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:

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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)?

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

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  • $\begingroup$ Oh, and $V$ is usually (if not always) proportional to $N$, so $\rho$ will always be fixed. $\endgroup$ – 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.

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  • $\begingroup$ 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! $\endgroup$ – Sha Vuklia Dec 11 '18 at 17:56

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