Application of Gibbs-Duhem: rationalization of ϕ∕x-curves

5.7.1 Application of Gibbs-Duhem: rationalization of \(\phi/x\)-curves

We now use the Gibbs-Duhem equation to calculate relations between \(\phi_1\) and \(\phi_2\) of a binary mixture. For the total derivatives we find

\begin{equation*} d\varphi_1 = \left(\frac{\partial \varphi_1}{\partial n_1} \right) dn_1 + \left(\frac{\partial \varphi_1}{\partial n_2} \right) dn_2 \quad \mbox{and} \quad d\varphi_2 = \left(\frac{\partial \varphi_2}{\partial n_1} \right) dn_1 + \left(\frac{\partial \varphi_2}{\partial n_2} \right) dn_2 \label{eq:dphi1_dphi2} \end{equation*}

(5.22)

Using \(\sum_i n_i d\varphi_i = 0\) we get

\begin{equation*} \begin{split} &\;\;\left[ \left(\frac{\partial \varphi_1}{\partial n_1} \right) dn_1 + \left(\frac{\partial \varphi_1}{\partial n_2} \right) dn_2 \right] n_1 + \left[\left(\frac{\partial \varphi_2}{\partial n_1} \right) dn_1 + \left(\frac{\partial \varphi_2}{\partial n_2} \right) dn_2 \right] n_2 =0\\ \Rightarrow &\;\; \left[ n_1 \left(\frac{\partial \varphi_1}{\partial n_1} \right) + n_2 \left(\frac{\partial \varphi_2}{\partial n_1} \right)\right] dn_1 + \left[ n_1 \left(\frac{\partial \varphi_1}{\partial n_2} \right)+ n_2 \left(\frac{\partial \varphi_2}{\partial n_2} \right)\right] dn_2 =0\\ \end{split} \label{eq:dphi1_plus_dphi2} \end{equation*}

(5.23)

which must be true for all \(dn_i\), so both terms of the sum must be zero, i.e.

\begin{equation*} \begin{split} \Rightarrow &\;\; n_1 \left(\frac{\partial \varphi_1}{\partial n_1} \right) + n_2 \left(\frac{\partial \varphi_2}{\partial n_1} \right) = 0 \quad \mbox{and} \quad n_1 \left(\frac{\partial \varphi_1}{\partial n_2} \right)+ n_2 \left(\frac{\partial \varphi_2}{\partial n_2} \right) =0\\ \Rightarrow &\;\; x_1 \left(\frac{\partial \varphi_1}{\partial n_1} \right) = - (1-x_1) \left(\frac{\partial \varphi_2}{\partial n_1} \right) \quad \mbox{and} \quad x_1 \left(\frac{\partial \varphi_1}{\partial n_2} \right) = - (1-x_1) \left(\frac{\partial \varphi_2}{\partial n_2} \right)\\ \end{split} \label{eq:dphi1_dphi2_2} \end{equation*}

(5.24)

This implies that the slopes of \(\varphi_1\) and \(\varphi_2\) have opposite signs. Choosing \(x_1 = 0.5\) we see that the absolute values of the slopes are equal. This is shown in Fig. 5.3 for 50 at% of the partial molar volume for water and ethanol.

With frame

Gibbs-Duhem equation