Warm air goes up....reason on a microscopic scale?

[GTrax]

On a practical level, I think Wrichik has a point. Please allow that I cannot manage a theoretical approach at the level of most in this company.

I was wondering about the microscopic reason warm air rises up, while cold air comes down. I am aware of the macroscopic reason - density changes. But what happens microscopically? Decrease in density means that the gas molecules are widely spaced out, but their mass remains the same. Then why does warm air go up?

The curious near equivalent is lower density gas, but not by reason of being warmer.
For me, I was impressed by the direct experience with Helium used in a vacuum furnace leak detection where a turbo-molecular pump feeding a mass spectrometer device used at the at a vacuum port when all is well pumped down. A small amount of helium is leaked from a wand probe into the air surrounding the vacuum chamber, and if there is even the smallest leak, the instrument squeals and displays the order of the leak.

Of course, if Helium is confined to a balloon, it readily displaces enough air for it to go straight on upwards by buoyancy. The situation seems completely different for letting the gas loose into the air. The amount was tiny, not even enough to partly inflate a party balloon, about 2 seconds of tiny puff through 3mm pipe into the room. I was struck by the speed the helium found it's way through the leak. Seemingly less than half a second to mix with the room air, through the leak several metres away, and into the instrument which howled immediately!

Getting curious, I then found that releasing a little Helium in the next door industrial unit (with connecting aperture) was similarly speedy.
After 4 short-lived "experiments", we found the unit would no longer properly detect leaks because of all the Helium hanging around, and that our helium releases had contaminated most of the space right out to beyond the car park outside. It took more than 20 minutes to disperse! Clearly it does quickly not "go upward" as like in a balloon, so maybe amid the mean free path movement, all there is to have it make it's way upward is the gravity effect of the heavier molecules around it going the other way. One wonders if a mix of Hydrogen and Helium in a tall jar would have Hydrogen end up at the top?

I get it that Wrichik was not talking about Helium, but instead warm air. The significant thing is that the warm air is only different from the other air by the fact it's molecules are moving faster. It still does all the mean free path mixing action. Does warm air "hang together" in some way to gain buoyancy from lower density, sort of like if it was in a ill-defined balloon? Here is where I lose it. I don't yet really see the real mechanism of an air convection current. I know it happens, because I have experienced how three quarters of a ton of sailplane can be strongly shoved upwards from a heat source (combine harvester) hundreds of feet below. In passing, I would just love to have some gadget capable of displaying such a rising air current!

 

[sophiecentaur]

Does warm air "hang together" in some way

The only way it can 'get away' is by diffusion. That is a much longer term effect than the macroscopic buoyancy effect, which is why clouds and thermals happen. Conduction in gases (the spreading of the heat) must be due to a combination diffusion of actual molecules and momentum transfer at the hot/cold boundary. The momentum would transfer faster than individual molecules. I wouldn't mind betting that experiments with isotopes or gases with similar molecular masses have been done to compare the two effects.

 

[Demystifier]

One additional remark. Once the less dense object (be it helium balloon or hot gas) goes up, it doesn't later fall down. Therefore the process is irreversible. On the other hand, the microscopic laws are reversible. This implies that the irreversible statistical law of entropy increase (which is valid only when a large number of degrees of freedom plays an important role) must take place in some form.

 

[sophiecentaur]

One additional remark. Once the less dense object (be it helium balloon or hot gas) goes up, it doesn't later fall down. Therefore the process is irreversible. On the other hand, the microscopic laws are reversible. This implies that the irreversible statistical law of entropy increase (which is valid only when a large number of degrees of freedom plays an important role) must take place in some form.

This has been my point. Diffusion will always be at work and, given time, the helium in a room would spread all round the volume. This process is a lot slower than the thermal conduction / convection process (hours or days). I think the Dalton Law of partial pressures would apply and you will arrive at a different pressure distribution for all the gases. The proportion of He molecules at the top would be greater than the proportion at the bottom. The ideal condition would require still air and no temperature gradients.

 

[Demystifier]

A new attempt, inspired by my remark above. In general, objects like to minimize their potential energy. Why? Contrary to a widespread opinion, this is not because force ${\bf F}=-\nabla V$ acts in the direction in which potential energy decreases. The Newton law $m\ddot{\bf x}={\bf F}$ is a reversible law, so by Newton law alone the potential energy may decrease as well as increase. The true reason why objects like to minimize their potential energy is dissipation. Objects like to give all its energy (both kinetic and potential) to the environment because environment has more degrees of freedom than the object, so if the initial energy of the object gets distributed among the environment degrees, then the total entropy gets larger.

