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
$$\int_{p_1}^{p_2}\frac{1}{p}\,{\mathrm d}p=\ln\frac{p_2}{p_1}.$$
That logarithm is at fault for a lot of the formulas in chemistry.
I find I have a surprisingly hard time computing $\Delta G$ for gas governed by the equation of state
$$\left(p + a\frac{n^2}{V^2}\right)\,(V – b\,n) = n\, R T,$$
where $a\ne 0,b$ are small constants.
What is $\Delta G$, at least in low orders in $a,b$?
One might be able to compute ΔG by an integral in which not V is the integrand.
Edit 19.8.15:
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.
Best Answer
I think shifting the variable of integration should work: \begin{align} \int_{p_1}^{p_2}dp~V & =\int_\text{state 1}^\text{state 2} d(pV)-\int_{V_1}^{V_2} dV~p \\ & = p_2V_2-p_1V_1-\int_{V_1}^{V_2} dV~\left[ \frac{nRT}{V-bn}-a\frac{n^2}{V^2} \right] \end{align}