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Why does carbon dioxide not sink in air if other dense gases do?

We evidently do not suffocate by carbon dioxide sinking to the bottom of the atmosphere and displacing oxygen and yet there are gases that do sink. This is commonly a problem in coal mines. Lower layers can fill up with gas that is unbreathable.

Here is a demonstration showing a 'boat' floating on sulphur hexafluoride.

Question

Given a mixture of two mutually non-reactive gases, what property determines whether the denser gas sinks to the bottom?

Hearth
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chasly - supports Monica
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7 Answers7

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Gases are all miscible. If initially separate and adjacent, they do not mix instantly, but once mixed (a process that occurs by molecular diffusion and is accelerated by macroscopic stirring or convection, just as for liquids), they do not spontaneously unmix.

During the time before substantial mixing occurs, gases behave somewhat like you may be picturing for immiscible liquids, e.g., water settling below oil. If a heavy gas is introduced into an environment in a pure or somewhat pure form (from some kind of reservoir), it will initially sink and displace lighter gases. This is a real danger with suddenly introduced carbon dioxide, but not with the carbon dioxide that has been in the atmosphere a long time.

In the mixed gas phase, the composition varies with height due to gravitational potential energy of different molecules, but all components are present at all heights, and on human scales the variation is small. In equilibrium, the vertical distance over which the density of a given gas changes substantially is termed the scale height, and is ~8 km for nitrogen, ~7 km for oxygen, and ~5 km for carbon dioxide (in Earth conditions). Even for sulfur hexafluoride it is ~1.5 km. As you go down, all components gradually become denser, the heaviest ones the fastest.

Moreover, as noted in a comment, the outdoor atmosphere is not in equilibrium but has a lot of turbulent motion, so even these gradual variations in composition that might be seen in controlled conditions are washed out in natural conditions.

nanoman
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  • This is the best answer, IMO. – Lawnmower Man Feb 01 '21 at 04:19
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    This is the answer I was looking for that calculates the characteristic height of segregation: kilometers. Too many answers and comments are treating the problem as a dichotomy, with demixing either absent or predominant and with much confusion about a sinking parcel of pure gas vs. unmixing of that gas from a mixture. – Chemomechanics Feb 01 '21 at 08:33
  • "re" lamination of two miscible fluids also requires a laminar flow regime (hence the name, actually.) Turbulence at any point reverses the trend. – Stian Feb 01 '21 at 08:50
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    The final paragraph gives the impression that the density of different molecular species varies with different scale heights in Earth's atmosphere. This is not true for the heterosphere (<100 km). All long-lived molecular species have essentially constant proportions in the heterosphere. Turbulence dominates over diffusion, so there is only a single scale height for all species. – jkej Feb 01 '21 at 09:40
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    what a horrific disaster (Lake Nyos) - thanks for that information ! – Fattie Feb 01 '21 at 15:32
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    Scale heights don't directly give the relevant ratios for concentration changes by height in a system where only diffusion and gravity compete. There would be notable gradients in atmospheric composition if they were the only factors. But there is little observed difference at heights of ~10km (see this for example. So something else must matter more than gravity or diffusion. – matt_black Feb 02 '21 at 16:17
  • @matt_black I think your main point is addressed in my final paragraph (added due to jkej's comment). I don't follow your first sentence -- it seems scale heights do "directly give...ratios...where only diffusion and gravity compete". Also I find it interesting and odd to describe diffusion as "competing" here. I guess you're picturing that if you could "turn off" diffusion, the atmosphere could remain completely stratified with heavy gases below light gases. But that wouldn't be any kind of thermal equilibrium (of which diffusion is an inherent aspect). In equilibrium we just use Boltzmann. – nanoman Feb 02 '21 at 17:33
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    @nanoman Yes you do address it but underweight its importance. Turbulence dominates. The other point is mainly about clarity. Calculating the expected gravity vs diffusion (my use of "competes" signifies that–even at equilibrium–gravity separates but diffusion mixes and the equilibrium is not perfect mixing) % composition at different heights for different gases would be a clearer way to show the effect you are talking about. – matt_black Feb 02 '21 at 17:49
  • @matt_black See my last comment on jamesqf's answer for why I am reluctant to say "turbulence dominates" in the context of this question even though it does dominate transport in the atmosphere. – nanoman Feb 02 '21 at 17:56
  • I'll point out that over time helium does diffuse to the upper atmosphere where it is lost to space. So if all helium release into the atmosphere were stopped, over a great period of time earth would lose all the helium that is currently in the atmosphere. – MaxW Feb 03 '21 at 06:01
  • @MaxW Technically, helium reaches the upper atmosphere through turbulent mixing, just like any other species released at ground level. Molecular diffusion is extremely slow in the lower atmosphere. But once helium reaches the heterosphere (>100 km) it stratifies to the top and when it reaches the exosphere it eventually gets lost through thermal escape. – jkej Feb 05 '21 at 20:13
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Because the strict separation of gases occurs only if initially we have them in their pure form. Then due to density difference of pure gas fluids, the denser falls down (buoyant force isn't strong enough to keep it at fixed height).

