Derivation of Maxwell's relations

  • Thread starter komodekork
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  • #1
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In thermodynamics one of the maxwell relations is:
[itex]
\left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial P}{\partial T} \right)_V
[/itex]

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

Answers and Replies

  • #2
834
2
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
1,082
25
In thermodynamics one of the maxwell relations is:
[itex]
\left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial P}{\partial T} \right)_V
[/itex]

So, what am I doing wrong?

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.
Helmholtz free energy specifically. The form he is trying to derive implies T & V are the parameters of the system.
 
  • #4
1,024
32
 

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