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The common explanation for why hot air rises is a macroscopic one. That is that the hot air is higher pressure, lower density, and therefore gravity affects it differently than the surrounding cold air. However, at a microscopic level, things are more confusing. Since, looking at a single hot molecule or cold molecule at the same altitude, the gravitational force is the same for each of them. My best guess to what is happening is that there's a very small density gradient where there is ever so slightly less dense molecules above the hot molecule compared to the cold molecule. This presumably is because of gravity (so it's gonna be very very small affect). Collisions will happen more frequently at the bottom than the top, therefore the gas molecules tend to rise. My intuition is the scale of this gradient would be way too small to explain why hot air rises, but can't think of any other reason?

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  • $\begingroup$ I believe this is a repeat of physics.stackexchange.com/questions/6329/… $\endgroup$ Jun 6 at 4:24
  • $\begingroup$ @SteveSaban , it is not a repeat. The explanation in that thread is only the macroscopic view. I'm looking for the microscopic view. $\endgroup$ Jun 6 at 4:47
  • $\begingroup$ Molecules scatter against each other, and the diffusion path is much smaller than the typical volume of air that we talk about. We thus can talk about layers of air pressung the adjacent layers. All this can be derived as a rigorous expansion in kinetic equation (hydrodynamic approximation) - see the answer to the question linked above. $\endgroup$ Jun 6 at 5:31

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You are getting a macroscopic explanation because this is a macroscopic effect.

Microscopically, air molecules move in random directions with approximately the speed of sound. (For mean thermal kinetic energy, use $\frac32 kT\approx\frac12 mv^2$.) If you gave a single projectile this speed, and neglected air resistance, it would arc up to an altitude of about ten kilometers. But the rule of thumb is that you can’t neglect air resistance if the mass of displaced air is comparable to the mass of the projectile, which means we absolutely can’t neglect interactions with the air when we are imaging the motion of air molecules. Instead, $10^4$ meters ends up being a good estimate for the scale height of the atmosphere.

The microscopic picture of the atmosphere is useful for distances shorter than or comparable to the mean free path of an air molecule, which is about 70 nm.

Suppose I light, then blow out, a match, so that air heated by the match carries wisps of smoke upwards at a few centimeters per second. An air molecule that has risen seven centimeters, a handspan, has moved about $10^6$ mean free paths, and has undergone tens or hundreds of millions of collisions with its neighbors in the second or so that it takes for the smoke to waft upwards. Each cubic millimeter of air contains something like $10^{16}$ air molecules. The centimeters-per-second drift of a wisp of hot smoke is a teeny, tiny correction to the speed-of-sound motion of the air molecules which carry it. The macroscopic number of collisions allow you to use statistics to make predictions about them in aggregate.

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  • $\begingroup$ From the link in this question: diffusion is slow. For a gas molecule in air to diffuse one meter away from its starting point takes about a day. Macroscopic convection is overwhelmingly more important for gas transport than microscopic diffusion. $\endgroup$
    – rob
    Jun 6 at 5:41
  • $\begingroup$ I guess my disconnect is I don't see the justification to treat hot air as a single entity. It seems inappropriate. The hot molecules aren't connected in some way. All I see is that there is a gravitational force for each individual particle. $\endgroup$ Jun 6 at 5:53
  • $\begingroup$ Gravitational and collisional forces. If the distance between two randomly-chosen air molecules changes by less than one millimeter per second (a reasonable estimate for random-walk diffusion), I think that “connected” may be a useful approximation. $\endgroup$
    – rob
    Jun 6 at 14:16
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One liter of hot air contains fewer air molecules than a liter of cold air, so it weighs less and will therefore rise if surrounded by colder air.

The reason the hot liter contains fewer molecules is that those molecules are moving randomly at higher speed, and are driven further apart by collisions between those molecules than in the case of cold air. Thus hot air is less dense than cold air.

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  • $\begingroup$ But this is what I refer to as the macroscopic explanation. You take the 1 liter of hot air as a single entity. But I'm interested in why an individual hot air molecule rises preferentially. You can no longer say that the hot molecule weights more than a cold molecule because they weigh the same. $\endgroup$ Jun 6 at 4:00
  • $\begingroup$ See Rob's explanation above. $\endgroup$ Jun 6 at 16:43
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There is a mismatch in the form that you are stating your question.

To explain the nature of that mismatch, let me discuss a particular classroom demonstration.

Take two drinking glasses, put them on top of each other, the openings facing each other, with a thin sheet acting as separator. (The thin sheet can for example be a piece of paper)

When the separator is removed the contents of the top and bottom glass are free to move vertically.

Let the top glass be filled with cold air, and let the bottom glass be filled with hot air. Let temperature probes be present in each glass.

In this case: When the separator is removed the time scale of macroscopic motion will be shorter than the time scale of molecular mixing. Sure, over time the hot air and cold air will mix, but, that takes tens of seconds. But in only a few seconds the stack of cold air on top of hot air will topple over. On a time scale of seconds: for the most part: heavier cold air will displace lighter hot air.


But we can change the setup to bring those two time scales together. Instead of drinking glasses we can set up two vacuum chambers. (But this being a thought demonstration we can pretend that we can remove a separator without loss of vacuum seal.)

In both chambers, make the density of air very low.
If in both chambers the density is very low then we can still have the same temperature difference (since temperature is defined as the average kinetic energy of the molecules), but there will be much weaker gravitational effect.

Under those circumstances the previous distinction between time scale of falling and time scale of mixing loses relevancy. Are the slower molecules falling or mixing? There is no meaningful distinction.


My point is:
as long as the circumstances are such that the time scale of gravitational effect is way shorter than the time scale of mixing:
Then the appropriate view is to understand the physics taking place from the macroscopic point of view.

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  • $\begingroup$ In principle, the microscopic perspective is the correct perspective because there is no reason for us to think of hot air as a single entity. In essence, I'm asking why a single hot air molecule preferentially rises as apposed to simply diffuse in all directions. $\endgroup$ Jun 6 at 4:45
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The system that you are describing has density and temperature gradients that favor the collective motion of heated air molecules upward. The density of air decreases with altitude creating a longer mean free path for hot air molecules traveling upward, the temperature of the atmosphere decreases with altitude creating thermodynamic conditions that favor the flow of heat upwards. So the microscopic motion of individual air molecules is influenced by collective behavior that is best described macroscopically.

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