Thought Experiment on the Second Law of Thermodynamics

[MikeW]

Hi,

I have a thought experiment which seems inconsistent with the 2nd Law of Thermodynamics, so there is probably a flaw in the idea somewhere.

Imagine a single molecule of a heavy gas (such as SF6) in a pipe 2.5 metres high and 100mm diametre. The pipe is upright and under the normal influence of Earth gravity (9.8N per Kg). The pipe is thermally insulated. The SF6 molecule has a room temperature (293K) velocity of 134 m/s and therefore translational kinetic energy of 134^2 * 0.5 * (146/NA/1000). If the molecule is on a path where it bounces off the bottom of the pipe and is heading for the top of of the pipe, without touching the walls, the translational kinetic energy will be reduced by gravity by 9.8 * 2.5m * (146/NA/1000) at the top. The molecule then collides with the pipe top, speeds up as it takes some energy from collision with atoms at the pipe top (on average). Then gains some speed on the way down due to gravity, then gives the extra energy (on average) to the pipe bottom atoms. My rough calculation is that this is 0.3K drop in temperature at the top of the pipe, from the bottom of the pipe, which would be measured with probe thermometers.

The photons emitted by electron energy state changes during collisions may or may not be significant.

Energy is conserved, but entropy is reduced. If there is a heat exchange (at top and bottom of pipe) connected to a convection loop, work can be extracted.

So:
1) What is the general flaw in the idea?
2) Will this be valid when bouncing off the walls of the pipe is part of the experiment?
3) Will this be valid when the pipe is full of SF6, and the macro effect of molecules colliding all the way up is included? (That is, will the speed of molecules, on average, reduce towards the top)

 

[andrewkirk]

Why do you expect the collision at the top to transfer energy from container to molecule and the opposite to happen at the bottom?

 

[MikeW]

Why do you expect the collision at the top to transfer energy from container to molecule and the opposite to happen at the bottom?

Hi Andrew, thanks for your question.

I assume when molecules/atoms collide, there is a random redistribution of energy. I assume that the system of the collision is the sum of both energies, and on average each gets half the sum. If on average the gas molecule is a little slower (at the top), then half the sum (on average) is greater than the current translational kinetic energy of the gas molecule. And at the bottom the oppersite (the gas molecule is a little faster).

 

[andrewkirk]

I assume when molecules/atoms collide, there is a random redistribution of energy. I assume that the system of the collision is the sum of both energies, and on average each gets half the sum.

It doesn't work that way. In an elastic collision both momentum and kinetic energy are conserved. If one molecule is free and the other is bound as part of a macroscopic solid, that means that almost all the energy will be retained by the free molecule, with an incredibly small bit of energy being transferred to the solid in which the other one is bound (which in the end is probably the entire Earth). That energy transfer will be so tiny that it cannot have any relevance given the unavoidable imperfections of the experimental setup, like the less than perfect heat-shielding of the side.

 

[MikeW]

It doesn't work that way. In an elastic collision both momentum and kinetic energy are conserved. If one molecule is free and the other is bound as part of a macroscopic solid, that means that almost all the energy will be retained by the free molecule, with an incredibly small bit of energy being transferred to the solid in which the other one is bound (which in the end is probably the entire Earth). That energy transfer will be so tiny that it cannot have any relevance given the unavoidable imperfections of the experimental setup, like the less than perfect heat-shielding of the side.

You are right, my two molecule surface system is a poor model, if all the atoms in container top/bottom are being considered. I assume you agree that a cold gas and warm container would affect each other; and that after a time the gas and container would reach the same temperature? That is: even with elastic collision, the gas is gaining some kinetic energy, and the container losing some. Re: "perfect heat-shielding", in the end I do not think this would be a problem, probably desired.

