Question 10.2.1: Using Azeotropic Data to Predict VLE of a Binary Mixture Ben...

a. The van Laar model. The starting point is the equilibrium relation

which, at the pressures here, reduces to

Since x_{ i }=y_{ i } at an azeotropic point, we have \gamma_{ i }=P / P_{ i }^{\text {vap }} \text {, so that at } x_{ B }=0.525

Using Eqs. 9.5-10, we obtain \alpha=0.125 \text { and } \beta=0.0919. Therefore,

\begin{aligned}\alpha &=\left(1+\frac{x_{2} \ln \gamma_{2}}{x_{1} \ln \gamma_{1}}\right)^{2} \ln \gamma_{1} \\\beta &=\left(1+\frac{x_{1} \ln \gamma_{1}}{x_{2} \ln \gamma_{2}}\right)^{2} \ln \gamma_{2}\end{aligned} (9.5-10)

\ln \gamma_{ B }=\frac{0.125}{\left[1+1.360\left(\frac{x_{ B }}{x_{ C }}\right)\right]^{2}} \quad \text { and } \quad \ln \gamma_{ C }=\frac{0.0919}{\left[1+0.736\left(\frac{x_{ C }}{x_{ B }}\right)\right]^{2}} (i)

The values of the activity coefficients for benzene (B) and cyclohexane (C) calculated from these equations are given in the following table and Fig. 1.

To compute the composition of the vapor in equilibrium with the liquid we use Eqs. 10.1-1b

x_{ i } \gamma_{ i }(T, P, \underline{x}) P_{ i }^{ vap }(T)=y_{ i } P (10.1-1b)

x_{ B } \gamma_{ B } P_{ B }^{ vap }=y_{ B } P \quad \text { and } \quad x_{ C } \gamma_{ C } P_{ C }^{ vap }=y_{ C } P (ii)

and Eq. 10.1-2b

\sum x_{ i } \gamma_{ i }(T, P, \underline{x}) P_{ i }^{ vap }(T)=P (10.1-2b)

x_{ B } \gamma_{ B } P_{ B }^{\text {vap }}+x_{ C } \gamma_{ C } P_{ C }^{ vap }=P (iii)

In these equations the vapor compositions, y_{ B } \text { and } y_{ C }, and the equilibrium pressure P are unknown (the equilibrium pressure is 1.013 bar only at x_{ B }=0.525). The solution is obtained by choosing a value of x_{ B }, \operatorname{using} x_{ C }=1-x_{ B }, and computing \gamma_{ B } \text { and } \gamma_{ C } from Eqs. i, and the total pressure from Eq. iii. The vapor-phase mole fractions are then computed from Eqs. ii. The results of this calculation are given in the table and Fig. 2.

b. Regular solution model. Since benzene and cyclohexane are nonpolar, and their solubility parameters are given in Table 9.6-1, the activity coefficients can be predicted using Eqs. 9.6-10. The results of this calculation are given in the table. The agreement between the correlation of the data using the van Laar model and the predictions (without reference to the experimental data) using the regular solution is good in this case.

\begin{array}{l}R T \ln \gamma_{1}=\underline{V}_{1} \Phi_{2}^{2}\left[\delta_{1}-\delta_{2}\right]^{2} \\R T \ln \gamma_{2}=\underline{V}_{2} \Phi_{1}^{2}\left[\delta_{1}-\delta_{2}\right]^{2}\end{array} (9.6-10)

Note

From Fig. 2 we see that for this mixture the compositions of the vapor and liquid in equilibrium are very close. An important implication of this is that it will be very difficult to separate benzene and cyclohexane from their mixture using distillation.