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I have seen many sources (for example this libretext) cite that the total differential of chemical potential is $$d\mu = V_m dP - S_m dT$$ where $V_m$ is the molar volume and $S_m$ is the molar entropy. The usual justification for this formula is that chemical potential is the same as partial molar Gibbs free energy, but I was hoping that someone could make this connection more clear. My understanding is that partial molar quantities are quite literally the rate of change of a certain quantity with respect to the number of moles of a substance (please correct me if this understanding is incorrect), so saying that molar Gibbs free energy and chemical potential are the same seems purely definitional and I don't see what conclusions can be drawn from that claim or how that can be used to derive this relationship.

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    $\begingroup$ It's Gibbs' not Gibb's. The chemical potential of a species tells you by how much the free energy changes if you alter the amount of that species while holding everything else constant. This is useful for predicting in which direction a reaction or process that alters the amount of the species will proceed. $\endgroup$
    – Buck Thorn
    Oct 10, 2023 at 6:11
  • $\begingroup$ Take a look at chem.libretexts.org/Courses/Millersville_University/… $\endgroup$
    – Ian Bush
    Oct 10, 2023 at 7:52
  • $\begingroup$ @IanBush It is the identical link as the OP refers to. :-) $\endgroup$
    – Poutnik
    Oct 10, 2023 at 7:57
  • $\begingroup$ "Purely definitional" is the best we could hope for generally, because it is not possible to derive nor proof definitions. $\endgroup$
    – Poutnik
    Oct 10, 2023 at 9:10
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    $\begingroup$ @IanBush You have omitted "molar", referring to the article. $\endgroup$
    – Poutnik
    Oct 10, 2023 at 10:47

1 Answer 1

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$$dG=VdP-SdT-\sum{\mu_in_i}$$So, $$\frac{\partial G}{\partial P}=V$$and $$\frac{\partial G}{\partial n_i}=\mu_i$$So, $$\frac{\partial^2 G}{\partial n_i \partial P}=\frac{\partial^2 G}{\partial P\partial n_i}=\frac{\partial V}{\partial n_i}=V_{m,i}=\frac{\partial \mu_i}{\partial P}$$

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  • $\begingroup$ And the same can be done for T and -S. $\endgroup$
    – Poutnik
    Oct 10, 2023 at 12:07
  • $\begingroup$ @Poutnik Yes. Thanks. I was hoping that the OP would realize that. $\endgroup$ Oct 10, 2023 at 12:51
  • $\begingroup$ Thank you, that makes a lot of sense about how you found the coefficients on $dP$ and $dT$. However, why don't we include the terms $dn_1,dn_2,\dots,dn_k$ (excluding $dn_i$) in the total differential of $\mu_i$? $\endgroup$ Oct 10, 2023 at 15:48
  • $\begingroup$ The equation assumes constant mixture composition. $\endgroup$ Oct 10, 2023 at 16:13
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    $\begingroup$ Did you mean $\sum_i\mu_idn_i$ in your first equation ? $\endgroup$
    – porphyrin
    Oct 10, 2023 at 17:57

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