# proportionality between Gibbs free energy and number of particles

The question contains a very common confusion about which independent variables a thermodynamic potential depends on.
**$G$ is not a function of $S,T,V,P$**. Actually the natural independent variables $G$ depends on can be obtained just looking at the differential: they are $T,P,N$. Each coefficient of the differential form $dG$ should be intended as a function of $T,P,N$, for a fluid one-component system.

Therefore, one would expect to have $\mu=\mu(P,T,N)$. However, for a normal thermodynamic system, thermodynamic potentials are expected to be extensive, i.e., in this context, $G$ is expected to be a homogeneous function of degree one of its extensive argument $N$. Formally, $$ G(T,P,\alpha N) = \alpha G(T,P,N) $$ should hold for all positive values of $\alpha$. Thus, its is enough to take $\alpha=1/N$ (allowed since $N>0$) to get $$ G(T,P,1) = \frac{G(T,P,N)}{N} $$ i.e. $ G(T,P,N) = N G(T,P,1) $, where $ G(T,P,1) $ has no dependence on N. On the other hand, $\mu = \left.\frac{\partial{G}}{\partial{N}}\right|_{T,P}$, and we arrive to the conclusion, since: $\mu=G(T,P,1)$ is clearly independent on $N$.

This derivation makes clear that a key ingredient to get the result is the extensiveness of the Gibbs free energy, which is granted for large (macroscopic) thermodynamic systems, but could fail for finite systems made by a small number of particles.

Any thermodynamic system has an equation of state, which in this case is of the form $f(P,V,T,N)=0$. Fixing $T$ and $P$ means that $V$ is completely determined by $N$.

The only way this reasoning would fail is if there was an equation of state that didn't involve the volume at all. I have never seen a physical system that had such an equation of state, and I strongly suspect such a system would be unphysical.