## Chapter 29

## Q. 29.1

An exhaust stream from a semiconductor fabrication unit contains 3 mol % acetone and 97 mol % air. In order to eliminate any possible environmental pollution, this acetone-air stream is to be fed to a mass transfer column in which the acetone will be stripped by a countercurrent, falling 293 K water stream. The tower is to be operated at a total pressure of 1.013 \times 10^5 \mathrm{~Pa}. If the combined Raoult–Dalton equilibrium relation may be used to determine the distribution of acetone between the air and the aqueous phases, determine

**(a)** the mole fraction of acetone within the aqueous phase, which would be in equilibrium with the 3 mol % acetone gas mixture.

**(b) ** the mole fraction of acetone in the gas phase, which would be in equilibrium with 20 ppm acetone in the aqueous phase.

## Step-by-Step

## Verified Solution

At 293 K, the vapor pressure of acetone is 5.64 \times 10^4 \mathrm{~Pa}.

**(a) ** By Raoult–Dalton law when y_A=0.03

(0.03)\left(1.013 \times 10^5 \mathrm{~Pa}\right)=x_A\left(5.64 \times 10^4 \mathrm{~Pa}\right)

\text { or } x_A=0.0539 \text { mole fraction acetone }

**(b)** 20 ppm acetone in solution

=\frac{20 \mathrm{~g} /(58 \mathrm{~g} / \mathrm{mol})}{999,980 \mathrm{~g}(18 \mathrm{~g} / \mathrm{mol})}

=6.207 \times 10^{-6} \frac{\text { mol acetone }}{\text { mol water }}

For the dilute solution, th mole fraction of acetone will be

x_A=\frac{86.027 \times 10^{-6} \mathrm{~mol} \text { acetone }}{1.0 \mathrm{~mol} \text { water }+6.027 \times 10^{-6} \mathrm{~mol} \text { acetone }}=6.207 \times 10^{-6}By Raoult-Dalton law

y_A P=x_A P_Ay_A\left(1.013 \times 10^5 \mathrm{~Pa}\right)=\left(6.207 \times 10^{-6}\right)\left(5.64 \times 10^4 \mathrm{~Pa}\right)

\text { or } y_A=3.45 \times 10^{-6} \text { mole fraction acetone }