It is said that the constant interaction between the molecules of a gas (in the form of collisions) acts as a randomising influence and prevents the gas molecules from settling. But given the force of gravity, won't the gas molecules settle at some point of time?

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    $\begingroup$ A mixture may stratify to some extent, depending on temperature. Is that what you mean? $\endgroup$
    – Jon Custer
    Apr 17, 2017 at 16:23
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    $\begingroup$ They do - it's called solidification. $\endgroup$
    – JMLCarter
    Apr 17, 2017 at 16:53
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    $\begingroup$ @JMLCarter its called condensation (gas -> liquid) or sublimation (gas -> solid). Solidification is liquid -> solid. $\endgroup$ Apr 18, 2017 at 12:49
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    $\begingroup$ @jml, yes, none takes it up because it refers to a different thing; phase change happens because of changes in the system's conditions (temperature etc), while the OP asks purely about the effect of gravity. $\endgroup$
    – Helen
    Apr 19, 2017 at 15:50
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    $\begingroup$ @Mindwin Last time I checked gas->solid phase transition was called deposition or desublimation, and sublimation was the opposite $\endgroup$
    – Sebi
    Apr 19, 2017 at 16:45

7 Answers 7


You are used to all collisions being somewhat lossy – that is, when you think of most collisions, a little bit of the kinetic energy is lost at each collision so the particles will slow down. If they are subject to gravity, they will eventually settle.

By contrast, the collisions between gas molecules are perfectly elastic – for a non-reactive gas (mixture), there is no mechanism by which the sum of kinetic energies after the collision is less than before. *

Even if an individual gas molecule briefly found itself at rest against the bottom of the container, the thermal motion of the molecules of the container would almost immediately give it a "kick" and put it back into circulation.

There is a theorem called the equipartition theorem that tells us that for each degree of freedom, the gas molecules will on average contain $\frac12 kT$ of energy. This is an average – individual molecules may at times have more or less. But the average must be maintained – and this means the gas molecules keep moving.

One way you can get the molecules of the gas to settle at the bottom of the container would be to make the walls of the container very cold – taking thermal energy away, the molecules will eventually move so slowly that the effect of gravity (and intermolecular forces) will dominate. That won't happen by itself – you need to remove the energy somehow.

To estimate the temperature you would need: for a container that is 10 cm tall, the gravitational potential energy difference of a nitrogen molecule is $mgh = 1.67\cdot 10^{-27}\ \mathrm{kg}\cdot 28 \cdot 9.8\ \mathrm{m\ s^{-2}}\cdot 0.1\ \mathrm m= 4.6\cdot 10^{-26}~\mathrm{J}$. Putting that equal to $\frac12 kT$ gives us a temperature of

$$T = \frac{2 m g h}{k} = 3.3\ \mathrm{mK}$$

That's millikelvin. So yes – when things get very, very cold, gravity becomes a significant factor and air molecules may settle near the bottom of your container.

* Strictly speaking, this is a simplification. With sufficient energy, some collisions can lead to electronic excitation and even ionization of the molecules. The de-excitation of these states can result in radiative "loss" of energy, but if the system is truly closed (perfectly isolated) the radiation will stay inside until it's re-absorbed. Still, this means that, at least for a little while, kinetic energy may appear to be "lost". Similarly, there are some vibrational modes for molecules that get excited at sufficiently high energy/temperature; in these modes, energy moves from "kinetic" to "potential" and back again – so that it is not "kinetic" for a little bit of the time.