 

[Nugatory]

An atom moving from the top of the pipe to the bottom does gain a very small amount of energy while an atom moving upwards loses the same amount. However, that doesn't lead to a temperature difference between the top and the bottom; at equilibrium the temperature is the same throughout while the pressure is greater (very slightly, unless we're working with a very tall pipe) at the bottom.

You can't extract useful work from that configuration, just as you can't extract work from the atmospheric pressure difference between sea level and a mountaintop.

 

[MikeW]

An atom moving from the top of the pipe to the bottom does gain a very small amount of energy while an atom moving upwards loses the same amount. However, that doesn't lead to a temperature difference between the top and the bottom; at equilibrium the temperature is the same throughout while the pressure is greater (very slightly, unless we're working with a very tall pipe) at the bottom.

You can't extract useful work from that configuration, just as you can't extract work from the atmospheric pressure difference between sea level and a mountaintop.

Hi Nugatory, Thank you for your comment.

My assumption was that temperature of a gas molecule was the combination of translational kinetic energy, vibration kinetic energy and rotational kinetic energy. If the translational kinetic energy has reduced due to the deceleration of gravity and the other kinetic energies have not increase, then the temperature of the gas molecule has dropped at the top (if only a little, may around 0.3K for SF6) before it touches the top. The thought experiment starts with a single molecule. With your mountain example the temperature has dropped along with the pressure.

 

[Nugatory]

With your mountain example the temperature has dropped along with the pressure.

That's because, unlike your hypothetical pipe, the Earth's atmosphere is not thermally insulated; the upper layers are radiating heat off into space. Put the entire mountain underneath an insulating dome, eliminate the effects of stray air currents, and that temperature gradient will go away.

 

[jartsa]

Hi,

I have a thought experiment which seems inconsistent with the 2nd Law of Thermodynamics, so there is probably a flaw in the idea somewhere.

Imagine a single molecule of a heavy gas (such as SF6) in a pipe 2.5 metres high and 100mm diametre. The pipe is upright and under the normal influence of Earth gravity (9.8N per Kg). The pipe is thermally insulated. The SF6 molecule has a room temperature (293K) velocity of 134 m/s and therefore translational kinetic energy of 134^2 * 0.5 * (146/NA/1000). If the molecule is on a path where it bounces off the bottom of the pipe and is heading for the top of of the pipe, without touching the walls, the translational kinetic energy will be reduced by gravity by 9.8 * 2.5m * (146/NA/1000) at the top. The molecule then collides with the pipe top, speeds up as it takes some energy from collision with atoms at the pipe top (on average). Then gains some speed on the way down due to gravity, then gives the extra energy (on average) to the pipe bottom atoms. My rough calculation is that this is 0.3K drop in temperature at the top of the pipe, from the bottom of the pipe, which would be measured with probe thermometers.

The photons emitted by electron energy state changes during collisions may or may not be significant.

Energy is conserved, but entropy is reduced. If there is a heat exchange (at top and bottom of pipe) connected to a convection loop, work can be extracted.

So:
1) What is the general flaw in the idea?
2) Will this be valid when bouncing off the walls of the pipe is part of the experiment?
3) Will this be valid when the pipe is full of SF6, and the macro effect of molecules colliding all the way up is included? (That is, will the speed of molecules, on average, reduce towards the top)

Trillion times a second all kinetic energy is removed from a pipe molecule in a similar process as some kinetic energy is removed from the uphill climbing SF6 molecule. If you take this into account, I believe the problem will be solved.

I'm being cryptic so that you can think about it.

 

[Khashishi]

The flaw is that you assume the molecule can just travel up and down the pipe without bumping into anything. There is a pressure gradient in the pipe. This exerts a force on the molecule as it moves up and down the pipe which counteracts gravity.

 

[MikeW]

That's because, unlike your hypothetical pipe, the Earth's atmosphere is not thermally insulated; the upper layers are radiating heat off into space. Put the entire mountain underneath an insulating dome, eliminate the effects of stray air currents, and that temperature gradient will go away.

