The GATE Grind

GATE 2026 CH – Question 39

Mass Transfer · Leaching, liquid-liquid extraction, drying, humidification and adsorption · 2 marks · Multiple choice

A volatile organic compound (VOC) is to be adsorbed from air onto a bed of activated carbon. The equilibrium capacity of activated carbon at the feed conditions is 0.4 grams VOC per gram of activated carbon. The column contains 4 grams of activated carbon per cm$^2$ of cross-section. The feed rate into the adsorber column is 0.2 grams VOC cm$^{-2}$ h$^{-1}$. Breakthrough time is defined as the time at which the concentration at the exit of the bed ($c$) reaches a value of $0.05c_0$, where $c_0$ is the concentration of VOC in the feed. The breakthrough time for the bed is 2.1 h. The area under $c/c_0$ curve between the initial and breakthrough times is 0.1 h. Which one of the following is the fraction of unused bed at breakthrough?

  1. 0
  2. 0.25
  3. 0.50
  4. 0.75

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (D) 0.75

Explanation

The time the bed would take to saturate completely is $t_{total} = \frac{4 \times 0.4}{0.2} = 8$ h. The effective time used at breakthrough is $t_b - \int_0^{t_b} \frac{c}{c_0}dt = 2.1 - 0.1 = 2.0$ h. The fraction of the bed that is used is $\frac{2.0}{8} = 0.25$, so the unused fraction is $1 - 0.25 = 0.75$.