If the gas column was isolated and kept at constant temperature, each individual gas species in it would eventually distribute along the whole column, with number density (concentration) decreasing with height $h$, according to function

$$ c(h) =c_0 e^{-\frac{mgh}{kT}}\tag{*} $$ where $m$ is mass of the molecule. This follows from the law of Boltzmann probability distribution: probability of state having energy $E$ is proportional to $e^{-\frac{E}{kT}}$.

Sharp separation of molecular-scale mixture into pure fluids does not happen. Eventually some lighter gases will reach even the greatest depth of the heavier gas and at each height some constant mixture proportion is established.

Molecular nitrogen has mass 28, while oxygen has mass 32, so (*) predicts that in equilibrium, oxygen density is varying more on the ground than the nitrogen density. So it is the oxygen that tends to "stay lower".

Ján Lalinský
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    Being pure in their initial form is not the issue. The question is how strong is the gravitational segregation compared to other factors. You don't show how strong that separation is weven in an ideal system where it competes with diffusion. You say some mixing occurs but don't state the equilibrium position given the average kinetic energy at room temperature. That's a big omission. – matt_black Feb 02 '21 at 15:32
  • @matt_black Being pure in their initial form is the major example on which people's expectation of gas separation is based. If you start with atmospheric mixture, gravity does nothing to alter it. – Ján Lalinský Feb 02 '21 at 16:16
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    True, but the way you describe the issue implies that pure gases will separate. No, they have been separated and will eventually mix. But the reason why they mix quickly is turbulence and diffusion will work over a longer period. You state both that different gases will separate (final paragraph) and won't separate (second paragraph) but don't describe the importance or relevance of the contradictory statements to the real world. – matt_black Feb 02 '21 at 16:26
  • But they will separate for a limited short term. Pure gases have been separated as you write, but they also will mostly separate again if put into same container with only a little mixing (macroscopic mixing is partially reversible). That's the well known example of CO2 poured into container in air. Although there is turbulent motion that makes diffusion of air into CO2 more intensive, if there is not too much of that, the gas turbulence will quiet down and gases will separate again due to difference in their density. – Ján Lalinský Feb 02 '21 at 20:46
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    That comment is just wrong. Gases in a motionless container with no turbulence will mix over time and will not separate. Diffusion is slow, bit it does the job and nothing will stop in a container even in the complete absence of turbulence. – matt_black Feb 03 '21 at 00:21
  • @JánLalinský: Are you referring to how, if two fluid phases are mixed macroscopically (but not microscopically), the various little pools of each phase can rejoin together, obeying bulk-flow dynamics (since they're not microscopically mixed yet), temporarily reversing the macroscopic mixing before microscopic mixing kicks in? – Nat Feb 03 '21 at 05:14
  • @matt_black I said "But they will separate for a limited short term." I agree that in the long term the diffusion will mix the gases irreversibly. – Ján Lalinský Feb 03 '21 at 12:54
  • @Nat Yes they can, because of the buoyancy difference and slowness of diffusion. Of course the boundary won't be perfect jump in density, there will be layer of mixture that is microscopically and irreversibly mixed due to diffusion happening before. – Ján Lalinský Feb 03 '21 at 12:56
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[edit] As stated in other answers gases in equilibrium are fully mixed by diffusion. The driving force for segregation or mixing, which in liquids is the difference in interaction between like and unlike molecules, is absent. In a gas, molecules only collide and do not stick together. For molecules with different weight in a gravitational field there just will be a different height profile. This is also why molecules heavier than air such cfk's are able to reach the ozone layer. Away from equilibrium gases can be segregated. Such segregation can be enhanced by gravity for gases of very different weight.

