# Relationship of the abundance of an isotope and the vapor pressure

Say I have a sample of some alkali metal - Rubidium for example. A sample of Rubidium consists of approximately 27% $^{87}$Rb and 73% $^{85}$Rb. Does the vapor pressure of each isotope individual necessarily have to be in the same proportion? More specifically, assuming I have enough Rubidium around that I have enough atoms of each isotope to reach the saturated vapor pressure, is there any physical reason why the ratio of the vapor pressures would match their relative abundances?

I guess this is coming from this perspective - the linear absorption spectrum for a sample of natural rubidium at a specific fine transition (D2 line - 5S1/2 ->5P3/2 if you are interested) looks like image below, the $^{87}$Rb peaks (outer) are clearly smaller than the $^{85}$Rb peaks (inner), and this amplitude ratio is directly related to their natural abundances. If the atoms are in a gas, what forces their vapor pressure to have the same ratio as their relative abundance? Along a similar line of questioning. A sample of Rb will naturally have some trace amount of Cs in it. The saturated number density of Cs at room temperature is ~$10^9$ atoms per cm$^3$, and I'm looking at a volume of say 1cm$^3$. If I have a droplet of 1 gram of Rb in the volume, which is .001 moles, even at .001% impurity I have $>10^{17}$ atoms of cesium in the sample, clearly enough to reach saturated density. -Does this imply that you can do spectroscopy on a volume with Rb and obtain the same spectrum as a volume with a droplet of Cs in it? If not, what prevents the Cs impurity in Rb to reach full saturation.

*Note: Perhaps I should make it clear : The linear absorption spectrum is proportional to the theoretical absorption (for a single atom) times the number density of atoms in the sample. The relative amplitudes of the peaks are therefore a proportional to the relative abundances of the two isotopes.