Adsorption of polar molecules On any given surface, is there more adsorption if the molecules in a vapor are polar? For example, for a given (approximate) molecular weight would one expect more adsorption onto a surface by the polar molecule?
 A: A nice simple model for adsorption on a surface is the Langmuir isotherm. We'll skip lightly over the fact that the real world is invariably more complicated than this and note that the Langmuir model gives the following equation for the fraction of the surface covered:
$$ \theta = \frac{KP_a}{1+KP_a} $$
where $P_a$ is the partial pressure of the adsorbate and $K$ is some rather vaguely defined equilibrium constant for the process:
$$ \text{Adsorbate} + \text{Surface} \rightarrow \text{Adsorbed molecule} $$
And in an equally hand waving way we expect the equilibrium constant $K$ to be related to a Gibbs free energy change for the adsorption process:
$$ K = e^{-\Delta G/RT} $$
The point of all this is that what matters is the free energy change for a molecule adsorbing on a surface, and for simple molecules where we don't worry too much about changes on rotational or conformational freedom this is dominated by the enthalpy of adsorption.
Which brings us eventually to your question. If your surface is polar, i.e. it is covered in dipolar groups, then it will interact more strongly with dipolar adsorbates than with non-polar adsorbates. So the more polar the molecule the greater the enthalpy of adsorption and the more strongly it will adsorb. If the surface is non-polar then the polarity of the adsorbate will make much less difference.
In the world around us the vast majority of surface are polar, so as a rule of thumb polar molecules will adsorb more strongly than non-polar ones.
