I am trying to find a way to find an expression got the isothermal compressibility ($\kappa_t$) and for the thermal expansion ($\alpha$) of a cristal that has an entropy function S(E,N,V). The entropy function is given in the problem. Using Maxwell relations and cyclical derivatives properties, I found: $$\kappa_T= \frac{1}{V} \left. \frac{\partial V}{\partial T} \right|_{P} \left. \frac{\partial V}{\partial S} \right|_{T}=\alpha \left. \frac{\partial V}{\partial S} \right|_{T}=\frac{\alpha}{\left. \frac{\partial S}{\partial V} \right|_{T}} $$ and $$\alpha= \frac{-1}{V} \left. \frac{\partial V}{\partial P} \right|_{T} \left. \frac{\partial S}{\partial V} \right|_{T}=\frac{1}{\kappa_T V^2} \left. \frac{\partial S}{\partial V} \right|_{T}$$

I can easily compute $\left. \frac{\partial S}{\partial V} \right|_{T}$ because S is given to me. The problem is that my answer seem kind of circular (you need $\alpha$ to express $\kappa_t$ and vice-versa. Is there any other approach that I could use to have a final expression in terms of variable instead of $\kappa_T$ and $\alpha$ ?


1 Answer 1


As far as I know, the quantities you are looking for are actually defined as

$$ \kappa_T=-\frac{1}{V} \left(\frac{\partial V}{\partial P}\right)_T$$

$$ \alpha=\frac{1}{V} \left(\frac{\partial V}{\partial T}\right)_P$$

Whcih make much more sense to me. See Callen, for example.

  • 1
    $\begingroup$ This was the starting point of my problem ( I should have mentioned it in my question). However, the expression that I have for S (E,V,N) doesn't make it possible to isolate V... That's why I believe that I need to express the $\kappa$ and $\alpha$ with different expressions (using maxwell relations and other properties) $\endgroup$
    – user526403
    Sep 6, 2018 at 13:02
  • $\begingroup$ Sure, that's the way, but Maxwell's relations replace derivatives. So, for example, you use $(\partial V/\partial T)_P=-(\partial S / \partial P ) _T$. That's replacing the derivative, yo don't have the product of two of them, or not that way. $\endgroup$
    – FGSUZ
    Sep 6, 2018 at 13:07

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