3
$\begingroup$

From Levine's Physical Chemistry 6th edition:

  1. In a closed system, $\mathrm dG = -S\,\mathrm dT + V\,\mathrm dP$ not only applies to constant composition but also reversible changes in composition (page 116).

  2. For an irreversible process: $\mathrm dG = -S\,\mathrm dT + V\,\mathrm dP + \mu\,\mathrm dN \lt -S\,\mathrm dT + V\,\mathrm dP$, thus $\mu\,\mathrm dN \lt 0$ (page 130).

For 1), why do we even bother with the chemical potential for closed systems if we can just find a reversible path in which 1) applies?

For 2), I thought that although $\mathrm dG = -S\,\mathrm dT + V\,\mathrm dP + \mu\,\mathrm dN$ can apply to irreversible processes, you had to calculate a reversible path. If so, that would mean that the $-S\,\mathrm dT$ on the left was calculated for a reversible path and the $-S\,\mathrm dT$ on the right was calculated for an irreversible path and they wouldn't cancel.

$\endgroup$
3
$\begingroup$

Q1. You won't need the chemical potential terms $\sum_{i,\alpha}\mu_i^\alpha\text{d}n_i^\alpha$ (because it equals zero) in using the differential relation $\text{d}G = -S\,\text{d}T+V\,\text{d}P$ in this case. Nevertheless, the chemical potentials are still relevant in establishing criteria for phase equilibrium: $\mu_i^\alpha = \mu_i^\beta$, etc.

Why should $\sum_{i,\alpha}\mu_i^\alpha\text{d}n_i^\alpha = 0$?

Reversible changes in composition imply that the process is in equilibrium, so that $\text{d}G = 0$. Assuming thermal and mechanical equilibrium and constant $T$ and $P$ we have $\text{d}T = \text{d}P = 0$, and the result follows by considering the (full) differential relation satisfied by $\text{d}G$.

Q2. Yes, you do need a reversible path. The expression on the right, if you follow its derivation carefully, does represent a reversible path. The inequality occurs because the chemical potential terms have not been accounted for. So the cancellation works out.

It seems, however, that we can derive the same result much more simply. Consider a system at thermal and mechanical equilibrium with constant $T$ and $P$, but not chemical equilibrium. Then $\text{d}G < 0$, since the Gibbs free energy is minimized at equilibrium. If $T$ and $P$ are constant $\text{d}T = \text{d}P = 0$, so all that's left in $\text{d}G$ are the chemical potential terms, which must thus be negative.

$\endgroup$
  • $\begingroup$ I thought that dG = 0 only for reversible processes involving constant T and P. $\endgroup$ – Yuki Aug 1 '18 at 1:40
  • $\begingroup$ @Yuki, can you quote the part of my response that you're concerned about? There are lots of dG's in my answer and I'm not sure what you're referring to. $\endgroup$ – a-cyclohexane-molecule Aug 1 '18 at 1:42
  • $\begingroup$ Thermal and mechanical equilibrium does not imply dT=dP=0 as both T and P can change reversibly. Also for 2), the expression on the right, -SdT+VdP, is definitely not a state function and thus for an irreversible process such that dG < -SdT+VdP it would not make sense to say that it is calculated for a reversible path. $\endgroup$ – Yuki Aug 1 '18 at 5:13
  • $\begingroup$ I think I understand your answer now. Because thermal and mechanical equilibrium is assumed for all equations, -SdT + VdP is always reversible both in the right and left side of dG = -SdT + VdP + udN < -SdT + VdP. In the original derivation of dG = -SdT + VdP + udN, when Levine said that this equation was only valid if you used a reversible path, he meant that specifically -SdT + VdP had to have a reversible path, not udN (which is valid for irreversible paths). $\endgroup$ – Yuki Aug 1 '18 at 5:50
  • $\begingroup$ @Yuki, (> Thermal and mechanical equilibrium does not imply dT=dP=0) sorry, you're right. I have updated my answer accordingly. And you've addressed the second part of your comment already. $\endgroup$ – a-cyclohexane-molecule Aug 1 '18 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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