Question

Which among the following gases is adsorbed to greater extent at similar conditions of temperature and pressure if the adsorbent remains same?

• A. $\text{N}_2$
• B. $\text{Cl}_2$
• C. $\text{H}_2$
• D. $\text{O}_2$

Hint:

To quickly rank the physical adsorption of non-polar gases, simply look at their molecular weight!
Higher Molecular Weight $\rightarrow$ Stronger van der Waals Forces $\rightarrow$ Higher Critical Temperature $\rightarrow$ Easier Liquefaction $\rightarrow$ Greater Adsorption.
Since $\text{Cl}_2$ ($71\text{ g/mol}$) is by far the heaviest among the choices, it wins instantly.

Answer 1

The Correct Option is B

Step 1: Understand the question.
Under the same temperature and pressure, on the same adsorbent, we must find which gas is adsorbed the most.
Step 2: Recall the key rule.
For physical adsorption, the gas that is easiest to liquefy sticks the most. Easy liquefaction means stronger forces between molecules.
Step 3: Note the molecules.
$H_2$, $N_2$, $O_2$ and $Cl_2$ are all non-polar, so they only have weak London dispersion forces.
Step 4: Link force strength to size.
Bigger and heavier molecules have larger electron clouds, which are easier to disturb, so their dispersion forces are stronger.
Step 5: Compare the masses.
The masses are about $H_2 = 2$, $N_2 = 28$, $O_2 = 32$, $Cl_2 = 71$ g/mol. Chlorine is the heaviest, so it has the strongest forces and the highest critical temperature.
Step 6: Choose the answer.
Chlorine is adsorbed the most, which is option 2. \[ \boxed{Cl_2} \]