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, 2017 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, 2017 at 23:50
  • $\begingroup$ That makes sense. However I seem to run into a bit of an ugly expression.. $\endgroup$ Mar 9, 2017 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, 2017 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, 2017 at 1:37

1 Answer 1


Firstly, assuming that you want the results for a reversible and adiabatic process where composition is constant throughout the process. (Evident from your expression for the ideal case)

$U$ is a state function depending on $V$ and $T$. Thus, we can write,

$$\mathrm dU=\left(\frac{\partial{U}}{\partial{V}}\right)_T\mathrm dV + \left(\frac{\partial{U}}{\partial{T}}\right)_V\mathrm dT \tag{1}$$

For ideal gases the value of $\left(\frac{\partial{U}}{\partial{V}}\right)_T$ is always zero and thus we can ignore it. But for non-ideal gases, it may or maynot be zero and thus we have to find an expression for it using the equation of state, $P=\frac{RT}{(V-b)}$ in this case.

Now, we know that for a reversible and adiabatic process where composition is constant throughout the process,

$$ \begin{align} \mathrm dU &= T\mathrm dS - P\mathrm dV\\ \left(\frac{\partial{U}}{\partial{V}}\right)_T &= T\left(\frac{\partial{S}}{\partial{V}}\right)_T - P \end{align}$$

Now, by Maxwell's relations we know that $\left(\frac{\partial{S}}{\partial{V}}\right)_T=\left(\frac{\partial{P}}{\partial{T}}\right)_V$. Thus,

$$\left(\frac{\partial{U}}{\partial{V}}\right)_T = T\left(\frac{\partial{P}}{\partial{T}}\right)_V - P \tag{2}$$

Now, it is easy to find $\left(\frac{\partial{P}}{\partial{T}}\right)_V$ with the help of equation of state.

$$\left(\frac{\partial{P}}{\partial{T}}\right)_V = \frac{R}{(V-b)}$$

Substituting this back in $(2)$,

$$ \begin{align} \left(\frac{\partial{U}}{\partial{V}}\right)_T &= \frac{TR}{(V-b)}-\frac{TR}{(V-b)}\\ \left(\frac{\partial{U}}{\partial{V}}\right)_T &=0 \end{align} $$

Surprisingly this results in ideal gas like behaviour. If we think about it in retrospect, it was expected. Because the $\left(\frac{\partial{U}}{\partial{V}}\right)_T$ term is related to the internal interactions of the gas molecules, which we still haven't considered. Notice, how there is no Van Der Walls' constant $a$ in the equation of state.

So, the answer to your question will be same as that for ideal gases, i.e,



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