I agree that the Earth atmosphere has many factors applying to it, which is why I was concentrating on the sealed pipe example, to establish the flaws in that idea without having to to take too many things into account.

 

[MikeW]

The flaw is that you assume the molecule can just travel up and down the pipe without bumping into anything. There is a pressure gradient in the pipe. This exerts a force on the molecule as it moves up and down the pipe which counteracts gravity.

Hi Khashishi, thanks for the comment. If you look at what I wrote, the thought experiment starts with a single molecule of a heavy gas that happens to be traveling straight up the pipe. So in that scope there is nothing to bump into and no pressure gradient. I was hoping the flaws in that could be isolated before moving onto the macro scope (where bumping and pressure gradients will be applicable).

 

[MikeW]

Trillion times a second all kinetic energy is removed from a pipe molecule in a similar process as some kinetic energy is removed from the uphill climbing SF6 molecule. If you take this into account, I believe the problem will be solved.

I'm being cryptic so that you can think about it.

Hi Jartsa, thank you for your comment.

I assume that you are meaning that the collision pipe molecule has transferred excess kinetic energy to its neighbouring pipe molecules via atomic bond vibration? Sorry if I am a little slow on this, does this mean you think there will be some kinetic energy/temperature reduction on the upward route, but it will not affect the top surface because it will dissipate very quickly?

 

[MikeW]

That's because, unlike your hypothetical pipe, the Earth's atmosphere is not thermally insulated; the upper layers are radiating heat off into space. Put the entire mountain underneath an insulating dome, eliminate the effects of stray air currents, and that temperature gradient will go away.

I find this a really interesting idea. To put this in the context of my thought experiment, there is only one molecule of nitrogen in the entire Earth atmosphere and no solar radiation or solar wind. Would you expect the molecule to reach escape velocity? or would you expected it to reach a maximum altitude where it would have no velocity (no kinetic energy/temperature) and then to fall back to earth?

 

[Nugatory]

Would you expect the molecule to reach escape velocity? or would you expected it to reach a maximum altitude where it would have no velocity (no kinetic energy/temperature) and then to fall back to earth?

It depends on the initial kinetic energy (and hence speed) of the molecule. If its initial speed is greater than escape velocity the molecule will escape and if it is less the molecule will fall back, just like a cannonball.

(no kinetic energy/temperature)

This might be a good time to point out that the concept of temperature only applies to a large number of particles; it's meaningless to talk of the temperature of a single molecule. Given a volume of gas at a given temperature I can calculate an average kinetic energy of the particles in the gas, but it doesn't work the other way. There's no temperature associated with a single particle with a given kinetic energy.

 

[Nugatory]

If you look at what I wrote, the thought experiment starts with a single molecule of a heavy gas that happens to be traveling straight up the pipe. So in that scope there is nothing to bump into and no pressure gradient. I was hoping the flaws in that could be isolated before moving onto the macro scope

The kinetic energy of the single molecule is higher when it is at the bottom of the pipe than when it is at the top. We can start with one molecule at rest at the top, let it fall to the bottom and collide inelastically with the bottom and some of that energy will indeed be transferred to the bottom of the pipe, increasing its temperature. So far, so good. However (and this is likely the flaw in your thinking), at that point the system is not in equilibrium because the bottom of the pipe is hotter than the top. If we wait a while the temperature difference between top and bottom will go away as the warmer bottom of the pipe transfers heat to the cooler top of the pipe.

Now we can move onto the multi-molecule case. We imagine starting with the molecules uniformly distributed throughout the pipe so that the density is uniform and there is no pressure gradient; achieving this state is a bit tricky but it can be done in principle. Because of the gravitational field, this is not a stable configuration. The molecules are going to move downwards under the influence of gravity, losing gravitational potential energy and gaining kinetic energy as they do, and eventually the system will reach a stable configuration with a slight pressure gradient from top to bottom and a net loss of potential energy (the center of mass of the system is lower because the density is no longer uniform).

