The activity coefficients of nonvolatile solutes cannot be determined by measuring the partial pressure of solute since it is too small.

Therefore, the vapor pressure on the solution (solvent pressure) $P_A$ is measured and with it the activity coefficient $\gamma_A$ is calculated based on the composition of the solution. Using the Gibbs-Duhem equation, the activity coefficient of the solvent is related to that of the solute $\gamma_B$.

We write the Gibbs-Duhem equation

\begin{equation} \sum_{i}n_id\mu_i=0 \end{equation}

We develop the equation for two components A and B.

Dividing by the total moles: $n_A+n_B$

Now we get $d\mu_A$ and $d\mu_B$

\begin{equation} \mu_A=\mu_{A}^{0}(T,P)+RTln\gamma_Ax_A\;\;\;\Rightarrow\;\;\;\; \mu_A=\mu_{A}^{0}(T,P)+RTln\gamma_A+RTlnx_A \end{equation}

Differentiating the chemical potential with respect to composition:

\begin{equation} d\mu_A=RTdln\gamma_A+ RT\frac{dx_A}{x_A} \end{equation}

\begin{equation} d\mu_B=RTdln\gamma_B+RT\frac{dx_B}{x_B} \end{equation}

Substituting both derivatives into the Gibbs- Duhem

\begin{equation} x_A\left(RTdln\gamma_A+RT\frac{dx_A}{x_A}\right)+x_B\left(RTdln\gamma_B+RT\frac{dx_B}{x_B}\right) \end{equation}

Dividing the equation by RT.

Given that $x_A+x_B=1$ differentiating $dx_A+dx_B=0$. This equation allows us to simplify the previous one

Solving

\begin{equation} dln\gamma_B=\frac{-x_A}{x_B}dln\gamma_A \end{equation}

Integrating between states 1 and 2, and using convention II.

\begin{equation} ln\gamma_{II,B,2}-ln\gamma_{II,B,1}=-\int_{1}^{2}\frac{x_A}{1-x_A}dln\gamma_ {II,A} \end{equation}

We set state 1, as pure solvent A, $x_A\rightarrow 1$, $x_B\rightarrow 1$, $\gamma_{II,B,1}\rightarrow 1$ and by therefore $ln\gamma_{II,B,1}=0$.

Under these conditions, the previous integral remains:

\begin{equation} ln\gamma_{II,B}=-\int_{1}^{x_A}\frac{x_A}{1-x_A}dln\gamma_{II,A } \end{equation}