= Entropy excess of a van der Waals gas
{title2=$\Delta S^{\mathrm{ex}}=R\ln(1-b/V)$}
For one mole with temperature-independent parameters $a,b$, the <equation of state> is $P=RT/(V-b)-a/V^2$. A <Maxwell relation> gives $(\partial S/\partial V)_T=R/(V-b)$, so the volume dependence is $R\ln(V-b)$. Matching the <ideal gas> entropy at infinite volume fixes the excess at the same $V,T$ to $R\ln(1-b/V)$. The attraction contributes $-a/V$ to the <Helmholtz free energy> but no excess entropy when $a$ is independent of temperature. A temperature-dependent attraction parameter would change this conclusion.
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