Why wouldn't the temperatures be different in this scenario

  • Thread starter Thread starter PaperProphet
  • Start date Start date
AI Thread Summary
In a vertical tube filled with gas, gravity affects the motion of gas molecules, causing them to gain energy when moving downward and lose energy when moving upward. Despite this, a temperature difference between the top and bottom of the tube does not occur because the system reaches thermal equilibrium, where no net heat flow exists. The principles of hydrostatic equilibrium dictate that pressure and density vary with height, but temperature remains constant in a stable system. The second law of thermodynamics prevents spontaneous temperature gradients from forming without external energy input. Therefore, while gravity influences molecular motion, it does not lead to a temperature difference in the gas-filled tube.
PaperProphet
Messages
18
Reaction score
0
Imagine a vertical tube filled with a gas. As the gas molecules move to the bottom with gravity, they gain energy and speed. As they randomly move upward against gravity, the molecules lose energy and speed. It seems like that would create a difference in temperature but obviously if the top of the tube was actually cooler than the bottom, it means a spontaneous, potentially useful, difference in energy was created so that can't be.

Why wouldn't the molecules in the tube create a difference in temperature between the top and the bottom of the tube?
 
Science news on Phys.org
Think about densities.
 
Bystander said:
Think about densities.

Thanks. Explanation?
 
Can you apply some thought/logic to Bystander's suggestion? We try to make people exercise their brains here...
 
russ_watters said:
Can you apply some thought/logic to Bystander's suggestion? We try to make people exercise their brains here...
I was actually hoping Bystander could expand upon his own suggestion.

I have a good physics background but I'll admit I don't have the answer to my question. I'm hoping someone can provide a cogent, clear, understandable explanation that even someone as dumb as myself can understand. Logically the molecules can't create a difference in temperature otherwise that would violate the laws of thermodynamics...but I don't know why they don't.
 
Consider all the conditions for equilibrium.
 
Bystander said:
Consider all the conditions for equilibrium.
Bystander, I assume you're trying to be helpful but that you don't know the answer either.

No pretenses here--I don't know the answer despite pondering it in depth. This isn't a homework question where I'm looking for hints--it's something I randomly thought about years ago and just can't figure out. I've given up and am hoping someone else can give a clear, unambiguous explanation. If you have (or anyone else has) an explanation, please share. Understandably, this isn't an easy question and you can see nobody else has come up with an answer either.
 
PaperProphet said:
As they randomly move upward against gravity, the molecules lose energy and speed.
Why do you assume their total energy is constant?
 
A.T. said:
Why do you assume their total energy is constant?
I'm not sure I follow.

It just seems that molecules going upward inside the tube would lose kinetic energy due to gravity and be 'colder' at the top of the tube while gaining kinetic energy due to gravity to be 'warmer' at the bottom of the tube. It would seem, in my mind, that molecules would continuously transfer heat energy from the top of the tube to the bottom of the tube. Obviously that can't be the case otherwise that would violate the second law of thermodynamics by spontaneously creating a heat source and a heat sink. It's a paradox that I can't solve and I'm hoping someone has a good explanation.
 
  • #10
A little background, I asked this question about the gas-filled tube maybe fifteen years ago on a message board and nobody knew the answer then either. It doesn't appear to be an easy question.
 
  • #11
PaperProphet said:
It just seems that...
That is not a physical argument. Check the assumptions you are making.
 
  • #12
A.T. said:
That is not a physical argument. Check the assumptions you are making.
Thank you for the hints, A.T.

Can I ask if you know the answer as to why there isn't a difference in temperature?
 
  • #13
This is just a reminder to everyone to keep the discussion focused on the original question, not on who is or isn't being respectful to whom. I have deleted some off topic posts accordingly.
 
  • #14
PaperProphet said:
Can I ask if you know the answer as to why there isn't a difference in temperature?

I suspect he does, but that's not really relevant. You say you have a "good physics background", but that's a general statement. Do you know the equations governing the equilibrium of a gas in a vertical tube in a gravitational field? There are at least two equations that are relevant, involving the temperature, density, and pressure. Have you studied any material of this type?
 
