The molar excess Gibbs free energy, $${{\text{G}}^{\text{E}}}$$, for a binary liquid mixture at T and P is given by, $$\left( {\frac{{{{\text{G}}^{\text{E}}}}}{{{\text{RT}}}}} \right)$$  = Ax₁x₂, where A is a constant. The corresponding equation for ln y₁, where y₁ is the activity co-efficient of component 1, is

Solution (By Examveda Team)

Given, $$\frac{{{G^E}}}{{RT}} = A{x_1}{x_2} \Rightarrow \frac{{n{G^E}}}{{RT}} = \frac{{A{n_1}{n_2}}}{n}$$
On partially differentiating it $$\left( {\frac{{\partial \frac{{n{G^E}}}{{RT}}}}{{\partial {n_1}}}} \right)T,P,{n_2} = \frac{{A{n_1}{n_2}}}{n} = A{x_2}^2$$
As $$\left( {\frac{{\partial \frac{{n{G^E}}}{{RT}}}}{{\partial {n_1}}}} \right)T,P,{n_2} = \frac{{\overline {n{G^E}_l} }}{{RT}} = ln{\gamma _1}$$
Hence $$ln{\gamma _1} = A{x_2}^2$$