= Solution
First solve the <equation of state> for the <volume (thermodynamics)>:
$$
V(T,p,N)=\frac{Nk_BT}{p}-NA(T).
$$
The <Gibbs free energy> has differential $dG=-S\,dT+V\,dp+\mu\,dN$. At fixed <temperature> and <particle number>, integrate $dG=V\,dp$ between the two pressures:
$$
\Delta G=\int_{p_0}^{p_1}\left[\frac{Nk_BT}{p}-NA(T)\right]dp.
$$
Consequently
$$
\boxed{\Delta G=Nk_BT\ln\frac{p_1}{p_0}-NA(T)(p_1-p_0).}
$$
Use $S=-(\partial G/\partial T)_{p,N}$, keeping both endpoint pressures fixed while differentiating the difference:
$$
\boxed{\Delta S=-Nk_B\ln\frac{p_1}{p_0}+NA'(T)(p_1-p_0).}
$$
Equivalently, the <Maxwell relation> $(\partial S/\partial p)_{T,N}=-(\partial V/\partial T)_{p,N}$ gives $-Nk_B/p+NA'(T)$, whose <pressure> <integral> is the same result. These are the <Gibbs free-energy change for a pressure-linear gas correction> identities. The physical path requires $p_0,p_1>0$ and positive <volume (thermodynamics)> throughout. The correction $A(T)$ has units of <volume (thermodynamics)> per particle, so each term in $\Delta G$ is an <energy> and each term in $\Delta S$ has units of <energy> per kelvin.
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