# proportionality between Gibbs free energy and number of particles

We have the thermodynamic identity for the Gibbs free energy (for a pure system): $$dG=-SdT+VdP+\mu dN.$$ Now if we keep $$T$$ and $$P$$ fixed, we get $$\mu=\left(\frac{\partial G}{\partial N}\right)_{T,P}.$$ Using the argument that $$T,P$$ are intensive quantities (which we keep fixed), while $$G$$ is extensive, if follows that $$\Delta G=\mu\Delta N$$. However, my book actually claims that $$G=\mu N.$$ Now, to me this seems odd, because $$G$$ still depends on $$S,T,V,P$$. I know that we're assuming that $$T$$ and $$P$$ are fixed, but I would think that we have $$G=\mu N+x,$$ where $$x$$ is some function of $$S,T,V,P$$. Why isn't this the case? I would think that we can't just set this $$x$$ to zero for every system, because that way we wouldn't be able to compare two systems where this $$x$$ would technically be different? Or can it be set to zero, and why?

Here is the relevant text from my book: EDIT

Hm, so I just looked at this question:

Prove that $$G=\mu N$$ and independance of $$\mu$$ on $$N$$

It seems then that it's $$\mu$$ which depends on $$T,P$$, and we're using the fact that $$G$$ is an extensive quantity. I guess that makes sense. What I do wonder about it how we know that the density stays fixed? Because that's what we need too. So I'm basically confused about the relation $$G(T,P,\alpha N)=\alpha G(T,P,N),$$ which should be the mathematical phrasing that $$G$$ is extensive. However, I don't understand why we can "forget" about the volume? We'd still need that $$\rho=N/V=\text{constant}. (1)$$ Where can we find the guarantee that this is the case, and why isn't it expressed in (1)?

• How can your substance have free energy if there are no molecules present? (N=0) – Chet Miller Dec 11 '18 at 12:58
• @ChesterMiller That's also a fair point. – Sha Vuklia Dec 11 '18 at 12:58

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.

• Oh, and $V$ is usually (if not always) proportional to $N$, so $\rho$ will always be fixed. – Sha Vuklia Dec 11 '18 at 12:53

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.

• Oh, that's so exactly what I needed. I'll keep the other answer as the "accepted one" (as it answered my edited question) - but many, many thanks for the insights you provided, and +1 of course! – Sha Vuklia Dec 11 '18 at 17:56