Now consider a vessel containing two gases, one with a lower density than the other. As we said above, the system wants to minimize its potential energy. Clearly, the configuration in which the gas with a lower density is up and gas with a larger density is down has a lower potential energy than the opposite configuration. But how does entropy increase in this process? The entropy increases due to interaction between the two gasses (for simplicity, we assume that the gasses do not mix), which takes place at the contact surface where the two gasses touch each other. And where does the initial potential energy go? It transforms into a heat, that is into an increase of temperature of both gasses due to a friction at the contact surface.

Finally, what if we have superfluids which do not have any friction? My theory above has an interesting testable prediction. If the theory is correct, then a superfluid with lower density immersed in a superfluid with larger density should be able to move both up and down. We can have a circulation of two fluids which never slows down. Can some experimentalist verify this prediction?

 

[DrDu]

You can't really have hot or cold molecules.

I do not quite agree on this.
Statistical mechanics is applicable to finite systems and you don't need large sample asymptotics to justify properties like temperature.
Hence I would assign the temperature of a container wherein the molecule is moving an with which it is in equilibrium to be a property also of the molecule.

 

[jbriggs444]

I do not quite agree on this.
Statistical mechanics is applicable to finite systems and you don't need large sample asymptotics to justify properties like temperature.
Hence I would assign the temperature of a container wherein the molecule is moving an with which it is in equilibrium to be a property also of the molecule.

"Equilibrium" with the container would indicate that the molecule within the container has a constant relationship with the container. The only way that can happen is if the molecule is at rest with respect to the container. This would lead to the conclusion that only molecules at rest with respect to the container can have a temperature and that their temperature is always zero. Surely this is not what you had in mind?

 

[DrDu]

"Equilibrium" with the container would indicate that the molecule within the container has a constant relationship with the container. The only way that can happen is if the molecule is at rest with respect to the container. This would lead to the conclusion that only molecules at rest with respect to the container can have a temperature and that their temperature is always zero. Surely this is not what you had in mind?

I don't mean mechanical but thermodynamical equilibrium.

 

[jbriggs444]

I don't mean mechanical but thermodynamical equilibrium.

Does that not beg the question? What do you mean by thermodynamic equilibrium for a single molecule within a container?

If one has a container at 100K and a molecule with a kinetic energy equivalent to 200K, is that molecule in thermodynamic equilibrium with the container? Is there enough information to tell?

 

[sophiecentaur]

As we said above, the system wants to minimize its potential energy.

That just means the lighter gas will end up in more concentration near the top and less at the bottom. But thermodynamics will ensure that the gases mix and have the same average KE in all locations, eventually.
Superfluids don't seem to follow the rules because it's a quantum effect. There are plenty of apparent examples of perpetual motion - fountains etc. so your suggestion would probably be reasonable. But it is not really relevant to the OP and subsequent discussion.

 

[Paul C]

As an atom of air (more energetic, i.e., warmer than its neighbors) rises it gives up energy as heat and over time stabilizes at an dynamic equilibrium. Atomic vibration is in and of itself not a qualification for motion in any particular direction, inasmuch as "rebounding" from more concentrated, to wit, "cooler" surroundings acts to increase the movement of atoms relative to one another. BTW, marvelous explanation https://www.physicsforums.com/threa...-microscopic-scale.937606/page-2#post-5927266 by Demystifier of a statistical rationale.

 

[MRBlizzard]

Try this:
1. What is the mean time between scattering / collisions for slightly warmer or cooler atoms?
2. How much is the difference between the height fallen (between collisions) between slightly warmer or cooler atoms?

How many times a second does the cooler atom drop a little bit. Multiply by the answer to #2.

Of course, there is a lot of scattering up and down. But conceptually, this is why cold atoms gather at the bottom, and warm atoms are at the top.

 

[DrDu]

Does that not beg the question? What do you mean by thermodynamic equilibrium for a single molecule within a container?

If one has a container at 100K and a molecule with a kinetic energy equivalent to 200K, is that molecule in thermodynamic equilibrium with the container? Is there enough information to tell?

For a single atom in a container of a given temperature, if we repeat this experiment many times, we would find a certain distribution of the energy of the atoms. I.e., our information about the atom will be described by a canonical density matrix with a certain temperature.

 

[256bits]

canonical density matrix

Isn't that what a thermometer measures with just one experiment using many atoms as a continuum.
An doing the whole calculation in one shot.

 

[DrDu]

Isn't that what a thermometer measures with just one experiment using many atoms as a continuum.
An doing the whole calculation in one shot.

Of course, but the question was whether it is admissible to use the concept of temperature also for a single atom.

 

[jbriggs444]

For a single atom in a container of a given temperature, if we repeat this experiment many times, we would find a certain distribution of the energy of the atoms. I.e., our information about the atom will be described by a canonical density matrix with a certain temperature.

So you are now saying that a single atom at a single time does not have a temperature.

