# 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.