The net downwards movement of the molecules will heat the bottom of the pipe, just as in the single molecules case. But also as in the single molecule case, the system is not in thermal equilibrium until the temperature equalizes throughout the entire volume. When it does, you end up with gas and pipe at a constant temperature slightly higher (because we did turn some potential energy into kinetic energy and then heat) than when we started.

(There's a simpler argument: You have a closed and thermally isolated system with neither heat sources nor heat sinks. Such a system cannot have a temperature gradient at equilibrium).

 

[jartsa]

Hi Jartsa, thank you for your comment.

I assume that you are meaning that the collision pipe molecule has transferred excess kinetic energy to its neighbouring pipe molecules via atomic bond vibration? Sorry if I am a little slow on this, does this mean you think there will be some kinetic energy/temperature reduction on the upward route, but it will not affect the top surface because it will dissipate very quickly?

Not quite. I mean the collisions are random.

Let's consider your scenario, but without gravity: It's quite simple: Gas molecule bounces around randomly, moving sometimes slowly, sometimes fast.

Now with gravity: Same as above, plus a downwards acceleration of the gas molecule.

And taking into account that randomness will make everything sensible. I promise.
So let me try it:

Let's say the gas molecule loses energy at the floor, then it loses energy at the ceiling, then it again loses energy at the floor, then it can't reach the ceiling. This is possible, so it happens sometimes.

But this is impossible: The gas molecule loses energy at the floor, then it loses energy at the ceiling, then it can't reach the floor.

So, if a gas molecule has low kinetic energy, it will most likely touch the floor next.

 

[MikeW]

It depends on the initial kinetic energy (and hence speed) of the molecule. If its initial speed is greater than escape velocity the molecule will escape and if it is less the molecule will fall back, just like a cannonball.

This might be a good time to point out that the concept of temperature only applies to a large number of particles; it's meaningless to talk of the temperature of a single molecule. Given a volume of gas at a given temperature I can calculate an average kinetic energy of the particles in the gas, but it doesn't work the other way. There's no temperature associated with a single particle with a given kinetic energy.

Fair comment; I will use 'kinetic energy' for the single molecule case, instead of temperature. I do not think it changes the concept of the single molecule thought experiment though.

 

[MikeW]

then it loses energy at the ceiling,

I think it would gain kinetic energy at the ceiling, do you think it would lose kinetic energy?

 

[MikeW]

However (and this is likely the flaw in your thinking), at that point the system is not in equilibrium because the bottom of the pipe is hotter than the top. If we wait a while the temperature difference between top and bottom will go away as the warmer bottom of the pipe transfers heat to the cooler top of the pipe.

I assume the pipe walls are the means for the warmer bottom to transfer to the top of the pipe, is there an another means? The thought experiment does not require pipe walls, the pipe was used to bring it back to a real world experiment. If you look at it in the single molecule of nitrogen/planet example and having the top at such an altitude that the molecule has lost half its vertical velocity (relative to the planet) by the time the top surface is reached (would it on average, gain an increase in kinetic energy as it bounced off the top surface?), this is the equivalent of the pipe without walls.

If we wait a while the temperature difference between top and bottom will go away as the warmer bottom of the pipe transfers heat to the cooler top of the pipe.

I agree this would happen at a given time rate in the pipe; but meanwhile the single molecule is bouncing up and down repeating the process of increasing kinetic energy at the bottom and reducing kinetic energy at the top. I think the rate of temperature transference in the pipe wall would affect the ability to observe something, rather than there would not be a kinetic energy difference.

Sorry for staying on just the single molecule case, but I would like to find the flaw in that case, as I have a thought process for scaling it up to the macro, that would be redundant if there is a flaw in the single molecule assumptions.

 

[Nugatory]

I but meanwhile the single molecule is bouncing up and down repeating the process of increasing kinetic energy at the bottom and reducing kinetic energy at the top.

