It's given that $P \equiv P(V,T)$ and $E \equiv E(V,T)$. These are known expressions.

If $S=-\frac{\partial F}{\partial T}$, $\boldsymbol{P}=-\frac{\partial F}{\partial V}$ and $F(V,T)= E(V,T)- TS(V,T)$, what will be an expression for $S$ using chain rules of partial differentiation?


1 Answer 1


$$-S= \frac{\partial (E-TS)}{\partial T} =\frac{\partial E}{\partial T} -\frac{\partial(TS)}{\partial T}=\frac{\partial{E}}{\partial T}-\left[T\frac{\partial S}{\partial T}+S\right]$$ using the usual product rule. Simplifying yields $$\frac{\partial S}{\partial T} = \frac 1T \frac{\partial E}{\partial T} \\ S=\int \frac 1 T \frac{\partial E}{\partial T} \ \partial T $$

  • 1
    $\begingroup$ Thanks! That was very helpful! $\endgroup$ Commented Dec 22, 2020 at 19:33

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