An important consideration in all this is "what is the temperature of anything that the gas can exchange energy with". That is not just the walls of the container (although their temperature is very important), but also the temperature of anything the gas "sees" – since every substance at non-zero temperature will be a black body emitter (some more efficiently than others), if the gas can exchange radiation with a cooler region, this will provide a mechanism for the gas to cool. And if the gas gets cold enough, gravity wins.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Apr 24, 2017 at 16:49
  • $\begingroup$ "collisions between gas molecules are perfectly elastic" - no they're not, as the particles interact EM waves are emitted caused by the acceleration of the charged particles. As per the Larmor formula. There's no requirement for there to be a significant amount of energy between the collisions in this. $\endgroup$
    – UKMonkey
    Jun 14, 2017 at 15:30
  • $\begingroup$ To support the conclusion, a typical centrifuge can get you 10,000 g acceleration but even then it would still be rendered insignificant above temperatures of 33K. If you could get the gases that cold in the first place, you would best be separating them by boiling points (fractional distillation) instead of by their molecular weights! $\endgroup$
    – gregsan
    Jan 13, 2018 at 6:34

In fact, particles in a box of gas are slightly denser at the bottom than they are at the top. In general, the probability of finding a particle with a total energy of $E$ is proportional to the Boltzmann factor: $$ P(E) \propto e^{-E/kT}. $$ In particular, the potential energy of a gas molecule is $mgh$, where $h$ is the height above some fixed point (the "floor" of the box, say.) If we consider the relatively probabilities of a particle to be found at the floor of a box ($h = 0$) versus being found at a height $h$ above the bottom of a box, we will have $$ \frac{P(h)}{P(0)} = \frac{e^{-mgh/kT}}{e^{0}} = e^{-mgh/kT}. $$ Thus, the densities of the gas molecules is lower at a height $h$ than it is near the floor, since they are less likely to be found at these heights.

The problem is that this factor is tiny for typical temperatures and masses of gas molecules. For the air in my office, we have $m \approx 32$ amu (the mass of an oxygen molecule, $h \approx 3$ m (the height from floor to ceiling), and $T \approx 293$ K. Plugging these all in, we get that the density of the air at the ceiling is 99.961% of the density at the floor.


Based on the various comments, I think it's worth noting that even molecules in a solid don't "settle" — they too are constantly in rapid motion.

The difference between a solid and a gas is that, in the solid, each molecule is confined to a small volume through interactions with nearby molecules.

When a solid appears to "settle", what's really happening is that all of the obvious macroscopic motion simply gets converted into making the individual molecules vibrate faster within their confines. (with some energy lost due to transferring the momentum to the surrounding air particles)


Classical physics.

The probability of finding a molecule at some height $h$ is proportional to the density $\rho(h)$, which can be calculated. A column of air from floor to ceiling weighs $\rho gh A$, where $\rho$ is the average density, $g$ is acceleration due to gravity =$9.8 m/sec^2$, $h$ is the height of the room $=3m$ and $A$ is the arbitrary area of the column. This weight is supported by a force $A dp$, where $dp$ is the pressure difference floor to ceiling, and for an ideal gas at constant temperature $T=293K$ this is als0 $RTd\rho$. Here $R$ is the gas constant $=287 J/kg^{-1}K^{-1}$ for air. So equating the force to the weight gives$$\frac{d\rho}{\rho}=\frac{3\times 9.8}{287\times293}=3.5\times 10^{-4}$$ and this gives the ratio of densities at ceiling and floor as 0.9996, just the same as the solution by Michael Seifert. Isnt physics wonderful?

How does this answer the question? The pressure exerted by a gas (on itself as well as on a wall) is caused by the random jostling of molecules. Loosely, it is the kinetic energy of the molecules in unit volume. There are slightly fewer molecules at the ceiling, but they are jostling just as vigorously because they are at the same temperature (and temperature = jostling per molecule). The diminishing pressure and the diminishing density just balance out. You can extend this argument to find the pressure and density all the way up through the atmosphere, although there is more to be accounted for then than when just considering the air in a room.

EXPANSION OF ANSWER There have been some good answers to this question, and some..not quite so good. Consider that the definition of a gas is that of a substance that will expand to fill any container that it is placed in. If the container is not closed, it will (slowly perhaps) leak out. Take some time to consider if this is a definition that you can agree with. It would follow from this definition that a gas will expand to fill any container you put it in, whether or not another gas is already there. It will just occupy the spaces

Now imagine that a container of sodium hexafluoride is placed in the middle of the floor and the stopper is removed. Within a few minutes almost all of the $SF_6$ will have replaced by air. In fact the percentages will be same in the container and in the room. Think carefully whether you accept this.

