How do you derive the relation

$$C_p-C_v= R\left(1 + \frac{2a}{RTV}\right)$$

for a gas obeying van der Waals equation of state? Any leads?

  • $\begingroup$ How do you know you can obtain an equation of this form by making certain assumptions if you don't know what those assumptions are? Wherever you found this formula should say what assumption they had to make for it to be true. $\endgroup$ – Tyberius Aug 9 '17 at 15:11
  • $\begingroup$ It is a question in a competitive exam, hence the doubt. $\endgroup$ – Arjun Gopal Aug 9 '17 at 16:41
  • 1
    $\begingroup$ Does the test question include any extra context? If so, you should include the text of the actual question. $\endgroup$ – Tyberius Aug 9 '17 at 17:52
  • $\begingroup$ No sir this is the only text. $\endgroup$ – Arjun Gopal Aug 10 '17 at 2:52
  • $\begingroup$ Problem 2 here seems to answer your question after a little algebra and I believe with the assumption that V>>b. $\endgroup$ – Tyberius Aug 10 '17 at 3:19

You could start with $C_V=(\partial U/\partial T)_V=T(\partial S/\partial T)_V$ and H and p instead of U and V for $C_p$ as appropriate. Then generate an expansion for S as $dS=(\partial S/\partial V)_T dV+(\partial S/\partial T)_VdT$ and differentiate wrt T. You should then get an expression in $C_V$ and $C_p$ plus other terms that you can find using the vdw equation.

  • $\begingroup$ Yes but since the vdw equation is not explicit in V, it does not boil down to the above expression. The above expression can be obtained using a specific assumption which is what I am after. $\endgroup$ – Arjun Gopal Aug 9 '17 at 13:12
  • $\begingroup$ You should find that the difference in heat capacity is $T(\partial P/\partial T)_V(\partial V/\partial T)_p$ so calculate the derivatives and keep going. It is messy. However, since a is in the answer given in the question a likely solution is that $b=0$. Making this assumption does simplify the answer a bit although its best usually to apply specific limits at the end of a calculation if possible. $\endgroup$ – porphyrin Aug 10 '17 at 6:49

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