What are analytical expressions for the isothermal change of Gibbs energy? The change of Gibbs energy at constant temperature and species numbers, $\Delta G$, is given by an integral $\int_{p_1}^{p_2}V\,{\mathrm d}p$. For the ideal gas law, $p\,V=n\,RT$, this comes down to
$$\int_{p_1}^{p_2}\frac{1}{p}\,{\mathrm d}p=\ln\frac{p_2}{p_1}.$$
We find this logarithm in many formulas in chemistry. 

I tried and failed to find a general formula for $\Delta G$ for the Van der Waals gas (see this question on the physics board). 

It makes me wonder:
Is there even another case of a chemical situation, other than the ideal gas, where $\int_{p_1}^{p_2}V(p)\,{\mathrm d}p$ or $\int_{V_1}^{V_2}p(V)\,{\mathrm d}V$ is known analytically?



My questions are mostly motivated by the desire to understand the functional dependencies of the chemical potential $\mu(T)$, that is essentially given by the Gibbs energy. For the ideal gas and any constant $c$, we see that a state change from e.g. the pressure $c\,p_1$ to another pressure $c\,p_2$ doesn't actually affect the Gibbs energy. The constant factors out in $\frac{1}{p}\,{\mathrm d}p$, resp. $\ln\frac{p_2}{p_1}$. However, this is a mere feature of the gas law with $V\propto \frac{1}{p}$, i.e. it likely comes from the ideal gas law being a model of particles without interaction with each other.
With regard to the Van der Waals equation, have you considered integrating by parts:  $V\mathrm dp=\mathrm d(pV)-p\mathrm dV$.  That can certainly be integrated analytically.  

The answer will be in terms of $V$, but that’s not a big limitation.