Question 4 of 6: Part B: Vapour–Liquid Equilibrium of Benzene–Toluene

Question 4 — Part B: Vapour–Liquid Equilibrium of Benzene–Toluene (25 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Benzene (1) and toluene (2) form a nearly ideal solution, so Raoult’s law applies: $P = x_1 P_1^{\text{sat}} + (1-x_1)P_2^{\text{sat}}$ and $y_1 = x_1 P_1^{\text{sat}}/P$, with the total pressure fixed at $P=101.32$ kPa. Pure-component vapour pressures are tabulated versus temperature (benzene boils at 80.0 °C, toluene at 110.1 °C, where each pure vapour pressure equals 101.32 kPa).

Find. (a) the equilibrium pairs $(x_1,y_1)$ at each temperature; (b) an empirical $y_1$–$x_1$ relation (relative volatility form).

Approach. At each tabulated temperature the total pressure must equal 101.32 kPa; a lever on the two pure vapour pressures gives the liquid composition $x_1=(P-P_2^{\text{sat}})/(P_1^{\text{sat}}-P_2^{\text{sat}})$, and Raoult’s law then gives $y_1$. Averaging the ratio $\alpha=P_1^{\text{sat}}/P_2^{\text{sat}}$ across the range yields a single-parameter $y$–$x$ curve.

1. Bubble-point lever for the liquid composition. Because $P=x_1P_1^{\text{sat}}+(1-x_1)P_2^{\text{sat}}=101.32$ kPa at every equilibrium temperature, $$x_1 = \frac{P - P_2^{\text{sat}}}{P_1^{\text{sat}} - P_2^{\text{sat}}}.$$ For example at 92.9... (worked point) $T=93.3$ °C: $x_1=(101.32-60.26)/(149.72-60.26)=0.459$.
2. Raoult’s law for the vapour composition. The vapour benzene fraction is $$y_1 = \frac{x_1 P_1^{\text{sat}}}{P}.$$ At $T=93.3$ °C, $y_1 = (0.459)(149.72)/101.32 = 0.678$. Repeating at every temperature gives the full VLE table:

3. Empirical relation via average relative volatility. The relative volatility $\alpha=P_1^{\text{sat}}/P_2^{\text{sat}}$ varies only mildly (2.58 at 82.9 °C down to 2.35 at 110 °C); its mean is $\bar\alpha=2.46$. The constant-$\alpha$ equilibrium curve then compactly represents the whole table: $$\boxed{y_1 = \frac{\bar\alpha\,x_1}{1+(\bar\alpha-1)x_1} = \frac{2.46\,x_1}{1+1.46\,x_1}}.$$ As a check at $x_1=0.5$ this gives $y_1=0.71$, matching the tabulated trend. That is part (b).