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 cm² of cross-section. The feed rate into the adsorber column is 0.2 g VOC cm⁻² h⁻¹. Breakthrough time is defined as the time at which the exit concentration (c) reaches 0.05 c₀, where c₀ is the feed concentration. The breakthrough time is 2.1 h. The area under the c/c₀ curve between the initial and breakthrough times is 0.1 h. Which one of the following is the fraction of unused bed at breakthrough?

Question

Accepted Answer

The correct answer is D, 0.75. GATE 2026 official key: (D). Using the breakthrough-curve method with the given area (0.1 h) and breakthrough time (2.1 h), the fraction of unused bed works out to about 0.75.