# Why does physisorption (physical adsorption) increase with ease of liquefaction?

This is what my textbook says about physisorption

The amount of gas adsorbed by a solid depends on the nature of gas. In general easily liquefiable gases (i.e. with higher critical temperature) are readily adsorbed. Thus $\pu{1g}$ of activated charcoal adsorbs more sulfur dioxide (critical temperature $\pu{630K}$) than methane (critical temperature $\pu{190K}$).

I couldn't figure out why this should be so. In fact, according to my reasoning it should be the opposite. A gas which is easily liquefiable has higher intermolecular forces (here van der Waals forces) which would bound the gas molecules together and reduce physisorption. Where is my reasoning wrong ?

It can be a probable answer but I am not sure about it.

You are correct to say that easily liquifiable gases have stronger intermolecular forces of attraction (van der Waals forces). As a result of which they are strongly bound to themselves. Consider physisorption, here, the forces between the adsorbate molecules and the adsorbent are weak van der Waals forces. Once the surface area of the adsorbent is completely occupied, the next layer of adsorbate molecules will be bound weakly. So if the gas molecules have stronger van der Waals force of attraction, they would be adsorbed readily in preference to gas molecules having weaker van der Waals forces of attraction.

Your textbook (which one is it? by the way) is correct: gases with higher critical temperature are easier to liquefy. You can confirm this by comparing boiling and critical points of different gases. For example:

• for CO$_2$ (high critical temperature, $T_c=304$ K) triple point is 220 K.
• for N$_2$ (low critical temperature, $T_c=126$ K) normal boiling point is 77 K.

This clearly means that CO$_2$ has stronger intermolecular forces than N$_2$. This also means that the forces between CO$_2$ and another molecule are stronger that between N$_2$ and the same molecule.

A surface is a (almost infinite, in the nanoscale) set of atoms. Many orders of magnitude denser than the gas phase. If $CO_2 - CO_2$ interactions are stronger than $N_2 - N_2$ interactions, also $CO_2 - surface$ interactions are stronger than $N_2 - surface$, whereas the gas phase interactions become negliglible due to its much lower density.

This is why we observe stronger adsorption interactions for molecules with high critical temperatures. You can see this in the plot attached from this source.

The thing is that we are talking about adsorption which is surface phenomena and we know gases with high critical temp are easily liquifiable which implies gas molecules can form strong van der waal forces but as forces on surface of gas are unbalanced then gas molecules tries to make van der waal forces with the surface of adsorbent [gas being adsorbate].And hence The amount of gas adsorbed by a solid depends on the nature of gas. In general easily liquefiable gases (i.e. with higher critical temperature) are readily adsorbed as the form large number of weak van der waal forces.

hope it helps!

The gas having lower critical temperature than room temperature will exist in gaseous form at room temperature. So to liquefy it we need to decrease the temperature.

While the gases with higher critical temperature will themselves be having strong van der Waal's attraction and thus they will get easily liquefied.

And as you said, easily liquefiable gases will get adsorbed more at the surface.