# How does physical adsorption of mixed gas occur on solid? (eg I mean how the gas molecules of compete each other for a slot on solid surface)

I am working on sorption phenomenon of gases (including inert gas) on coal seam surface.

For a mixture of 2 gases at a certain pressure, carbon dioxide and methane, for instance, $\ce{CO2}$ is definitely adsorbed quicker and more comparing to methane. How come? I try to explain by Langmuir Isotherm and IAS model but it doesn't help much. For Langmuir isotherm, the affinity of gases to solid is determined by "b=1/PL" which comes from experiment. Langmuir isotherm assumes that all gas have equal access to the solid surface; however, this is untrue in reality.

Can you help me explain it in the thermodynamics point of view or molecular forces please? ($\ce{CH4}$, $\ce{CO2}$ are both non-polar gases)

• CO2 is polar, in that its bonds are polar. It is just that its overall dipole moment happens to be 0. – Ivan Neretin Apr 12 '17 at 4:52

## 2 Answers

Your supposition that gases do not have equal access to teh surface does not seem to me to be correct.

At the surface physisorption and chemisorption can occur. The former is due to inter-molecular interaction with the gas molecule and the surface molecules so clearly will depend on the nature of both of them. In chemisorption a chemical bond is formed. The physisorption occurs at longer range and has a smaller potential well (as measured by potential energy vs distance from surface) than chemisorption. (See Atkins & DePaula 'Physical Chemistry' for a figure.)

The gases approach the surface at random and the number of collisions of each will be in proportion to their partial pressure. What happens on collision depends, as mentioned above, on the strength of the intermolecular interaction of gas and surface and so depends on what they are. At higher temperature the collision energy is greater than at lower ones and so one would expect less physisorption as the collision energy will be greater than that of most available physisorption sites.

It is also possible for multiple layers to be formed (as in BET isotherm) and for a physisorbed molecule to diffuse around on the surface until it leaves again (by random thermal activation) of finds a more stable site or becomes chemisorbed.

As to your particular case we can only surmise but $\ce{CO2}$ is far more polarisable than methane so should have the larger dispersion forces so larger induced dipole - induced dipole intermolecular interaction and so preferentially occupy the surface. Both gases have transient dipoles due to molecular vibrations of which the $\ce{CO2}$ should be the larger. This may also be important.

You are talking about a porous solid. This is very different to a flat surface where adsorption of all molecules can occur without any barrier. To be adsorbed in the pores, the molecules need to be transported from the gas bulk to inside the pores, this process can be very slow depending on the diffusion rate.

By measuring an adsorption isotherm, you just get the amount adsorbed at equilibrium, that is, after all the molecules entered inside the pores.

If your solid has very narrow pores, say, micropores, then the size of the pore is very near the size of your molecules, and adsorption will be slow. In your case, the size of the $\ce{CO2}$ and $\ce{CH4}$ is very different, $\pu{0.33 nm}$ for $\ce{CO2}$ and $\pu{0.38 nm}$ for $\ce{CH4}$. So the diffusion inside the pores is much more restricted for $\ce{CO2}$, then it is expected to be slower. Probably your sample have many narrow micropores that result in this behavior.