Question

The correct order of adsorption of the following gases on the surface of charcoal is

• A. \(III \gt II \gt I\)
• B. \(III \gt I \gt II\)
• C. \(II \gt I \gt III\)
• D. \(II \gt III \gt I\)

Hint:

The adsorption of gases on solids generally increases with increasing critical temperature and ease of liquefaction: \[ \text{More easily liquefied gas} \Rightarrow \text{Greater adsorption} \] Typical order: \[ SO_2 \gt NH_3 \gt CO_2 \gt CH_4 \gt H_2 \]

Answer 1

The Correct Option is A

Step 1: Recall the factors affecting adsorption of gases.
The extent of adsorption of gases on a solid surface depends mainly on: (i) Critical temperature (ii) Ease of liquefaction (iii) Molecular size and polarity Gases that are more easily liquefied are adsorbed more strongly.

Step 2: Compare the nature of the given gases.
Among the given gases: \[ H_2 \] is a very light, non-polar gas with very low critical temperature.
\[ CH_4 \] has a higher molecular mass and is more easily liquefied than \(H_2\).
\[ SO_2 \] is a polar gas with a much higher molecular mass and a very high critical temperature. It is therefore the most easily liquefied among the three gases.

Step 3: Use the adsorption trend.
Adsorption generally increases with increasing critical temperature: \[ SO_2 \gt CH_4 \gt H_2 \] because \[ T_c(SO_2) \gt T_c(CH_4) \gt T_c(H_2) \]

Step 4: Convert into the notation given in the question.
Given: \[ I = H_2 \] \[ II = CH_4 \] \[ III = SO_2 \] Therefore, \[ SO_2 \gt CH_4 \gt H_2 \] becomes \[ III \gt II \gt I \]

Step 5: Final conclusion.
Hence, the correct order of adsorption on charcoal is \[ \boxed{III \gt II \gt I} \] Therefore, option (1) is correct.