12.3: Binary Mixture in Equilibrium with a Pure Phase
This section considers a binary liquid mixture of components A and B in equilibrium with either pure solid A or pure gaseous A. The aim is to find general relations among changes of temperature, pressure, and mixture composition in the two-phase equilibrium system that can be applied to specific situations in later sections.
In this section, \(\mu\A\) is the chemical potential of component A in the mixture and \(\mu\A^*\) is for the pure solid or gaseous phase. We begin by writing the total differential of \(\mu\A/T\) with \(T\), \(p\), and \(x\A\) as the independent variables. These quantities refer to the binary liquid mixture, and we have not yet imposed a condition of equilibrium with another phase. The general expression for the total differential is \begin{equation} \dif(\mu\A/T) = \bPd{(\mu\A/T)}{T}{p,x\A}\!\dif T + \bPd{(\mu\A/T)}{p}{T,x\A}\!\difp + \bPd{(\mu\A/T)}{x\A}{T,p}\!\dx\A \tag{12.3.1} \end{equation} With substitutions from Eqs. 9.2.49 and 12.1.3, this becomes \begin{equation} \dif(\mu\A/T) = -\frac{H\A}{T^2}\dif T + \frac{V\A}{T}\difp + \bPd{(\mu\A/T)}{x\A}{T,p}\dx\A \tag{12.3.2} \end{equation}
Next we write the total differential of \(\mu\A^*/T\) for pure solid or gaseous A. The independent variables are \(T\) and \(p\); the expression is like Eq. 12.3.2 with the last term missing: \begin{equation} \dif(\mu\A^*/T) = -\frac{H\A^*}{T^2}\dif T + \frac{V\A^*}{T}\difp \tag{12.3.3} \end{equation}
When the two phases are in transfer equilibrium, \(\mu\A\) and \(\mu\A^*\) are equal. If changes occur in \(T\), \(p\), or \(x\A\) while the phases remain in equilibrium, the condition \(\dif(\mu\A/T) = \dif(\mu\A^*/T)\) must be satisfied. Equating the expressions on the right sides of Eqs. 12.3.2 and 12.3.3 and combining terms, we obtain the equation \begin{equation} \frac{H\A-H\A^*}{T^2}\dif T - \frac{V\A-V\A^*}{T}\difp = \bPd{(\mu\A/T)}{x\A}{T,p}\dx\A \tag{12.3.4} \end{equation} which we can rewrite as \begin{gather} \s{ \frac{\Delsub{sol,A}H}{T^2}\dif T - \frac{\Delsub{sol,A}V}{T}\difp = \bPd{(\mu\A/T)}{x\A}{T,p}\dx\A } \tag{12.3.5} \cond{(phases in} \nextcond{equilibrium)} \end{gather} Here \(\Delsub{sol,A}H\) is the molar differential enthalpy of solution of solid or gaseous A in the liquid mixture, and \(\Delsub{sol,A}V\) is the molar differential volume of solution. Equation 12.3.5 is a relation between changes in the variables \(T\), \(p\), and \(x\A\), only two of which are independent in the equilibrium system.
Suppose we set \(\difp\) equal to zero in Eq. 12.3.5 and solve for \(\dif T/\dx\A\). This gives us the rate at which \(T\) changes with \(x\A\) at constant \(p\): \begin{gather} \s{ \Pd{T}{x\A}{\!p} = \frac{T^2}{\Delsub{sol,A}H} \bPd{(\mu\A/T)}{x\A}{T,p} } \tag{12.3.6} \cond{(phases in} \nextcond{equilibrium)} \end{gather} We can also set \(\dif T\) equal to zero in Eq. 12.3.5 and find the rate at which \(p\) changes with \(x\A\) at constant \(T\): \begin{gather} \s{ \Pd{p}{x\A}{T} = -\frac{T}{\Delsub{sol,A}V}\bPd{(\mu\A/T)}{x\A}{T,p} } \tag{12.3.7} \cond{(phases in} \nextcond{equilibrium)} \end{gather}
Equations 12.3.6 and 12.3.7 will be needed in Secs. 12.4 and 12.5.