Assume for the moment that the molecule is initially at rest when it's dropped from the top of the pipe (this assumption simplifies the analysis - we'll eliminate it in a moment). On each bounce, the molecule transfers a bit of its kinetic energy to the bottom of the container to heat it, so won't rebound to the same height as on the previous bounce and will have less energy for the next bounce. Because we dropped the molecule from the top, and it won't reach that height on the subsequent bounce, the molecule never bounces off the top of the pipe - all of its initial potential energy ends up as heat added to the bottom and there is no interaction with the top.

This is not a process that we can use to transfer energy from the top of the pipe to the bottom. We had this one molecule, it had some amount of potential energy, we turned that potential energy into the equivalent amount of heat energy at the bottom, and we're done. The top of the pipe didn't interact with the molecule, so it clearly didn't lose any energy in the process. A convection loop between the top and bottom cannot extract any more work that the initial potential energy of the molecule because that's all the energy there is to extract.

Now suppose that the molecule starts with some non-zero speed and kinetic energy. It will be moving faster when it hits the bottom so can leave some energy there as heat, yet still rebound with enough kinetic energy to reach the top. It slows as it climbs against gravity but still has some speed and kinetic energy when it reaches the top, so it bounces and starts back down again. But when it bounces, it again deposits some of its kinetic energy as heat so now we're heating the top, and the heads back down with less speed and kinetic energy than it had on the way up. Eventually all the kinetic and potential energy is transferred to the pipe, and just as before, that's all the energy we had. The only difference is that some of that energy was used to heat the top as well; there's still no transfer from top to bottom.

For most reasonable assumptions about the initial speed of the molecule and the fraction of energy transferred to the end of the pipe with each bounce, we will end up with the bottom receiving slightly more heat than the top (it's a good exercise to calculate this: parameters are initial height and speed of the molecule, and the fraction of the kinetic energy converted to heat on each bounce). However, that's not an ongoing process from which we can extract work, and it's not a transfer of energy from the top to the bottom. It's just a system that started in a non-equilibrium state.

 

[MikeW]

Assume for the moment that the molecule is initially at rest when it's dropped from the top of the pipe (this assumption simplifies the analysis - we'll eliminate it in a moment). On each bounce, the molecule transfers a bit of its kinetic energy to the bottom of the container to heat it, so won't rebound to the same height as on the previous bounce and will have less energy for the next bounce. Because we dropped the molecule from the top, and it won't reach that height on the subsequent bounce, the molecule never bounces off the top of the pipe - all of its initial potential energy ends up as heat added to the bottom and there is no interaction with the top.

This is not a process that we can use to transfer energy from the top of the pipe to the bottom. We had this one molecule, it had some amount of potential energy, we turned that potential energy into the equivalent amount of heat energy at the bottom, and we're done. The top of the pipe didn't interact with the molecule, so it clearly didn't lose any energy in the process. A convection loop between the top and bottom cannot extract any more work that the initial potential energy of the molecule because that's all the energy there is to extract.

I agree, in this situation, which is different to the one I am describing, there will be no interaction with the top, PROVIDING the bottom is near 0 K so it cannot initially add energy to the gas molecule. The molecule in your set-up would have velocity around 3.5m/s at bottom for a 2.5m pipe, which is extremely low.

It slows as it climbs against gravity but still has some speed and kinetic energy when it reaches the top, so it bounces and starts back down again. But when it bounces, it again deposits some of its kinetic energy as heat so now we're heating the top

I think we are nearing the critical part in this; which maybe my misunderstanding of the interaction with the top. I believe that at the top, the gas molecule will GAIN kinetic energy on average, because gravity has reduced its upward velocity component (extreme single molecule air example: 330m/s at bottom, top surface at height where gravity has slowed molecule to 20m/s, touches top, gains energy from 293K top surface [assuming sealed dome, no temperature gradient]). Do you not think a relatively warmer surface, would on average increase the kinetic energy of the gas molecule?