N$_2$ and O$_2$ have a mass of 28 and 32 a.u. respectively. At room temperature and in unstable atmospheric conditions this difference is too small for gravitational segregation. The same is true for CO (28 a.u.). CO$_2$ however weighs 44 a.u. and can replace air in stagnant conditions, such as in mines and wells. Besides that, I know of only one fatal incident, when a supersaturated crater lake in Cameroon suddenly released a large amount of CO$_2$. SF$_6$ is quite heavy weighs 140 a.u. so it stays gravitationally segregated for longer. In poorly ventilated places radioactive radon (222 and 220 a.u.) emerging from concrete, poses a risk.

my2cts
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  • Thanks for your answer. I have changed my title to CO2 because I hadn't realised that Nitrogen is lighter than Oxygen! It doesn't affect my main question and I don't think it affects your answer materially. Hope that's OK. – chasly - supports Monica Jan 31 '21 at 13:57
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    So are you saying that if I have, say, an ordinary cube-shaped room 3 m on a side, and I fill it with a thoroughly blended mixture of 50% air and 50% SF$_6$, then the SF$_6$ will spontaneously settle to the bottom of the room? I didn't think gases worked that way; what I thought is that in an equilibrium situation, each gas distributes itself throughout the container in the same way that it would if it were the only gas present. So, I would expect the mixture to remain nearly uniform. (And if the room had air in the top half and SF$_6$ in the bottom half, I'd expect them to spontaneously mix.) – Tanner Swett Jan 31 '21 at 20:58
  • @TannerSwett You are right. In the ende the gases mix. I corrected my statement on SF$_6$ accordingly. – my2cts Jan 31 '21 at 22:49
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    In large quantities, CO₂ does indeed sink in air, as demonstrated in the Lake Nyos disaster… – gidds Jan 31 '21 at 23:53
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    @gidds It did not sink, it just emerged from the water and streamed down the hill. – my2cts Feb 01 '21 at 00:21
  • -1: No matter how "stagnant" the conditions, nitrogen/oxygen/carbon dioxide that are already mixed will not noticeably separate unless you look over several kilometers vertically. – nanoman Feb 01 '21 at 00:52
  • @TannerSwett, I would expect your mixture of 50% air and 50% SF6 to develop a vertical concentration gradient if it remained totally undisturbed for several hours. That would be an interesting experiment to conduct. – David White Feb 01 '21 at 01:19
  • @DavidWhite Over a 3 m height, the relative concentration of air versus SF${}_6$ would vary by only ~0.1%. – nanoman Feb 01 '21 at 01:32
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    @my2cts From the linked article: “The gas cloud initially rose at nearly 100 kilometres per hour (62 mph) and then, being heavier than air, descended onto nearby villages”.  Is sinking different from descending? – gidds Feb 01 '21 at 01:41
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    "N2 and O2 have a mass of 28 and 32 a.u. respectively, and it is" There isn't a singular antecedent for "it". – Acccumulation Feb 01 '21 at 04:56
  • @Acccumulation Yeah, that dates to when nitrogen was the main subject of the question -- "it" refers to nitrogen. – nanoman Feb 01 '21 at 05:00
  • This answer misses the point. The question isn't whether a large amount of dense gas will flow for a time, it is is why does this not happen spontaneously under atmospheric conditions? Why, in other words, is it not the equilibrium or normal behaviour? – matt_black Feb 02 '21 at 15:36
  • @matt_black The post has been edited but the answer could be more complete. I'l make an edit. – my2cts Feb 02 '21 at 22:33
  • Surely all the danger from radioactive radon is in the radiation (quite common where I live). I have never heard of there being any danger of suffocation. – Auspex Feb 03 '21 at 13:07
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The answers so far all seem to have overlooked the most obvious reason: heavier gasses don't generally settle to the bottom of the atmosphere because the atmosphere is continually being stirred. It's that thing called weather.

Even inside closed rooms, you have temperature gradients, people moving around, HVAC systems designed to keep the air mixed. Maybe if you could create an enclosure without such things, you could measure a concentration gradient, but I doubt that many such exist in real life.

