# Diffusion of gases in the atmosphere

Suppose that the atmosphere is composed of 21% $O_2$ and 78% $Kr$ (instead of $N_2$). Since the density of $Kr$ is greater than the density of $O_2$, the lower atmosphere (where we live) should be deprived of $O_2$. Am I right? I know that diffusion has a role here, but is there a temperature where diffusion can't overcome the fact that $O_2$ should be in the upper atmosphere?

• Why should the less-dense gas sink? – user10851 Jul 6 '13 at 20:47
• Ok now it's a good question (even more so assuming it is about the real situation, not purely diffusion). Many similar principles apply to this related question about the ocean. A good answer here will deal not only with diffusion, but also convection, since parts of the atmosphere (like the troposphere we know and love) are unstable to convection. – user10851 Jul 6 '13 at 21:02
• @metacompactness We define homework slightly loosely around here. From the tag-wiki: "Applies to questions of primarily educational value - not only questions that arise from actual homework assignments, but any question where it is preferable to guide the asker to the answer rather than giving it away outright." Which to me seems to fit this kind of question. It's definitely an interesting question with some interesting answers -- being tagged as homework doesn't mean it's a bad thing! – tpg2114 Jul 6 '13 at 21:08
• @ChrisWhite My first inclination is that it will be fairly well mixed because the atmosphere is extremely turbulent due to both the terrain and temperature gradients caused by surface heating. Turbulence usually trumps diffusion almost everywhere... If I feel like putting off writing my code and papers, I'll come up with an answer :) – tpg2114 Jul 6 '13 at 21:10
• The troposphere experiences powerful convective mixing which is why water, carbon-dioxide, argon and molecular nitrogen and oxygen stay mixed, but what little helium and molecular hydrogen there is in the atmosphere does layer out (and then slowly bleeds off). – dmckee --- ex-moderator kitten Jul 6 '13 at 21:51

If you assume a constant temperature atmosphere and constant homogeneous gravitational field, you get the height dependence of the different molecular or atomic gas species concentrations n easily using the equilibrium Boltzmann energy distribution $n∝\exp(-E/kT)$, where $E=mgh$ is the potential energy of a gas molecule with mass $m$, $g$ is the gravitational acceleration, $h$ is the height above ground, $k$ is the Bolzmann constant, and $T$ is the absolute temperature. So, in this idealized picture, you need no consideration of (thermo-)diffusion processes, convection, turbulence, and the like, which would be a scientific major project to take into account. This shows that the heavier the molecule the faster is the exponential decrease in concentration with height. This means that also the concentration ratio of $Kr$ to $O_2$ decreases exponentially with height. If you assume that the percentages given for $O_2$ and $Kr$ are those of the total atmospheric content, then it follows by integration of these distributions that the ratio of $Kr$ to $O_2$ concentrations at ground level is higher than the ratio following from the total atmospheric percentages.