Thank you for your patience in dealing with my questions.

 

[jartsa]

Let's consider a group of very energetic molecules with temperature T and a Maxwellian distribution of speeds.

What happens if we subtract a very small energy E from each molecule, assuming no subtraction produces a negative energy? The distribution becomes a very slightly non-Maxwellian one, and the temperature drops very slightly.

When we successfully subtract the same energy from each molecule, we increase order in the group of molecules, or decrease disorder. That is not forbidden, except in physics when the number of molecules is very large and disorder does not increase somewhere else.

EDIT: The above paragraph is wrong, some extra order is required before the subtraction, so that we can successfully subtract the same energy. So the order does not necessarily increase.

Note that the larger the number of molecules, the more unjustified the assumption that no energy becomes negative.

EDIT: This relevant, because a molecule climbing a distance h loses energy $m*g*h$, which is constant if m and g and h are constants.

 

[jartsa]

I think it would gain kinetic energy at the ceiling, do you think it would lose kinetic energy?

Yes, the molecule would gain kinetic energy at the ceiling - in your story about the molecule.

While in my story the molecule gained kinetic energy at the floor - at the end of the story when the molecule was out of energy near the floor. I guess I left the story just a little bit unfinished.

So now then the whole story: There are times when heat goes from the floor to the ceiling, and there are times when heat goes from the ceiling to the floor. In other words random fluctuations in the floor-ceiling system are occurring, and the molecule is serving as a conductor of heat.

 

[MikeW]

I agree, in this situation, which is different to the one I am describing, there will be no interaction with the top, PROVIDING the bottom is near 0 K so it cannot initially add energy to the gas molecule. The molecule in your set-up would have velocity around 3.5m/s at bottom for a 2.5m pipe, which is extremely low

Correction: 7m/s at bottom for a 2.5m pipe

 

[MikeW]

So now then the whole story: There are times when heat goes from the floor to the ceiling, and there are times when heat goes from the ceiling to the floor. In other words random fluctuations in the floor-ceiling system are occurring, and the molecule is serving as a conductor of heat.

I agree, for shorter lengths of pipe that, due to the distribution of speeds for a gas molecule, sometimes the single gas molecule will heat the top. But if you look at the probability distribution of speeds, it is no longer a Maxwell–Boltzmann distribution at top and bottom. If I have a pipe 1 metre long, with a gas molecule at the top (heading down) with Maxwell–Boltzmann speed distribution, the distribution when it reaches the bottom (before touching surface) cannot be a Maxwell–Boltzmann speed distribution because the probability of the gas molecule having velocity less than 4.42m/s is zero. If it is a 1000 metres long pipe, the probability of the gas molecule having velocity less than 140m/s is zero. The peak in the probability speed distribution would go the other way for the upward gas molecule. So while there is randomness, the speed probability distribution is different bottom and top (for single gas molecule, before touching surface). After touching surface, the speed distribution maybe a Maxwell–Boltzmann distribution, but the energy difference (before to after) has gone to/come from the surface.

 

[jartsa]

If it is a 1000 metres long pipe, the probability of the gas molecule having velocity less than 140m/s is zero. The peak in the probability speed distribution would go the other way for the upward gas molecule. So while there is randomness, the speed probability distribution is different bottom and top (for single gas molecule, before touching surface). After touching surface, the speed distribution maybe a Maxwell–Boltzmann distribution, but the energy difference (before to after) has gone to/come from the surface.

Yes those special molecules heat the bottom. A molecule that jumps million meters up will indeed heat the floor when landing back on it.

Now let me ask: Where did a molecule that has enough energy to jump 1000 km upwards most likely get that energy?

 

[MikeW]

Yes those special molecules heat the bottom. A molecule that jumps million meters up will indeed heat the floor when landing back on it.

