14.9: The Dependence of Chemical Potential on Other Variables
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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
\[{\mu }_A={\mu }_A\left(P,T,n_1,n_2,\dots ,n_A,\dots {,n}_{\omega }\right) \nonumber \]
for which the total differential is
\[d{\mu }_A={\left(\frac{\partial {\mu }_A}{\partial T}\right)}_PdT+{\left(\frac{\partial {\mu }_A}{\partial P}\right)}_TdP+\sum^{\omega }_{j=1}{{\left(\frac{\partial {\mu }_A}{\partial n_j}\right)}_{PT}dn_j} \nonumber \]
Recalling the definition of the chemical potential and the fact that the mixed second-partial derivatives of a state function are equal, we have
\[{\left(\frac{\partial {\mu }_A}{\partial T}\right)}_P={\left(\frac{\partial }{\partial T}\right)}_P{\left(\frac{\partial G}{\partial n_A}\right)}_{TP}={\left(\frac{\partial }{\partial n_A}\right)}_{TP}{\left(\frac{\partial G}{\partial T}\right)}_P=-{\left(\frac{\partial S}{\partial n_A}\right)}_{TP}=-{\overline{S}}_A \nonumber \] Similarly,
\[{\left(\frac{\partial {\mu }_A}{\partial P}\right)}_T={\left(\frac{\partial }{\partial P}\right)}_T{\left(\frac{\partial G}{\partial n_A}\right)}_{TP}={\left(\frac{\partial }{\partial n_A}\right)}_{TP}{\left(\frac{\partial G}{\partial P}\right)}_T={\left(\frac{\partial V}{\partial n_A}\right)}_{TP}={\overline{V}}_A \nonumber \] Thus, the total differential of the chemical potential for species \(A\) can be written as
\[d{\mu }_A=-{\overline{S}}_AdT+{\overline{V}}_AdP+\sum^{\omega }_{j=1}{{\left(\frac{\partial {\mu }_A}{\partial n_j}\right)}_{PT}dn_j} \nonumber \]
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
\[d{\mu }_{\alpha }=-{\overline{S}}_{\alpha }dT+{\overline{V}}_{\alpha }dP+{\left(\frac{\partial {\mu }_{\alpha }}{\partial n_{\alpha }}\right)}_{PT}dn_{\alpha }+{\left(\frac{\partial {\mu }_{\alpha }}{\partial n_{\beta }}\right)}_{PT}dn_{\beta } \nonumber \] and \[d{\mu }_{\beta }=-{\overline{S}}_{\beta }dT+{\overline{V}}_{\beta }dP+{\left(\frac{\partial {\mu }_{\beta }}{\partial n_{\alpha }}\right)}_{PT}dn_{\alpha }+{\left(\frac{\partial {\mu }_{\beta }}{\partial n_{\beta }}\right)}_{PT}dn_{\beta } \nonumber \]
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
\[{\left(\frac{\partial {\mu }_{\alpha }}{\partial n_{\alpha }}\right)}_{PT}={\left(\frac{\partial {\mu }_{\beta }}{\partial n_{\alpha }}\right)}_{PT}={\left(\frac{\partial {\mu }_{\alpha }}{\partial n_{\beta }}\right)}_{PT}={\left(\frac{\partial {\mu }_{\beta }}{\partial n_{\beta }}\right)}_{PT}=0 \nonumber \]
Hence,
\[-{\overline{S}}_{\alpha }dT+{\overline{V}}_{\alpha }dP=-{\overline{S}}_{\beta }dT+{\overline{V}}_{\beta }dP \nonumber \]
and the rest of the derivation follows as before.