That is: You can have a temperature for a bunch of atoms at one time. Or a temperature for a single atom over a long time. But you can't have a temperature for a single atom at a single time.

I would further remark that if you define the temperature of an atom in terms of its long term average behavior in the context of a container then you do not need the atom. Just measure the temperature of the container.

 

[DrDu]

I am not saying that a single atom at a single time has no temperature. If my knowledge is so that it is in a canonical mixed state, the state is basically defined by its temperature and vice versa.

I would further remark that if you define the temperature of an atom in terms of its long term average behavior in the context of a container then you do not need the atom. Just measure the temperature of the container.
[\QUOTE]
That's basically the zeroth law of thermodynamics.

 

[jbriggs444]

I am not saying that a single atom at a single time has no temperature. If my knowledge is so that it is in a canonical mixed state, the state is basically defined by its temperature and vice versa.

What operational test can you propose to determine whether an atom at a particular time is or is not in a canonical mixed state?

That hypothetical atom with 200K worth of kinetic energy in a 100K container -- how do you tell what its temperature is? If you can propose no test at a point in time then I maintain that it has no temperature at a point in time.

 

[DrDu]

It is an ensemble property. It is the same as when I say that a dice has a probability of 1/6 to show a one before throwing it one time.

 

[jbriggs444]

It is an ensemble property. It is the same as when I say that a dice has a probability of 1/6 to show a one before throwing it one time.

So again, you are agreeing that a single molecule at a single time does not have a measurable temperature.

You have posed the definition of temperature so that the molecule only has a temperature if it is in thermodynamic equilibrium with its container. And you have clarified the definition so that it is a measure of a statistical distribution of states the molecule might find itself in. Under those conditions, the "temperature of a molecule" is not really an attribute of the molecule at all. It is, if it exists at all, an attribute of the container.

 

[DrDu]

It is good practice to speak of e.g. hot and cold neutrons. I really don't see a fundamental difference to the temperature concept of atoms and molecules.

 

[jbriggs444]

It is good practice to speak of e.g. hot and cold neutrons. I really don't see a fundamental difference to the temperature concept of atoms and molecules.

Let me try to argue from a different direction and possibly find some common ground.

Suppose that we stick a temperature probe into a small volume of gaseous nitrogen. We obtain a measured result of 100K. Strictly speaking, this is a statistical measurement. We only sampled from a small volume of nitrogen in a large container. So there is some statistical uncertainty in our result. Making up numbers, a result of 100K might be consistent with thermodynamic temperatures between 99.999K and 100.001K at a 90% confidence level.

Of course, there are mundane experimental uncertainties as well. Our instruments may only be good to the plus or minus 0.01K. In this situation we would be comfortable ignoring the statistical uncertainty associated with sampling error. It is down in the noise. One can talk about the sampled temperature as if it were identical to the actual temperature to within the mundane experimental uncertainty.

Now we attempt to probe the temperature by measuring the kinetic energy of a single molecule. We obtain a measured result that corresponds to a temperature of 200K. At the 90% confidence level, what is the range of temperatures that the container might have?

Eyeballing the Stephan Boltzman distribution, it looks like the 90% confidence interval will cover a range from approximately half the mean to approximately double the mean. Roughly speaking then, at the 90% confidence level, a sampled molecule at a kinetic energy of 200K is consistent with temperatures between 100K and 400K.

So I can agree that a single molecule [or neutron] does have a temperature. But that temperature is only measurable to within plus or minus a factor of two at the 90% confidence level.

 

[Rap]

I think it is helpful to look at how much a particle "comes down" between collisions (on average). The amount a particle "comes down" is (1/2)g t^2 where g is the acceleration due to gravity and t is the time between collisions. If t is greater for lower temperature, that would explain it.

t=L/v where L is the distance between molecules and v is the velocity of the particle. L is proportional to n^(1/3) where n is density. v is velocity of the particle, proportional to T^(1/2) where T is the temperature (because T is proportional to average energy which is (1/2)mv^2). Using the ideal gas law, P=nRT where P is pressure, R is the gas constant. Pressure is practically constant (pressure differences are equalized at the speed of sound and I'm assuming the column is not high enough to produce noticeable pressure differences) so we can say that n is proportional to 1/T. So t is proportional to n^(1/3)/T^(1/2) or 1/T^(5/6) and the amount the particles "come down" is proportional to 1/T^(5/3). The higher the temperature, the less they "come down" between collisions, and the low-temperature low-velocity particles wind up near the bottom.

I don't mean to imply a particle has a temperature, so please append "on average" to any statement that is deemed offensive.

 

[SirCurmudgeon]

Interesting (at least to me) wet air (air saturated with water vapor) is lighter that dry air.
Wet air goes up...

