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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

Y_A P=x_A P_A

 

(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 \text { g acetone }}{999,980 \text { g water }}

 

=\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_A

 

y_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 }