  • #15
PeterDonis said:
I suspect he does, but that's not really relevant. You say you have a "good physics background", but that's a general statement. Do you know the equations governing the equilibrium of a gas in a vertical tube in a gravitational field? There are at least two equations that are relevant, involving the temperature, density, and pressure. Have you studied any material of this type?

Thanks Peter. The reason I asked is because I would like to know whether the person giving me hints has the answer or not. I don't mind discussing the topic since I'm obviously very interested in the topic...but I just want to make sure that someone trying to 'mentor' me has more answers than I do before I re-explore the angles I've already explored.

I certainly understand that gas pressure increases under gravity but I don't know any formulae which indicate if or how temperature would spontaneously change near the top and bottom of a gas-filled tube. In fact, the second law of thermodynamics dictates that it shouldn't. Still, molecules are affected by gravity and a molecule moving downward with gravity should gain additional kinetic energy and be 'hotter'.

Do you have any thoughts on whether or not molecules at the top of a gas filled tube would be cooler than molecules near the bottom?
 
  • #16
PaperProphet said:
Do you have any thoughts on whether or not molecules at the top of a gas filled tube would be cooler than molecules near the bottom?

This isn't a matter of "thoughts", it's a matter of physics, and the answer can be worked out from physical principles. That is what the other posters in this thread have been trying to get you to do.

Here are the key physical principles involved. We are assuming that the system is in equilibrium, i.e., nothing is changing with time.

(1) There must be no net heat flow anywhere in the gas. That implies, as you have assumed, that there cannot be any variation of temperature anywhere in the tube; the entire tube must be at the same temperature. The question then is, how does this relate to the effects of gravity? So the next thing we need to look at is, what are the effects of gravity?

(2) The effect of gravity is that the gas inside the tube must be in hydrostatic equilibrium. That means, if we consider a small parcel of gas at a given height ##h## in the tube, with a small thickness ##dh##, there must be a pressure difference from the bottom to the top of the parcel, and this pressure difference must be just sufficient to balance the weight of the gas above the small parcel we are looking at. If we work out what this means, taking the limit as the thickness ##dh## goes to zero, we get the equation

$$
\frac{dP}{dh} = - \rho g
$$

where ##dp / dh## is the rate of change of pressure with height, ##\rho## is the density of the gas, and ##g## is the acceleration due to gravity (we assume this is constant everywhere in the tube). The fact that ##dp / dh## is negative means that pressure decreases with height.

(3) Also, it is important to realize that the pressure, density, and temperature of the gas are not independent; they are related by an equation of state. For this case, we can assume that the equation of state is that of an ideal gas, which is

$$
P = \rho R T
$$

where ##P## is the pressure, ##\rho## is the density, ##R## is a physical constant called the "gas constant", which will depend on the chemical constitution of the gas, and ##T## is the temperature.

If we assume that ##T## is constant, the above equations tell you that ##P## and ##\rho## will vary with height in the tube. Can you see how they will vary? And can you see how that explains how the temperature can be constant in the presence of gravity?
 
  • #17
PeterDonis said:
If we assume that TTT is constant, the above equations tell you that PPP and ρρ\rho will vary with height in the tube. Can you see how they will vary? And can you see how that explains how the temperature can be constant in the presence of gravity?
Peter, I can see you believe you've answered my question. I can see you've shown in detail how pressure changes with height given a constant temperature. Keep in mind that your equations don't exclude the possibility that temperature can vary from the top to the bottom. In fact, you could put a heater on the bottom of the hypothetical tube and a similar set of equations will solve for the change in pressure. In other words, you can assume T varies and nothing would break...other than the equation which assumes T is static.

I know I'm not entitled to an answer and I'm not trying to be mean, but I'm specifically asking why the temperature *wouldn't* change due to the effect of gravity pulling on the molecules.
 
  • #18
There is a non-zero thermal conductivity; there is zero heat flow at equilibrium. Therefore, zero temperature gradient, QED.
 
  • Like
Likes gracy
  • #19
PaperProphet said:
molecule moving downward with gravity should gain additional kinetic energy
Why?
 
  • #20
PaperProphet said:
I can see you believe you've answered my question.

