= Gibbs free-energy change for a pressure-linear gas correction
{title2=$\Delta G=Nk_BT\ln(p_1/p_0)-NA(T)(p_1-p_0)$}
For the <equation of state> $V=Nk_BT/p-NA(T)$, integrate $(\partial G/\partial p)_{T,N}=V$ along an isotherm. This gives the displayed <Gibbs free energy> difference. Its negative <temperature> <derivative> at fixed endpoint pressures gives $\Delta S=-Nk_B\ln(p_1/p_0)+NA'(T)(p_1-p_0)$. The physically admissible branch must retain positive <volume (thermodynamics)>.
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