6
$\begingroup$

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:

enter image description here

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)?

$\endgroup$
  • $\begingroup$ How can your substance have free energy if there are no molecules present? (N=0) $\endgroup$ – Chester Miller Dec 11 '18 at 12:58
  • $\begingroup$ @ChesterMiller That's also a fair point. $\endgroup$ – Sha Vuklia Dec 11 '18 at 12:58
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.