Question 18.8: A batch of crude pentane contains 15 mol percent n-butane an......
A batch of crude pentane contains 15 mol percent n-butane and 85 percent n-pentane. If a simple batch distillation at atmospheric pressure is used to remove 90 percent of the butane, how much pentane would be removed? What would be the composition of the remaining liquid?
The final liquid is nearly pure pentane, and its boiling point is 36°C. The vapor pressure of butane at this temperature is 3.4 atm, giving a relative volatility of 3.4. For the initial conditions, the boiling point is about 27°C, and the relative volatility is 3.6. Therefore, an average value of 3.5 is used for {}\alpha_{A B}.
Basis: 1 mol feed
n_{0,A}=0.15\,(\mathrm{butane})\qquad n_{A}=0.015\qquad n_{0B}=0.85\,(\mathrm{pentane})
From Equation (18.72)
n_{A}=x n (18.72)
{\frac{n_{B}}{0.85}}=0.1^{1/3.5}=0.518\qquad n_{B}=0.518(0.85)=0.440
n=0.44+0.015=0.455\,\mathrm{mol}\qquad x_{A}=\frac{0.015}{0.455}=0.033