Analyzing the Gibbs Energy of a Mixture to Determine Whether It Is an Ideal Mixture
Experimentally (as will be described in Sec. 10.2) it has been found that the Gibbs energy for a certain binary mixture has the form
\underline{G}_{\operatorname{mix}}(T, P, \underline{x})=\sum_{ i =1}^{2} x_{ i } \underline{G}_{ i }(T, P)+R T \sum_{ i =1}^{2} x_{ i } \ln x_{ i }+a x_{1} x_{2} (**)
where a is a constant. Determine whether this mixture is an ideal mixture.
For an ideal mixture Eqs. 9.3-1a and b must be satisfied at all conditions. To see if this is so, we start with the relation between the Gibbs energy and volume and use the equation above to obtain
\bar{H}_{ i }^{ IM }(T, P, \underline{x})=\underline{H}_{ i }(T, P) (9.3-1a)
\underline{V}_{\operatorname{mix}}=\left(\frac{\partial \underline{G}_{ mix }}{\partial P}\right)_{T, N}=\sum_{ i =1}^{2} x_{ i }\left(\frac{\partial \underline{G}_{ i }}{\partial P}\right)_{T, N}=\sum x_{ i } \underline{V}_{ i }(T, P)
Therefore,
\bar{V}_{1}(T, P)=\left.\frac{\partial\left(N \underline{V}_{\text {mix }}\right)}{\partial N_{1}}\right|_{T, N_{2}}=\frac{\partial}{\partial N_{1}}\left(N_{1} \underline{V}_{1}(T, P)+N_{2} \underline{V}_{2}(T, P)\right)_{T, N_{2}}=\underline{V}_{1}(T, P)
Similarly, \bar{V}_{2}(T, P)=\underline{V}_{2}(T, P), so Eq. 9.3-1b is satisfied. We now move on to Eq. 9.3-1a by starting with
\bar{V}_{ i }^{ IM }(T, P, \underline{x})=\underline{V}_{ i }(T, P) (9.3-1b)
\left.\frac{\partial}{\partial T}\right|_{P}\left(\frac{\underline{G}}{T}\right)=\frac{1}{T}\left(\frac{\partial \underline{G}}{\partial T}\right)_{P}-\frac{\underline{G}}{T^{2}}=\frac{1}{T}(-S)-\frac{(\underline{H}-T \underline{S})}{T^{2}}=-\frac{\underline{H}}{T^{2}}
or
\underline{H}=-\left.T^{2} \frac{\partial}{\partial T}\right|_{P}\left(\frac{\underline{G}}{T}\right)
Therefore,
\begin{aligned}\underline{H}_{\text {mix }} &=-\left.T^{2} \frac{\partial}{\partial T}\right|_{P}\left[\sum x_{ i } \frac{G_{ i }}{T}+R \sum x_{ i } \ln x_{ i }+\frac{a}{T} x_{1} x_{2}\right] \\&=-T^{2}\left[\sum x_{ i }\left(-\frac{\underline{H}_{ i }}{T^{2}}\right)-\frac{a}{T^{2}} x_{1} x_{2}\right]=\sum x_{ i } \underline{H}_{ i }+a x_{1} x_{2}\end{aligned}
Then
\begin{aligned}\bar{H}_{1}(T, P, \underline{x}) &=\left.\frac{\partial}{\partial N_{1}}\right|_{T, P, N_{2}} N \underline{H}_{ mix }=\left.\frac{\partial}{\partial N_{1}}\right|_{T, P, N_{2}}\left(N_{1} \underline{H}_{1}+N_{2} \underline{H}_{2}+\frac{a N_{1} N_{2}}{N_{1}+N_{2}}\right) \\&=\underline{H}_{1}(T, P)+a\left[\frac{N_{2}}{N_{1}+N_{2}}-\frac{N_{1} N_{2}}{\left(N_{1}+N_{2}\right)^{2}}\right] \\&=\underline{H}_{1}(T, P)+a\left(x_{2}-x_{1} x_{2}\right)=\underline{H}_{1}(T, P)+a x_{2}\left(1-x_{1}\right) \\&=\underline{H}_{1}(T, P)+a x_{2}^{2}\end{aligned}
Similarly, \bar{H}_{2}(T, P, \underline{x})=\underline{H}_{2}(T, P)+a x_{1}^{2}.
Therefore, Eq. 9.3-1a is not satisfied, and the mixture described by Eq. ** above is not an ideal mixture.
Comment
Clearly, for this mixture \underline{G}^{ ex }=a x_{1} x_{2} \text { and } \underline{A}^{ ex }=a x_{1} x_{2}. Since the excess Gibbs energies of mixing are not zero, this also shows that the mixture cannot be ideal.