Then you see incorrectly, because I didn't quite answer your question. I only stated the key physical principles that lead to the answer. I specifically left out part of the reasoning, in the hope that you would work it out for yourself.

PaperProphet said:
your equations don't exclude the possibility that temperature can vary from the top to the bottom. In fact, you could put a heater on the bottom of the hypothetical tube and a similar set of equations will solve for the change in pressure.

Obviously if there is a heater present, then gravity is not the only effect. So saying that the equations allow ##T## to vary if a heater is present does not mean ##T## will vary if a heater is not present.

One of the assumptions I made actually requires that there is no heater. (One of the things I left out was not explicitly pointing out this fact.) Can you see which one?

PaperProphet said:
I'm specifically asking why the temperature *wouldn't* change due to the effect of gravity pulling on the molecules.

You should be able to work that out for yourself with the information you've already been given, just as you worked out the fact that the information I gave you was not quite a complete answer.
 
  • #21
PaperProphet said:
Imagine a vertical tube filled with a gas. As the gas molecules move to the bottom with gravity, they gain energy and speed. As they randomly move upward against gravity, the molecules lose energy and speed. It seems like that would create a difference in temperature but obviously if the top of the tube was actually cooler than the bottom, it means a spontaneous, potentially useful, difference in energy was created so that can't be.

Why wouldn't the molecules in the tube create a difference in temperature between the top and the bottom of the tube?

The molecules in the tube create a difference in temperature between the top and the bottom of the tube, by falling down and heating up. That seems quite clear to me, particularly if I think about a gas with low density.

When there is difference in temperature between the top and the bottom of the tube we can conduct the heat back up, I don't see a problem there. The bottom of the tube loses the extra heat, and the the top of the tube also gets cooler when we conduct heat away from the bottom. The tube cools when we conduct heat away from it, when it has cooled enough we can not conduct heat away from it anymore.
 
  • #22
Peter, PV=nRT (and variations) don't constrain temperature. I won't be able to work out any answer based on what you've given me. If you complete your line of reasoning with the change in pressure, I'm confident you'll realize the mistake in your reasoning. I know you genuinely believe you've given me tools to get an answer and I thank you for trying.
 
  • #23
jartsa said:
The molecules in the tube create a difference in temperature between the top and the bottom of the tube, by falling down and heating up. That seems quite clear to me, particularly if I think about a gas with low density.

When there is difference in temperature between the top and the bottom of the tube we can conduct the heat back up, I don't see a problem there. The bottom of the tube loses the extra heat, and the the top of the tube also gets cooler when we conduct heat away from the bottom. The tube cools when we conduct heat away from it, when it has cooled enough we can not conduct heat away from it anymore.
Thanks, Jartsa!

You're the first person here who was able to see (or at least acknowledge) the effect of gravity on the molecules. However keep in mind that if the effect was really there, there would be a potentially useful heat source and heat sink implying free energy. Obviously that can't be the case. It seems like a paradox.
 
  • #24
jartsa said:
The bottom of the tube loses the extra heat, and the the top of the tube also gets cooler when we conduct heat away from the bottom.

In other words, in your model, there is continuous heat flow through the tube, and heat exchange between the tube and the outside environment. My understanding of the OP was that there should be no heat flow or heat exchange--that the tube is to be considered as an isolated system in thermal equilibrium with no heat exchange with the outside.
 
  • #25
PaperProphet said:
PV=nRT (and variations) don't constrain temperature.

Not by themselves, no. But that wasn't the only assumption I made. Look at item #1 in post #16. What does it say? And does what it says match the scenario you were envisioning? (Also see my previous post in response to jartsa.)
 
  • #26
PaperProphet said:
You're the first person here who was able to see (or at least acknowledge) the effect of gravity on the molecules.

You are mistaken. Everyone in this thread has acknowledged the effect of gravity on the molecules. The equation for hydrostatic equilibrium that I gave you explicitly includes the effect of gravity on the molecules. If there were no gravity, there would be no pressure gradient.
 