Though such things do happen. When a heavier gas like CO2 is released in a low spot, it does take some time to mix with the rest of the atmosphere. See for instance the Lake Nyos disaster: https://en.wikipedia.org/wiki/Lake_Nyos_disaster

jamesqf
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    Stirring contributes, but is not the main reason. The static equilibrium distributions of different gases are already essentially uniform on the vertical scale of human structures. – nanoman Feb 01 '21 at 04:10
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    @nanoman: A point, though I'd put it the other way around, with stirring being the more important factor in the real world. But it should be noted that there is a real world case where gasses are separated by "gravity": the centrifugal enrichment of uranium. https://en.wikipedia.org/wiki/Zippe-type_centrifuge – jamesqf Feb 01 '21 at 17:42
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    @nanoman I'd argue that stirring is the main reason because stirring is much faster than diffusion. Even if we didn't have stirring, diffusion would do the job but it would do it very slowly. Turbulence is a fast way of mixing so, I'd argue, that makes it more important. – matt_black Feb 02 '21 at 15:41
  • @jamesqf (and matt_black) I'm speaking of steady-state statistics. It seems we're addressing three scenarios, one unphysical hypothetical (A) no diffusion, so complete stratification, and two physical hypotheticals (B) thermal equilibrium (no stirring) and (C) nonequilibrium steady state with diffusion and stirring. If we have to talk about unphysical (A), I prefer to compare it to physical rather than have another unphysical like "stirring but no diffusion" (which actually means no steady state). Now by "stirring is not the main reason" I mean that (B) is most of the way from (A) to (C). – nanoman Feb 02 '21 at 17:50
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    @nanoman: And I'd argue that B is only a small fraction of the way from A to C. Certainly casual observation seems to confirm this. Consider the example of air pollution trapped by an inversion layer. Gasses should diffuse through the interface at the same rate as anywhere else, but the diffusion is not readily apparent. A strong wind, though, will break up the inversion and disperse the pollutants. – jamesqf Feb 02 '21 at 23:28
  • That's a good observation; I think we're talking about different things. As mentioned, I'm assuming a steady state. Because your inversion scenario is not a steady state, I don't consider it an instance of (B). However, your viewpoint is appropriate if we ask why the real, unsteady atmosphere mostly remains close to a steady state. For the nature of the steady state, diffusion is essential; for the rate of approach to the steady state, stirring has a dominant effect. ... – nanoman Feb 03 '21 at 06:28
  • ... It's similar to discussing what controls energy dissipation in turbulence. From one defensible viewpoint, the dissipation is entirely due to molecular viscosity because an inviscid fluid cannot dissipate. From another, it's mainly due to the macroscopic motion because this is what sends energy cascading to be dissipated (nearly independent of viscosity at large but finite Reynolds number). – nanoman Feb 03 '21 at 06:32
  • @nanoman The atmosphere is homogeneous to much higher altitudes than what molecular diffusion steady state would explain. For most atmospheric processes molecular diffusion is completely irrelevant and the molecular diffusivity can be set to zero in the models. It just changes the Kolmogorov microscale. – Vladimir F Героям слава Feb 03 '21 at 09:50
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$\mathrm{CO}_2$ does sink in air if it's suitably pure. In particular, you can take a flask of $\mathrm{CO}_2$ and pour it out like a liquid onto something burning, like a candle, and you can watch it smother it.

So why then doesn't the ambient $\mathrm{CO}_2$ all sink? The answer is in the "suitably pure" bit above. You see, $\mathrm{CO}_2$, like any gas in a gas-gas mix, will undergo diffusion as its molecules, bumping around due to their thermal motion, manage to bump their way through and in between molecules of other gases like nitrogen and oxygen that are the dominant constituents of Earth's air. As a result, over time, a $\mathrm{CO}_2$ blanket resting on the ground will slowly expand upwards, becoming mixed with other gases, until eventually it dissolves into the rest of the air. Moreover, once the mixing dilutes it enough that its density is no longer higher than the other gases, it will then rise as expected, rapidly concluding the mixing process as turbulence forms and finishes it off.

Ambient $\mathrm{CO}_2$, then, is in effect, already in its "maximally dissolved" state, so at an approximate equilibrium (modulo factors, such as the incessant pumping of it by manmade factories and power plants, and other fires, but the significance of these depends on the timescale in question). Thus it doesn't just come out of the air.

