Thought Experiment on the Second Law of Thermodynamics

[PeterDonis]

Not a group of molecules, a single SF6 molecule.

A single molecule doesn't have a velocity distribution function, which is what the M-B distribution is supposed to model in this connection.

Initial M-B Distribution assumed at bottom for upward vector

Distribution of what? There's only one molecule. It doesn't have a velocity distribution. See above.

If what you are trying to model is a probability distribution for a single molecule having a given velocity, why do you think a M-B distribution is the right one?

looses velocity due to gravity

As others have pointed out, you are assuming that the molecule has no collisions as it rises. But your claims about temperature only have meaning to begin with for a large number of molecules, not a single molecule, in which case the single molecule you are focusing on will have many collisions as it rises. Your analysis does not take that into account.

different distribution at top BEFORE touching top surface.

"Different distribution", yes. But that does not just mean "a M-B distribution with a different peak" (even if we assume your initial assumption of a M-B distribution at the bottom is correct, which it might not be, and that the molecule has no collisions as it rises, which it will--see above).

Write the M-B distribution for a system with a single degree of freedom as follows (where I have assumed the temperature $T$ is in energy units, i.e., Boltzmann's constant $k$ is 1):

$$
f(E) = C \frac{E}{T^{2/3}} \exp \left( - \frac{E}{T} \right)
$$

where $C$ is a combination of constant factors that I won't bother to write down explicitly. Your assumption is that at the bottom, there is some $T = T_b$ such that the distribution satisfies this equation, i.e., we have

$$
f_b(E) = C \frac{E}{T_b^{2/3}} \exp \left( - \frac{E}{T_b} \right)
$$

Now we shift $E$ by a constant amount, the potential energy difference $\Delta U$ from the bottom to the top. Your claim is that the resulting distribution will also satisfy the above equation at some lower temperature $T_t$. But that can't possibly be the case, because of the factor $E / T^{2/3}$ in front of the exponential. That is, in order for your claim to be true, $T_t$ would have to satisfy two mutually inconsistent conditions: it would have to be equal to $T_b - \Delta U$ (to satisfy the argument of the exponential) but also equal to $\left( T_b - \Delta U \right)^{2/3}$ (to satisfy the factor in front of the exponential). That is impossible, so your claim must be false. A similar argument applies when going from top to bottom.

 

[MikeW]

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.

Hi Ron, thanks for your comment. Have thought about it quite a bit.

I agree that for coldest cases, there is no interaction with the top, I think this is the same point Jartsa is making in comment #29. I think when considering energy transfer, the change in speed distribution (top to bottom or bottom to top) is important; as the cold case gas molecules have no interaction with top, they cannot change the energy balance.

(graph version 2, might change if I spot more errors)

This is an attempt at what I think the distributions will be for the gas molecule arriving at the top and bottom surfaces, after leaving the other surface with M-B Distribution. I have used 1000m pipe to show separate distributions for clarity, I think it will be similar for shorter pipes, just less spread. You can see that the typical speed for the gas molecule arriving at the top is lower than the typical speed for the general M-B Distribution. You can also see that the typical speed for the gas molecule arriving at the bottom is higher than the typical speed for the general M-B Distribution. I think this would mean that at the top, energy would be transferred from the top surface to the gas molecule (on average). And at the bottom, energy would be transfer from the gas molecule to bottom surface (on average).

I think even with elasticity (the speed distribution after bounce being clustered around the speed before bounce), the energy transfers as described, would hold; only with a slower rate of energy transfer.

 

[MikeW]

A single molecule doesn't have a velocity distribution function, which is what the M-B distribution is supposed to model in this connection

I think a single gas molecule can have a velocity distribution over time, if interacting with a container.

even if we assume your initial assumption of a M-B distribution at the bottom is correct

I do not know the precise distribution when velocity of gas molecule is affect only by container, not other gas molecules, but M-B seemed a good starting point.

"Different distribution", yes. But that does not just mean "a M-B distribution with a different peak" (even if we assume your initial assumption of a M-B distribution at the bottom is correct, which it might not be, and that the molecule has no collisions as it rises, which it will--see above).

No collisions, please read entire thread to see the context in this.

But your claims about temperature only have meaning to begin with for a large number of molecules

Which claim on temperature? Do you mean the temperature of the bottom surface or temperature of bottom surface?

I did not want to bring distributions into this, but other commenters asserted that the distribution for the SINGLE gas molecule leaving the bottom with M-B Distribution upon arriving at the top (without touching) would have the SAME M-B Distribution. I disagreed.

It might help you if you read the whole thread, and tried to understand this is a thought experiment on a single molecule.

 

[mfb]

Something is still wrong with the distribution. Did you take into account that even high-energetic molecules might not have enough energy to reach the top?
If you want to consider this as 1-dimensional problem, you cannot take Maxwell-Boltzmann out of the box, the 1-dimensional speed distribution looks different.

I think a single gas molecule can have a velocity distribution over time, if interacting with a container.

That still doesn't give it a temperature.

 

[MikeW]

Something is still wrong with the distribution. Did you take into account that even high-energetic molecules might not have enough energy to reach the top?

Hi MTB, I think this has being address in answering Ron's comment

That still doesn't give it a temperature.

The temperature used in the formula is that of the surface that the gas molecule interacts with

 

[mfb]

Hi MTB, I think this has being address in answering Ron's comment

Well, something is wrong for sure. If you plot the 3D speed, the yellow curve should start at 0, if you plot the 1D-speed, the blue curve should not start at 0.

 

[PeterDonis]

I think a single gas molecule can have a velocity distribution over time, if interacting with a container.

Its velocity can change with time, but that's not what "velocity distribution" means, and you certainly can't assume that a single molecule's velocity over time will look like a M-B distribution. A single molecule's velocity over time will obey precise equations of motion, and the M-B distribution formula is not a solution of those equations. A "velocity distribution" is only used in the first place when you have so many molecules that you can't track each one individually and you have to use statistical methods.

No collisions, please read entire thread to see the context in this.

I have read the entire thread. My comments (not just the one about collisions) are based on the entire thread.

Which claim on temperature?

The one you made in the OP:

My rough calculation is that this is 0.3K drop in temperature at the top of the pipe, from the bottom of the pipe

That's what sparked all of this discussion in the first place--everybody else in this thread except you believes that in equilibrium there will be no temperature difference between the top and bottom of the pipe.

It might help you if you read the whole thread, and tried to understand this is a thought experiment on a single molecule.

It might help if you read all of the posts in this thread explaining why the concepts you are trying to use don't even apply to a single molecule, and why, if you insist on just looking at a single molecule, your claim is obviously false.

The argument in your OP, if you insist on taking it as an argument about a single molecule, can be reduced to the claim that work can be continuously extracted from a ball that inelastically bounces at the top and bottom of a container. It should be obvious that this claim is false: you can extract a finite amount of work, but only a finite amount, which will be less than the energy you originally put into the ball to start it bouncing, in accordance with the second law.

 

[MikeW]

Well, something is wrong for sure. If you plot the 3D speed, the yellow curve should start at 0, if you plot the 1D-speed, the blue curve should not start at 0.

The 'arriving at top' line does not start at zero probability because the gas molecule is slowed on the way up, this changes the distribution. The distributions might not be precisely right, the bigger picture of this is if there is ANY consistent shift in typical velocity (what ever the initial distribution), then there is an energy transfer (on average) to/from the surface to the gas molecule. That is my point.

 

[PeterDonis]

Thread closed for moderation.

Edit: the thread will remain closed, there is just no sensible way to talk about the temperature of a single molecule.