# 14.9: The Dependence of Chemical Potential on Other Variables


The chemical potential of a substance in a particular system is a function of all of the variables that affect the Gibbs free energy of the system. For component $$A$$, we can express this by writing

${\mu }_A={\mu }_A\left(P,T,n_1,n_2,\dots ,n_A,\dots {,n}_{\omega }\right)$

for which the total differential is

$d{\mu }_A={\left(\frac{\partial {\mu }_A}{\partial T}\right)}_PdT+{\left(\frac{\partial {\mu }_A}{\partial P}\right)}_TdP+\sum^{\omega }_{j=1}{{\left(\frac{\partial {\mu }_A}{\partial n_j}\right)}_{PT}dn_j}$

Recalling the definition of the chemical potential and the fact that the mixed second-partial derivatives of a state function are equal, we have

${\left(\frac{\partial {\mu }_A}{\partial T}\right)}_P={\left(\frac{\partial }{\partial T}\right)}_P{\left(\frac{\partial G}{\partial n_A}\right)}_{TP}={\left(\frac{\partial }{\partial n_A}\right)}_{TP}{\left(\frac{\partial G}{\partial T}\right)}_P=-{\left(\frac{\partial S}{\partial n_A}\right)}_{TP}=-{\overline{S}}_A$ Similarly,

${\left(\frac{\partial {\mu }_A}{\partial P}\right)}_T={\left(\frac{\partial }{\partial P}\right)}_T{\left(\frac{\partial G}{\partial n_A}\right)}_{TP}={\left(\frac{\partial }{\partial n_A}\right)}_{TP}{\left(\frac{\partial G}{\partial P}\right)}_T={\left(\frac{\partial V}{\partial n_A}\right)}_{TP}={\overline{V}}_A$ Thus, the total differential of the chemical potential for species $$A$$ can be written as

$d{\mu }_A=-{\overline{S}}_AdT+{\overline{V}}_AdP+\sum^{\omega }_{j=1}{{\left(\frac{\partial {\mu }_A}{\partial n_j}\right)}_{PT}dn_j}$

To illustrate the utility of this result, we can use it to derive the Clapeyron equation for equilibrium between two phases of a pure substance. In Chapter 12, we derived the Clayeyron equation using a thermochemical cycle. We can now use the total differential of the chemical potential to present essentially the same derivation using a simpler argument. Letting the two phases be $$\alpha$$ and $$\beta$$, the total differentials for a system that contains both phases becomes

$d{\mu }_{\alpha }=-{\overline{S}}_{\alpha }dT+{\overline{V}}_{\alpha }dP+{\left(\frac{\partial {\mu }_{\alpha }}{\partial n_{\alpha }}\right)}_{PT}dn_{\alpha }+{\left(\frac{\partial {\mu }_{\alpha }}{\partial n_{\beta }}\right)}_{PT}dn_{\beta }$ and $d{\mu }_{\beta }=-{\overline{S}}_{\beta }dT+{\overline{V}}_{\beta }dP+{\left(\frac{\partial {\mu }_{\beta }}{\partial n_{\alpha }}\right)}_{PT}dn_{\alpha }+{\left(\frac{\partial {\mu }_{\beta }}{\partial n_{\beta }}\right)}_{PT}dn_{\beta }$

Since equilibrium between phases $$\alpha$$ and $$\beta$$ means that $${\mu }_{\alpha }={\mu }_{\beta }$$, we have also that $$d{\mu }_{\alpha }=d{\mu }_{\beta }$$ for any process in which the phase equilibrium is maintained. Moreover, $$\alpha$$ and $$\beta$$ are pure phases, so that $${\mu }_{\alpha }$$ and $${\mu }_{\beta }$$ are independent of $$n_{\alpha }$$ and $$n_{\beta }$$. Then

${\left(\frac{\partial {\mu }_{\alpha }}{\partial n_{\alpha }}\right)}_{PT}={\left(\frac{\partial {\mu }_{\beta }}{\partial n_{\alpha }}\right)}_{PT}={\left(\frac{\partial {\mu }_{\alpha }}{\partial n_{\beta }}\right)}_{PT}={\left(\frac{\partial {\mu }_{\beta }}{\partial n_{\beta }}\right)}_{PT}=0$

Hence,

$-{\overline{S}}_{\alpha }dT+{\overline{V}}_{\alpha }dP=-{\overline{S}}_{\beta }dT+{\overline{V}}_{\beta }dP$

and the rest of the derivation follows as before.

This page titled 14.9: The Dependence of Chemical Potential on Other Variables is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Paul Ellgen via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.