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Everyone must have heard that balloons are filled with helium, and the fact associated with it that helium gas is light and light gases always go upward.

There comes a question to mind: if the molar mass of $\ce{CO2}$ is greater than that of $\ce{O2}$ and $\ce{N2}$, then why doesn't $\ce{CO2}$ occupy the lower layer of the atmosphere, since it is heavier than $\ce{O2}$ and $\ce{N2}$, as in the case of balloons, where helium being light rise upwards.

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    $\begingroup$ Proportionally, CO2 is slighter more prevalent at the surface than high up. But, as you may have noticed outside, there is a lot of activity in our atmosphere, from winds to large scale updrafts (and downdrafts) that keep the atmosphere pretty well mixed over long length scales. $\endgroup$
    – Jon Custer
    Oct 17, 2016 at 18:37
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    $\begingroup$ Related: Why does ozone, being heavier than air, not settle down? $\endgroup$
    – Mithoron
    Oct 17, 2016 at 22:52
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    $\begingroup$ Related: Chlorofluorocarbons rising $\endgroup$
    – user7951
    Nov 1, 2016 at 7:17

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That's because of two reasons. One is entropy, the ultimate force of chaos and disorder. Sure, gases would like to be arranged according to their density, but even above that, they would like to be mixed, because mixing creates a great deal of entropy. If you prevent the mixing, then they would behave just as you expected. Indeed, a balloon filled with $\ce{CO2}$ would drop right to the floor and remain there.

On the other hand, if you allow mixing, light gases wouldn't necessarily go upward. Just pierce that balloon with helium, and... Well, you won't actually see much, but in fact, the helium will disperse in the atmosphere and remain there. True, its contents in the upper layers would be somewhat higher, but only somewhat. It is not like a layer of pure helium floating atop all those $\ce{O2}$ and $\ce{N2}$.

The importance of entropy is by no means limited to gases. Think of all that salt in the oceans. Salt is much more dense than water; wouldn't it just drop to the ocean floor? Well, no, it rather wouldn't.

The other reason is the constant action of winds and currents mentioned by Zhe. They stir the atmosphere (or sea water, for that matter) and make it even more uniform than it might have been otherwise.

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Here's how you can get a quantitative handle on this. Suppose you have a tall column of two gases with molecular weights $M_1$ and $M_2$ in a cylindrical container of height $L$ and cross sectional area $A$, and you have the amount of substances $n_1$ and $n_2$ of each gas, respectively, in the container. Suppose that the gases can be treated as ideal gases, and that the container is maintained at constant temperature. When the system reaches equilibrium, what is the extent of the gravitational separation of the two gases?

Since you are dealing with an ideal gas mixture, each gas can be treated as if it is the only gas present. Each gas must satisfy the local static equilibrium condition: $$\frac{\mathrm{d}p}{\mathrm{d}z}=-\rho g, $$ where $z$ is the elevation from the bottom of the container, $p$ is the partial pressure of a given species at elevation $z$, and $\rho$ is the local density of the species at elevation $z$. From the ideal gas law: $$\rho=\frac{pM}{RT},$$ where $M$ is the molecular weight. If we combine these two equations, we obtain: $$\frac{\mathrm{d}p}{\mathrm{d}z}=-\frac{pMg}{RT}$$ Integrating this equation between $z = 0$ and $z = L$ yields: $$\frac{p(L)}{p(0)} =\frac{\rho(L)}{\rho(0)}=\exp{\left(-\frac{MgL}{RT}\right)}$$ So, comparing species $1$ and $2$, $$\frac{(\rho_1(L)/\rho_1(0))}{(\rho_2(L)/\rho_2(0))}=\exp{\left(-\frac{(M_1-M_2)gL}{RT}\right)}$$ As an example, for the case of $L = \pu{1 km}$, $M_1 = \pu{44 g/mol}$, $M_2 = \pu{28 g/mol}$, and $T = \pu{298 K}$, the ratio of the densities at $z = 0$ compared to $\pu{1 km}$ differ by only about $6\%$.

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  • $\begingroup$ That all looked very convincing until the final calculation. Should not the units of L be metres? 16*9.8*1000/8.314/298 > 60, implying diffusion doesn't prevent complete separation over 1000 metres of still air. Suggests CO2 mixing requires wind: scienceblogs.com/illconsidered/2010/10/is-co2-well-mixed Another discussion: researchgate.net/post/… Reliable sources would be better... $\endgroup$ Aug 12, 2022 at 20:46
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    $\begingroup$ @CedricKnight. I never implied that convective transport was not involved. $\endgroup$ Aug 12, 2022 at 22:10
  • $\begingroup$ OK. I was hoping you would clear up my 'factor of 1000' confusion by clarifying the answer. The units of g are metres, so why is L in km? Your answer derives the scale height of the atmosphere (about 8 km), but M₁ and M₂ are described as 'molecular weights' (shouldn't that be molecular masses?) in g/mol. To use entirely SI units, these would be expressed in kg/mol and then L would be in metres. So the 6% appears correct, but seemed doubtful in context of the second link I posted (where sulfur hexafluoride flows like an invisible liquid initially, before diffusing). $\endgroup$ Aug 13, 2022 at 10:09
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The primary reason why you don't get separation of gases in the atmosphere by density is that the atmosphere (especially the lower atmosphere) is turbulent. The second reason, but one that is unimportant in the lower atmosphere, is diffusion.

If you mix a bunch of gases in a jar, diffusion will ensure that they are thoroughly mixed. So diffusion can cause a lot of mixing. When the atmosphere is thin, then diffusion is competing with gravity and some (small) degree of separation by density does occur (but this is trivial in the lower atmosphere: the air at the top of the tallest mountain has essentially the same mix of gases as the air at sea level). But in the lower atmosphere this is unimportant and the dominant force is turbulent mixing (or as it is more commonly know to non-chemists, wind). Given the amount of wind and the constant turbulent motion of air, things get mixed quickly and stay well mixed. Only in the very rarefied upper atmosphere does the battle between diffusion and gravity lead to any separation between light and dense gases.

If you are in a large closed space where wind doesn't exist, then things will still mix, but a lot more slowly. Some industrial accidents, for example, are caused because some gases or vapours will accumulate in the lower parts of a vessel pushing out the oxygen and air. If you don't follow good safety procedure this can rapidly asphyxiate you (good practice would involve fully ventilating the space allowing turbulent mixing to do the job of mixing up the dense gases with everything else). Equally, if you filled a closed vessel with helium, it could accumulate in the upper parts of the vessel and have the same effect at least in the short term before diffusion evened out the mix). But if you partly filled the vessel with air and another gas and left it long enough, the two would be mixed eventually.

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There's certainly a preference for carbon dioxide to be lower, but I think you're forgetting that the outside world isn't a static mass of air. There are a lots of convection currents. Just think how much a major storm like would have mixed up the gases.

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