The example I gave was 1000 metres NOT 1000 Kilometres. I agree you would need a very high temperature in the bottom surface to obtain the velocity (for gas the molecule) needed to reach a 10^6 metre height.

Now let me ask: Where did a molecule that has enough energy to jump 1000 km upwards most likely get that energy?

(1000km or 1000m?)
INITIALLY the gas molecule may (depending on initial velocity) get energy when it bounces off the bottom surface to an upward direction. Gravity then reduces the kinetic energy on the upward path. The gas molecule then gets some more energy (on average) when it bounces off the top surface to a downward direction. Gravity increases the kinetic energy on the downward path.
After that, the gas molecule would deposit some energy to the bottom surface (on average), and would receive some energy (on average) from the top surface.

 

[jartsa]

Take a bunch of molecules with Maxwell-Bolzman distribution. Remove every molecule whose energy is lower than E, then reduce every remaining molecule's energy by E, the result of those operations is a bunch of molecules with the same distribution as originally, and the same temperature as originally.

Just do the maths if you doubt that. (I personally have not done the maths)

Now what does a gravity well do to molecules bouncing around in the well? It allows molecules with energy higher than E to escape, and reduces those molecule's energy by E.

 

[Ron Maimon]

This is a great thought experiment, and the thing you are neglecting is that not all the time does the atom reach the top. There is a small percentage of collisions at the bottom where the atom doesn't get enough energy to reach the top at all, it simply falls back down. These cases are those with next to zero kinetic energy in the atom, these are the "coldest cases", and you have to condition the energy transfer calculation on the atom reaching the top at all, which means throwing away the coldest fraction with the smallest kinetic energy. Once you do the conditioning, the distribution of speeds at the top is the same as at the bottom, despite this being surprising.

 

[PeterDonis]

This is what I think the distributions will be for the gas molecule arriving at the top and bottom surfaces.

Have you actually calculated this? Or are you just waving your hands?

 

[MikeW]

Have you actually calculated this? Or are you just waving your hands?

I used the Maxwell–Boltzmann distribution function. I adjusted the probabilities for the upward and downward case due to gravity (+/- 140m/s for 1000m). As no gas molecule traveling upward with velocity less than 140m/s can reach the top, the probabilty for this was scaled to keep the area under the graph line to 1. I hope I have implemented it correctly.

 

[PeterDonis]

I adjusted the probabilities for the upward and downward case due to gravity (+/- 140m/s for 1000m).

In other words, you're assuming that a group of molecules that start out with an M-B distribution at some altitude, and then rise or fall under gravity, will still have an M-B distribution at the new altitude, with the peak shifted. Can you justify that assumption? (I see at least two problems with it.)

 

[MikeW]

I used the Maxwell–Boltzmann distribution function. I adjusted the probabilities for the upward and downward case due to gravity (+/- 140m/s for 1000m). As no gas molecule traveling upward with velocity less than 140m/s can reach the top, the probabilty for this was scaled to keep the area under the graph line to 1. I hope I have implemented it correctly.

Sorry, there is an error in the graph to do with the spread, forgot to do the energy adjust for initial speed (ie reduction/gain less than 140m/s as intital velocities increase). Will change and repost. Will affect, spread, not general idea.

 

[MikeW]

In other words, you're assuming that a group of molecules that start out with an M-B distribution at some altitude, and then rise or fall under gravity, will still have an M-B distribution at the new altitude, with the peak shifted. Can you justify that assumption? (I see at least two problems with it.)

Hi Peter, thank you for your comment.

As initially stated: Not a group of molecules, a single SF6 molecule.

Not at some altitude:
a) Initial M-B Distribution assumed at bottom for upward vector, reduces velocity due to gravity, different distribution at top BEFORE touching top surface.
a) Initial M-B Distribution assumed at top for doward vector, increases velocity due to gravity, different distribution at bottom BEFORE touching bottom surface.