The_Sympathizer
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    Last sentence: No, the equilibrium distribution of CO${}_2$ with height in the atmosphere is independent of the presence of "other air molecules". Those molecules were keeping the CO${}_2$ down temporarily in the case of a "blanket resting on the ground". After mixing, they are not keeping it up -- they have no effect (except a little bit due to turbulence). – nanoman Feb 01 '21 at 02:20
  • @nanoman: removed. – The_Sympathizer Feb 01 '21 at 02:21
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    Purity isn't the issue except that a large amount of a pure gas takes time to become mixed. And the explanation of diffusion is confused. – matt_black Feb 02 '21 at 15:38
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Dense gases can and do sink but two factors prevent that in the atmosphere

Carbon dioxide an other dense gases and vapours do sink. This is the cause of many industrial accidents where vessels fill with some inert gas (carbon dioxide or methylene chloride vapour have killed people entering closed vessels without care and attention, for example and volcanic lakes suddenly emitting carbon dioxide have cause major natural disasters). But this does not happen in the open atmosphere to any large extent.

There are two reasons why the atmosphere is well mixed: diffusion and turbulence.

Ideal gases (and, under normal circumstances, ideality is a good approximation for atmospheric constituents) gases are entirely miscible. One molecule doesn't know what other molecules are doing and there is no strong mechanism to separate them. Diffusion alone will, ultimately, mix the gaseous components in a vessel together thoroughly. But diffusion is slow which is why natural disasters like Lake Nyos can happen and industrial accidents can happen in closed vessels. On very large scales diffusion should compete with gravity to give a concentration gradient but this is not observed in the lower atmosphere and would never affect human-scale experiments.

The lower atmosphere sees another factor that mixes gases faster: weather. Turbulent mixing operates much faster than diffusion and is very obvious on any windy day. Turbulent mixing dominates the lower atmosphere to the extent that it is a major topic of human conversation in some countries. The atmosphere is like a very large vessel that is strongly stirred, thoroughly mixing its components. This mixing force is far stronger than gravity.

If diffusion and gravity were the only factors, we would see a concentration gradient on large scales with denser gases being less common on top of tall mountains. But we don't see that effect in the lower atmosphere because turbulence is far more important. The air composition at the top of Everest is the same as at sea level (apart from being a lot less dense).

So the intuition that dense gases should separate due to gravity is correct but that effect is dominated by diffusion and turbulence. On a planetary scale there is some separation due to gravity but this effect is small and not notable even on top of the tallest mountain. On a human scale you can beat both effects with careful experiments (like filling a vessel with sulfur hexafluoride). But only under conditions where the air is still and, even then, diffusion will eventually mix the heavy gas with the air in the room, though slowly.

matt_black
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Something that the other answers seem to be getting at, but don't explicitly state, is that there is not a sharp separation between "gasses that mix" and "gasses that don't mix". If one gas is one hundred times as dense as the other, there will very little mixture. If one is 1% denser, there will be a lot of mixture. If we could continuously vary the ratio from 1.01 to 100, the amount of mixture would continuously vary from "a lot" to "very little". And it varies exponentially, so the amount of mixture for a ratio of 2 is a lot more than twice the mixture for a ratio of 4.

Also, if we do some dimensional analysis on the sea level pressure of 101,325 N/m^2, we can substitute in N = kg m/s^2 and get 101,325 kg/(m s^2). Divide by the gravitational acceleration and we get a bit more than 10,000 kg/m^2. Divide by the density of air at STP and we get about 10 km. Multiply by the fraction of atmosphere that is CO2 and we get 4m. So very roughly speaking (I did a lot of rounding and swept some complications under the rug), there's only four meter's worth of carbon dioxide in the atmosphere. You would need nearly half of that to be at the ground for a standing adult of average height to be asphyxiated.

Acccumulation
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    "If one gas is one hundred times as dense as the other, there will very little mixture" -- incorrect. Radon is about 100 times heaver than hydrogen. In equilibrium in a 3 m tall room, hydrogen and radon mix uniformly with only ~0.3% difference in composition between floor and ceiling. In equilibrium, neither gas "knows" the other is present! Each simply develops a density profile based on its own scale height. – nanoman Feb 01 '21 at 06:25
  • It helps to realize that the forces of two molecules bouncing off each other are much larger than the gravity force on either. That's because the molecule-molecule distance in a collision is nanometer-scale. Earth's gravity is caused by atoms on average 6000 kilometers away. That's a 12 orders of magnitude difference. – MSalters Feb 01 '21 at 13:23
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    Also, the 4 meters height makes no sense. The Dutch are tall, but not that tall. There are plenty of areas in the Netherlands that are several meters below sea level, but they're not filled with CO2. And trees typically absorb their CO2 via leaves at heights well over 4 meters. – MSalters Feb 01 '21 at 13:26