frac{\partial A(T,P)}{\partial T}$

Last edited: Apr 8, 2015
5. Apr 8, 2015

### BvU

Hello Persephone, welcome to PF !

Could it be as simple as N is constant, hence $U(S, V, N) = U(S, V)$ ?

6. Apr 8, 2015

Thanks BvU!