  • Like
Likes russ_watters and gracy
  • #27
PeterDonis said:
You are mistaken. Everyone in this thread has acknowledged the effect of gravity on the molecules. The equation for hydrostatic equilibrium that I gave you explicitly includes the effect of gravity on the molecules. If there were no gravity, there would be no pressure gradient.
Yes, I can see how you pointed out the change in pressure vs. height. Sure, an isolated system is fine. I'm just saying that there is no constraint on temperature with PV=nRT. There's nothing to prevent a molecule from working its way to the bottom and picking up kinetic energy or vise versa.

Would it be easier for you to think of a single molecule in a tube?? The molecule can freely move to the top, losing kinetic energy, and on the next leg the molecule can move all the way to the bottom of the tube, having kinetic energy added to it. With a molecule in flight, it will lose kinetic energy as it moves against gravity and gain kinetic energy as it moves with gravity.
 
  • #28
PaperProphet said:
an isolated system is fine.

Ok, good.

PaperProphet said:
I'm just saying that there is no constraint on temperature with PV=nRT.

But that isn't the only equation. You keep on ignoring the other constraints.

PaperProphet said:
There's nothing to prevent a molecule from working its way to the bottom and picking up kinetic energy or vise versa.

Really? Have you actually thought this through? Including what happens when that molecule collides with other molecules?

PaperProphet said:
Would it be easier for you to think of a single molecule in a tube?

A single molecule is not a gas. Can you see what the key difference is between the two? (I implicitly told you what it was just above.) And how does that difference affect your reasoning about what happens to a molecule as it falls or rises within the tube?
 
  • #29
PaperProphet said:
The molecule can freely move to the top, losing kinetic energy, and on the next leg the molecule can move all the way to the bottom of the tube, having kinetic energy added to it.

Even this example, as unrelated as it is, can still be instructive. What happens to the molecule at the bottom and top of the tube? And how does that square with your assumption that the tube is isolated, exchanging nothing with its environment? (For example, could I extract work from the tube with one molecule in it?)
 
  • #30
Under conditions of equilibrium, there are just as many molecules moving upward as downward in any measurable volume of gas; i. e., no net fluid flow. Moreover, statistical mechanics tells us that the speed distribution function is the same for both arms of the Z-axis, even though there are differences in molecular number density throughout the tube due to gravity. Every upward molecule in a given volume that starts with speed v will lose speed Δv in going distance d. Every downward molecule in that same volume that starts with speed v will gain that same speed Δv in going that same distance. Since gas temperatures measure only the kinetic energies of translation of the molecules, no change in the gas temperature will occur due to gravity-induced acceleration or deceleration.
 
  • #31
Peter, if you have an answer, please spell it out for me and show me I'm wrong. I understand your need to give me partial answers so I can figure it out for myself...but I can't see it. I only see your formula which shows how you try to lock temperature with PV=nRT. I can see your confidence but I think you're as wrong as you believe I am. When I posted this somewhere fifteen years ago, I got a lot of that...but the other party always found a reason to break contact once I asked for anything more than 'hints'.
 
  • #32
klimatos said:
Under conditions of equilibrium, there are just as many molecules moving upward as downward in any measurable volume of gas; i. e., no net fluid flow. Moreover, statistical mechanics tells us that the speed distribution function is the same for both arms of the Z-axis, even though there are differences in molecular number density throughout the tube due to gravity. Every upward molecule in a given volume that starts with speed v will lose speed Δv in going distance d. Every downward molecule in that same volume that starts with speed v will gain that same speed Δv in going that same distance. Since gas temperatures measure only the kinetic energies of translation of the molecules, no change in the gas temperature will occur due to gravity-induced acceleration or deceleration.
Klimatos, I definitely believe you're the closest. It's something along those lines. It comes down to a Maxwell-Boltzmann distribution...if gravity were included as a variable. The delta-v definitely occurs...and the average energy should certainly be the same...but those molecules which are at a different Z somehow aren't at a different average energy.
 
  • #33
PaperProphet said:
I only see your formula which shows how you try to lock temperature with PV=nRT.

Did you read the rest of my post #16? Did you see the formula for hydrostatic equilibrium? Have you tried to work out anything using it?

PaperProphet said:
I think you're as wrong as you believe I am.

In other words, you think it's actually possible to get free energy from a vertical tube of gas? Or you think you have a different resolution to the "paradox"? I'm going to assume it's not the former, since you seem to consider that you have indeed posed a "paradox". But then, if it's the latter, what is your resolution?