Now look to see which answers can explain this, not forgetting what the definition of a gas is.

Last point. if you cool a gas sufficiently it becomes a liquid. At this temperature the molecules no longer behave independently but form transient bonds, and the result is a liquid that has a definite volume and settles at the bottom of any container. Cool it some more and the bonds become permanent although not quite rigid. Then we call it a solid.


Gas molecules dont settle because they have too much kinetic energy. As time goes on, that energy (heat) dissipates, and the molecules do 'settle.' When this occurs they become a liquid or a solid. When gas 'settles' as you describe, it is no longer a gas. However, its worth noting that dense gases do 'settle' beneath less dense gases in much the way you describe, although they are still gases and the molecules do take up volume they would not in solid or liquid form.

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    $\begingroup$ What do you mean by a dense gas? Heavier molecules? In that case they just move slower. All molecules in a mixture have the same kinetic energy at a given temperature. $\endgroup$
    – Philip Roe
    Apr 17, 2017 at 22:35
  • $\begingroup$ @PhilipRoe yes I meant that it has heavier molecules, as with Sulfur Hexafluoride compared to oxygen, or oxygen compared to helium and so forth. $\endgroup$ Apr 18, 2017 at 20:05
  • $\begingroup$ Then see my expanded answer $\endgroup$
    – Philip Roe
    Apr 19, 2017 at 15:15
  • $\begingroup$ @PhilipRoe Unless I am missreading your expansion, it is wrong. You can in fact open a container with sulfur hexafluoride to the air and it will not be displaced by air. $\endgroup$ Apr 19, 2017 at 21:00
  • $\begingroup$ Do you agree that a gas, by definition, expands to fill any available volume? $\endgroup$
    – Philip Roe
    Apr 20, 2017 at 13:47

Yes, without a heat source the molecules in a gas will cool and settle in a gravitational field.

Collisions between molecules are only elastic insofar as they conserve overall energy. Energy can still be transferred between translational kinetic energy and the vibrational and rotational energy levels of molecules, or if the gas is hot enough, into electronic transitions in atoms. These excitations can then de-excite radiatively and that radiation can escape from the gas.

However a gas in thermal equilibrium will not settle (beyond the small settling effect correctly identified by Michael Seifert) because the energy lost by radiation will be balanced by input from whatever container the gas is held in, in the form of collisions with the walls and absorption of radiation from the walls.

If you have no container - for example a molecular cloud in outer space - then the settling process can take place. Molecular clouds can reach temperatures as low as the microwave background, but even at this temperature, may have sufficient internal energies and turbulence to support themselves against total collapse. Others don't and that's how stars form

  • $\begingroup$ Say if I have a box of gaseous molecules and I leave it on a table on Earth. Won't gravity somehow accelerate the molecules towards it and increase the energy of the molecules in my box? Or is it that no change in the energy of the gaseous molecules in my box is observed because as many molecules are accelerated as are decelerated. (Sorry if this sounds stupid) $\endgroup$ Apr 22, 2017 at 17:09
  • $\begingroup$ @KunalPawar That effect is covered by Michael Seifert's answer. Energy is conserved. Compare the KE of a molecule with mgh. Unless they are comparable, the effect is small. $\endgroup$
    – ProfRob
    Apr 22, 2017 at 17:28
  • $\begingroup$ @Rob. Answers to this question are muddled because of ambiguity in the original question of the word settle. Does it mean stratify? In that case the answer is that they do, by reaching hydrostatic equilibrium. Clearly some people have taken it to mean pool like a liquid, and I think that is the intent of the original question. The behavior of a self-gravitating gas cloud is not relevant here. Nor is loss by radiation under collisions because in thermal equilibrium this is exactly compensated by energy from outside the system. The extensive theory of gas dynamics ignores this completely $\endgroup$
    – Philip Roe
    Apr 23, 2017 at 16:52
  • $\begingroup$ @PhilipRoe The answer to this question needs to make exactly those statements or caveats; not assume them as some sort of obvious truth. Real systems contain non-ideal gases in non-closed systems. The question is indeed somewhat loose. It does not say that the gas is ideal; in thermal equilibrium ; or isothermal for example. $\endgroup$
    – ProfRob
    Apr 23, 2017 at 17:36
  • $\begingroup$ @RobJeffries. To preserve our sanity we usually omit types of energy that are of no interest (like rest mass, for example). For most (real!) situations we do not mislead ourselves by thinking of the collisions as elastic. $\endgroup$
    – Philip Roe
    Apr 24, 2017 at 2:03

