Derivation of Maxwell's relations

  • Context: Graduate 
  • Thread starter Thread starter komodekork
  • Start date Start date
  • Tags Tags
    Derivation Relations
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 9K views
komodekork
Messages
17
Reaction score
0
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?
 
Physics news on Phys.org
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.
 
komodekork said:
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?

Muphrid said:
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.