Previous, I proved that


I've been asked to derive a similar expression for a gas that obeys the equation of state


I started with the first law and made the assumption that

$$\mathrm dU=\frac{\partial{U}}{\partial{V}} + \frac{\partial{U}}{\partial{T}} + \frac{\partial{U}}{\partial{n}}$$

I know that for an ideal gas, we can say that $\frac{\partial{U}}{\partial{V}}$ goes to zero... but would this also be true for my given equation of state?

Making that assumption, I ended up with something like


Any suggestions would be immensely appreciated.

  • $\begingroup$ First, you need to qualify your derivatives better. if you say $\partial{U}\{V}$, what are you keeping constant? It is important to know because that defines the result of your derivative. Second, du/dv)T goes to zero only for an ideal gas. This term is a measure of the intermolecular forces. For other equations of states the result will be different $\endgroup$ Mar 8 '17 at 23:11
  • $\begingroup$ For an arbitrary gas, $$\left(\frac{\partial U}{\partial V}\right)_T=T\left(\frac{\partial P}{\partial T}\right)_V-P$$ $\endgroup$ Mar 8 '17 at 23:50
  • $\begingroup$ That makes sense. However I seem to run into a bit of an ugly expression.. $\endgroup$ Mar 9 '17 at 1:30
  • $\begingroup$ @PedroO'Verde , that makes sense.. however my expression seems a pit ugly.. so let's say I say that $ \left(\frac{\partial{T}}{\partial\bar{V}}\right) = \pi_T $, I end up having to integrate the expression $$ C_\bar{V}= \frac{-RT}{(\bar{V}-b)} - \pi_T $$ $\endgroup$ Mar 9 '17 at 1:35
  • $\begingroup$ if you calculate the above derivative it should give you 0. Remember that V is constant so the term $\frac{R}{(V-b)}$ is a constant. $\endgroup$ Mar 9 '17 at 1:37

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