Gibbs free-energy change for a pressure-linear gas correction ($\Delta G=N{k}_{B}T\ln (p_1/p_0)-NA(T)(p_1-p_0)$)

For the equation of state $V=N{k}_{B}T/p-NA(T)$, 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 $\Delta S=-N{k}_B\ln (p_1/p_0)+N{A}^{\prime }(T)(p_1-p_0)$. The physically admissible branch must retain positive volume.