# Derivation of Maxwell's relations

1. Oct 21, 2012

### komodekork

In thermodynamics one of the maxwell relations is:
$\left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial P}{\partial T} \right)_V$

When I try to derive it from $dU = TdS - PdV$ i get:
$T = \left( \frac{\partial U}{\partial S} \right)_V$
$P = -\left( \frac{\partial U}{\partial V} \right)_S$
$\left( \frac{\partial T}{\partial V} \right)_S = \frac{\partial}{\partial V}\left( \frac{\partial U}{\partial S} \right)_V = \frac{\partial}{\partial S}\left( \frac{\partial U}{\partial V}\right)_S = -\left( \frac{\partial P}{\partial S} \right)_V$
I then multiply with $\frac{\partial S}{\partial T}$,
$\frac{\partial S}{\partial T} \left( \frac{\partial T}{\partial V} \right)_S = \frac{\partial S}{\partial T} \left( -\frac{\partial P}{\partial S} \right)_V$
$\left( \frac{\partial S}{\partial V} \right)_S = -\left( \frac{\partial P}{\partial T} \right)_V$
So, what am I doing wrong?

2. Oct 21, 2012

### Muphrid

I don't think you can derive that Maxwell relation from the internal energy differential. Try enthalpy, Gibbs free energy, or Helmholtz free energy instead.

3. Oct 21, 2012

### Jorriss

Helmholtz free energy specifically. The form he is trying to derive implies T & V are the parameters of the system.

4. Oct 22, 2012

### andrien

multiplying with $$\frac{\partial S}{\partial T}$$
$$\frac{\partial S}{\partial T} \left( \frac{\partial T}{\partial V} \right)_S = \frac{\partial S}{\partial T} \left( -\frac{\partial P}{\partial S} \right)_V$$
this is not correct.
https://docs.google.com/viewer?a=v&q=cache:WMre38mL-A4J:www.eng.cam.ac.uk/~cnm24/files/EngineeringIIAThermodynamicsandPowerGeneration/DataSheet.pdf+perfect+differential&hl=en&gl=in&pid=bl&srcid=ADGEESi3BrYVyWNJBlodQNUPnaT2iMRWkftfNB0cfHBdfkrpC3_IuqNag5ERONO5roYB4xOBo2LzTOpZ7PAwAF8zcVMHsQPPKAnpxfPqxVoW3JLEjJ5V75ZxuYvNKUVJD1YGL8Yi8Dbg&sig=AHIEtbR_6nyPiPXxjMc7VLSZsAjrLICLQQ

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