 

[NFuller]

Instead of a kinetic approach, I would try with a statistical approach. In general, a particle has a kinetic energy (due to motion) and a potential energy (due to gravitational force acting on the particle). In statistical physics there is an equipartition theorem, which, loosely speaking, says that energy likes to be distributed equally in all forms. So if particle has a lot of kinetic energy (corresponding to high temperature), then it also likes to have a lot of potential energy (and hence likes to go up). The only question is - why do particles like to distribute energy equally? That's because it maximizes entropy. When there is a lot of particles, then there is more phase space available for equidistributed energy than for all energy distributed in one form only.

I think this is a pretty good answer except that the equipartition theorem only applies at equilibrium. The difficulty in explaining why hot air rises at the microscopic level, is that this system is not at equilibrium. The temperature is not uniform through the gas and not all of the particles have the same energy. Eventually the system will reach equilibrium, at which point we can apply equilibrium statistical mechanics, but this after the warm gas has risen and come to the same temperature of the cool gas. It may be possible to develop a kinetic model for the motions of the warm and cool gas but I doubt it would be easy.

 

[sophiecentaur]

Interesting (at least to me) wet air (air saturated with water vapor) is lighter that dry air.
Wet air goes up...

Yes. The standard description and explanation really doesn't help. Air is described as a sort of SPONGE which, 'of course', would get heavier when full of water. But air is a mixture of gases. H₂O molecules are much lighter than N₂ and O₂ molecules so a volume of 'wet air' will contain more light molecules than a volume of 'dry air' (Same pressure and temperature) so it will float in air.

 

[Brendan Graham]

It's satisfactory.

What happens to the gas which is cool? It needs less room. That's why it comes down?

Assuming the base of the enclosing vessel is fixed, both hot and cool gas molecules under an external gravitational field have a fixed floor beyond which they cannot travel. Rather than coming down, more appropriately, molecules of a cooler gas just do not reach up (against the force of gravity) as far as those that are hotter therefore the probability distribution of finding cooler gas molecules near the floor is greater.

 

[Demystifier]

I think this is a pretty good answer except that the equipartition theorem only applies at equilibrium. The difficulty in explaining why hot air rises at the microscopic level, is that this system is not at equilibrium. The temperature is not uniform through the gas and not all of the particles have the same energy. Eventually the system will reach equilibrium, at which point we can apply equilibrium statistical mechanics, but this after the warm gas has risen and come to the same temperature of the cool gas. It may be possible to develop a kinetic model for the motions of the warm and cool gas but I doubt it would be easy.

How about my post #40, where I further elaborated my explanation? You are of course right that equipartition theorem is only valid in the equilibrium, but my explanation is based on the idea that the system evolves towards the equilibrium. In other words, the hot gas goes up because that makes the system closer to the equilibrium.

 

[Ilythiiri]

In passing, I would just love to have some gadget capable of displaying such a rising air current!

http://www.instructables.com/id/Schlieren-Imaging-How-to-see-air-flow/

Here you go (:

 

[Dr. Courtney]

I'm not sure buoyancy has a microscopic explanation.

 

[Zaya Bell]

You need to first understand that gravitation acts on molecules, and they have pressure. For hotter object, the spaces between them increases, thus reducing pressure, while the cold molecules' spaces decreases and pressure increases. Thus if the cold molecules were up they would apply more pressure downward and displace the hotter molecules upward. It's analogous to putting your hand into a water container. The higher pressure (F/A) of your hand displaces the lower pressure of the water molecules and your hand now occupies the space the water did.
Got it?

 

[jbriggs444]

gravitation acts on molecules, and they have pressure.

Individual molecules do not have temperature or pressure. Groups of molecules have temperature and pressure. If one invokes the emergent properties of temperature and pressure to describe convection, one has abandoned the microscopic explanation and gone for a macroscopic explanation.

Not that there is anything wrong with that.

 

[Zaya Bell]

Individual molecules do not have temperature or pressure. Groups of molecules have temperature and pressure. If one invokes the emergent properties of temperature and pressure to describe convection, one has abandoned the microscopic explanation and gone for a macroscopic explanation.

Not that there is anything wrong with that.

They sure do. They are not point particles which means they have area. Thus P=mg/A. Even a photon has pressure.

 

[jbriggs444]

They sure do. They are not point particles which means they have area. Thus P=mg/A. Even a photon has pressure.

A single photon does not have pressure either. And a single molecule does not have force. Or a meaningful momentum flux.

 

[Zaya Bell]

A single photon does not have pressure either. And a single molecule does not have force. Or a meaningful momentum flux.

A single photon does not have pressure either. And a single molecule does not have force. Or a meaningful momentum flux.

I don't think you know what you're saying. Check this up anyway.
https://www.photonworld.de/nc/en/magazine/featurearticle/the-power-of-photons/fseite/1.html