Let's assume the collisions are elastic. The average influence of gravity on the momenta of the particles is zero in the vertical direction of motion of the molecules. The particles' vertical components of their momenta are changed by the same amount (on the average), but opposite in direction, so gravity doesn't influence the vertical moment composed of all the particles' vertical momenta.

The horizontal components of the particles' momenta are all changed into a downward direction by gravity (between collisions). So gravity enhances the momenta of the molecules (because the vertical change of all the particles' momenta is zero on the average). So the momentum of each molecule is increased (let's not consider the equipartition theorem), which means the gas is getting warmer, wich is no surprise if gravity is acting on it: the horizontal velocity of an object moving in horizontal direction, with zero velocity in the vertical direction (let's not forget the changes in vertical momenta in the gas caused by gravity are on the average cancelled out) is always changed downwards by gravity, giving the object a higher momentum, and thus $E=\frac {p^2} 2m$.

So in the ideal condition that the collisions are elastic, the gas is actually being heated up by the influence of gravity. Off course the collisions are not elastic, so at each collision, the molecules lose energy, and this loss is bigger than the increased energy of the molecules by gravity. We can conclude that the loss in energy due to collisions between the gas particles is equal or bigger than the gain in energy by gravitation. This would happen if the sun weren't there to constantly heat the gas of the atmosphere, which isn't actually the case so the atmosphere is constantly heated by the sun. With the result that the atmospheric gas won't settle.

  • $\begingroup$ " the collisions are not elastic" ... there are a lot of answers here, but it would be nice to see an answer expand upon the elastic vs inelastic thing. If it wasn't elastic, where would the energy go? It probably has to be elastic because exciting an electronic transition (or even more extreme, a nuclear excitation) is too much energy compared to the average kinetic energy. It looks like the only option would be some collision energy could go into vibrational energy of a molecule. Maybe the key point is the energy would not be dissipated and would still be available in further collisions. $\endgroup$
    – PPenguin
    Apr 18, 2017 at 17:04
  • $\begingroup$ @PPenguin: We know that the molecules dont fall to the floor, despite having had billions of years to do so. So they dont lose energy. They do not all have equal amounts of energy, but when they collide their energy is just shared differently, never lost. Our best way of relating this to familiar events that we have words for is to say that they experience perfectly elastic collisions. Energy can go into the vibrational mode if the gas is hot enough, but a molecule that picks up vibrational enenergy in one collision can lose it in the next. $\endgroup$
    – Philip Roe
    Apr 18, 2017 at 19:12
  • $\begingroup$ @PhilipRoe Makes sense, and is also why I am uncomfortable with this answer claiming "Of course the collisions are not elastic, so at each collision, the molecules lose energy". $\endgroup$
    – PPenguin
    Apr 18, 2017 at 20:47
  • $\begingroup$ Imagine all the gas molecules are in a container somewhere in space and the gas has a certain temperature. Don't you think the container with the gas will cool down (you can see it as a black body)? And so will the atmosphere of the earth if the sun weren't there to heat it up every day. The collisions between molecules maybe don't excite electrons inside the molecule, but at the moment of collision, there are accelerations of charges. $\endgroup$ Apr 19, 2017 at 7:25
  • $\begingroup$ So what does this container look like? If its not really well insulated, the gas will first liquefy and then freeze. Where are you trying to go? $\endgroup$
    – Philip Roe
    Apr 19, 2017 at 15:13

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