GATE 2024 CH – Question 40
A gas stream containing 95 mol% CO$_2$ and 5 mol% ethanol is to be scrubbed with pure water in a counter-current, isothermal absorption column to remove ethanol. The desired composition of ethanol in the exit gas stream is 0.5 mol%. The equilibrium mole fraction of ethanol in the gas phase, $y^*$, is related to that in the liquid phase, $x$, as $y^* = 2x$. Assume CO$_2$ is insoluble in water and neglect evaporation of water. If the water flow rate is twice the minimum, the mole fraction of ethanol in the spent water is
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Correct answer: (B) 0.0126
Explanation
Use mole ratios. The gas has $G_s = 0.95$ per mole of feed, with $Y_1 = \frac{0.05}{0.95} = 0.05263$ and $Y_2 = \frac{0.005}{0.995} = 0.005025$. The minimum liquid rate has the exit liquid in equilibrium with the inlet gas: $y = 0.05$ gives $x = 0.025$, so $X^* = \frac{0.025}{0.975} = 0.02564$. Then $L_{s,min} = \frac{0.95(0.05263 - 0.005025)}{0.02564} = 1.764$. The actual rate is twice this, $L_s = 3.527$, so $X_1 = \frac{0.95 \times 0.047605}{3.527} = 0.01282$ and $x_1 = \frac{X_1}{1 + X_1} = 0.0127$, which is 0.0126.