What is the value of $$\left(\frac{\partial U}{\partial T}\right)_{\!P} - \left(\frac{\partial U}{\partial T}\right)_{\!V}$$ for an ideal gas and for a van der Waals gas?

($U$ refers to the molar internal energy.)


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Here I am going to use $U$ to refer to the total internal energy. If you want the molar internal energy ($U_\mathrm m \equiv U/n$) just divide your final result by $n$.

In another of my answers I have proven (equation $(14)$) that

$$\left(\frac{\partial U}{\partial T}\right)_{\!p} = \left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial T}\right)_{\!p} + \left(\frac{\partial U}{\partial T}\right)_{\!V}$$

The proof up until there is completely general, so this applies for both an ideal and a van der Waals gas.

Furthermore here I have shown that (combining equations $(6)$ and $(8)$)

$$\left(\frac{\partial U}{\partial V}\right)_{\!T} = T\left(\frac{\partial p}{\partial T}\right)_{\!V} - p$$

Putting these together gives you

$$\begin{align} \left(\frac{\partial U}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} &= \left(\frac{\partial V}{\partial T}\right)_{\!p}\left[ T\left(\frac{\partial p}{\partial T}\right)_{\!V} - p\right] \end{align}$$

All that's left is to evaluate the partial derivatives. The ideal gas case is very easy; since $p = nRT/V$, $(\partial p/\partial T)_V = nR/V$ and the term in square brackets goes to zero.

For the van der Waals gas you need to work at those partial derivatives which are a bit tedious, but should be doable, especially if you employ implicit differentiation. For example, if you take the van der Waals equation of state

$$\left(p + \frac{an^2}{V^2}\right)(V - nb) = nRT$$

and differentiate with respect to $T$ throughout, keeping $V$ constant, one gets

$$\begin{align} \left[\left(\frac{\partial p}{\partial T}\right)_{\!V} + 0\right](V - nb) + \left(p + \frac{an^2}{V^2}\right)(0) &= nR \\ \left(\frac{\partial p}{\partial T}\right)_{\!V} &= \frac{nR}{V - nb} \end{align}$$

and likewise, barring any careless mistakes, I obtained

$$\left(\frac{\partial V}{\partial T}\right)_{\!p} = nR\left(p - \frac{an^2}{V^2} + \frac{2abn^3}{V^3}\right)^{-1}$$

Maybe somebody has a less tedious way (and I hope so), but this is all I can offer you myself.


For the ideal gas, you can get to the result a little quicker if you know the equipartition theorem. Using the equipartition theorem, one can show that $$U=\frac{\mathrm{d.o.f}}{2}nRT=U(T)$$ where d.o.f is the degrees of freedom of the molecule. This says that for an ideal gas, U is just a function of temperature, so $$\left(\frac{\partial U}{\partial T}\right)_P=\left(\frac{\partial U}{\partial T}\right)_V \to \left(\frac{\partial U}{\partial T}\right)_P-\left(\frac{\partial U}{\partial T}\right)_V=0$$

From statistical mechanics, we can use the partition function of the van der waal gas to show that $$U=\frac{\mathrm{d.o.f}}{2}nRT-\frac{an^2}{V}$$

Now with a direct expression for the energy, we can solve using the relation: $$\left(\frac{\partial U}{\partial T}\right)_P-\left(\frac{\partial U}{\partial T}\right)_V=\left(\frac{\partial U}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P} + \left(\frac{\partial U}{\partial T}\right)_{V}-\left(\frac{\partial U}{\partial T}\right)_V=\left(\frac{\partial U}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}$$ with $$\left(\frac{\partial U}{\partial V}\right)_{T}=\frac{an^2}{V^2}$$ $$\left(\frac{\partial V}{\partial T}\right)_{\!p} = nR\left(p - \frac{an^2}{V^2} + \frac{2abn^3}{V^3}\right)^{-1}$$

If the energy of the van der waal gas is taken as a known result (or derived as shown by juanrga in this answer on Physics SE), you only have to evaluate two derivatives, though similar to orthocresol's answer, it is a bit tedious to compute $\left(\frac{\partial V}{\partial T}\right)_{\!p}$

  • $\begingroup$ What does "d. o. f" mean? $\endgroup$ – aventurin Sep 12 '17 at 16:37
  • $\begingroup$ @aventurin Degrees of Freedom of the molecule $\endgroup$ – Tyberius Sep 12 '17 at 16:38

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