Four gases, A, B, C and D have critical temperatures 5.3, 33.2, 126.0 and 154.3K respectively. For their adsorption on a fixed amount of charcoal, the correct order is :

We are given four gases A, B, C, and D with critical temperatures 5.3 K, 33.2 K, 126.0 K, and 154.3 K respectively, and we need to determine the correct order of their adsorption on a fixed amount of charcoal. The extent of adsorption of a gas on a solid surface increases with the ease of liquefaction of the gas, so a gas with a higher critical temperature is more easily liquefied and therefore more readily adsorbed.

The critical temperatures of the four gases are:
A = 5.3 K
B = 33.2 K
C = 126.0 K
D = 154.3 K

Arranging these in decreasing order of critical temperature (and hence decreasing order of adsorption) gives
$$D \, (154.3 \, K) \; > \; C \, (126.0 \, K) \; > \; B \, (33.2 \, K) \; > \; A \, (5.3 \, K)$$

Therefore, the correct order of adsorption is D > C > B > A.

The correct answer is Option C.