I assume your answer to the question I just asked is going to be that you don't have a resolution, and that's why you're coming here asking. But if that's the case, how can you possibly think I'm wrong?

I don't mind giving more information, but I'm curious to see your answer to the above. It strikes me that you are expending a lot of effort in thinking I and others in this thread are wrong, but very little effort in actually figuring out what the right answer is.
 
  • #34
PaperProphet said:
those molecules which are at a different Z somehow aren't at a different average energy.

That's right. Can you think of some other variable besides the kinetic energy of an individual molecule, that affects the average kinetic energy of all the molecules at a given height, and which might change with height?
 
  • #35
"Molecular flow" which you have now explicitly invoked is different from "bulk flow" which was implicit in the use of the word "randomly"
PaperProphet said:
randomly move upward against gravity, the molecules lose energy and speed.
in your OP. "Molecular flow" involves ballistic trajectories; "bulk flow" does not.
 
  • #36
PeterDonis said:
In other words, in your model, there is continuous heat flow through the tube, and heat exchange between the tube and the outside environment. My understanding of the OP was that there should be no heat flow or heat exchange--that the tube is to be considered as an isolated system in thermal equilibrium with no heat exchange with the outside.

Well I just thought that there would be a noticeable temperature difference. But now I remember there is no noticeable temperature difference, although intuition says otherwise.

Somehow steam from water has the same temperature as the water, although every molecule loses kinetic energy when leaving the liquid. And somehow group of gas molecules escaping a gravity well has the same temperature as the gas in the well.
 
  • #37
Peter, I don't know the answer but I do know the right answer centers around the Maxwell-Boltzmann distribution if that somehow accounted for gravity. I can tell you you're just not going to get there with PV=nRT. I appreciate your effort...but your solutions are just not correct no matter how hard you press. I don't know if Klimatos knows the answer either...but I do know his answer is somewhere on or around the ballpark.
 
  • #38
jartsa said:
Somehow steam from water has the same temperature as the water, although every molecule loses kinetic energy when leaving the liquid.

The molecules that escape from the water and form the steam have higher than average kinetic energy, in order to break the inter-molecular bonds. If there is no heat source to keep the water at a constant temperature, the water that remains behind will cool down due to evaporation (since the molecules that remain will have lower than average kinetic energy).

Similar remarks apply to molecules escaping from a gravity well; they are preferentially the ones with higher than average kinetic energy, in order to achieve escape velocity. And again, if there is no heat source present, the gas that remains behind will be cooler.

Neither of these cases are really good analogies for the present discussion, because they involve open systems, and the OP specifically was talking about a closed, isolated system.
 
  • #39
PaperProphet said:
I can tell you you're just not going to get there with PV=nRT.

Why do you keep talking only about that equation, when I've said repeatedly that it is not the only equation that applies?

That said, did you read post #34? What variables determine the average kinetic energy of the molecules at a given height in the gas tube? Obviously the sum of kinetic energies over all molecules at that height (or more precisely, all molecules within a small interval of height from ##h## to ##h + dh##) is one; what is the other?

PaperProphet said:
your solutions are just not correct no matter how hard you press.

I haven't given any "solutions". I've asked you questions. Which you have repeatedly not answered, even though they're simple ones. The one above is about as simple as I can make it; the answer should be obvious. Can you answer it? Just the simple answer is all I'm looking for at this point.
 
  • #40
PeterDonis said:
Why do you keep talking only about that equation, when I've said repeatedly that it is not the only equation that applies?

That said, did you read post #34? What variables determine the average kinetic energy of the molecules at a given height in the gas tube? Obviously the sum of kinetic energies over all molecules is one; what is the other?
I haven't given any "solutions". I've asked you questions. Which you have repeatedly not answered, even though they're simple ones. The one above is about as simple as I can make it; the answer should be obvious. Can you answer it? Just the simple answer is all I'm looking for at this point.
Sure, the simple answer to your question is the number of molecules at a height, in addition to the total kinetic energy at a height, can be used to calculate the average kinetic energy at that height. Do you feel that's the solution?
 
  • #41
PaperProphet said:
the number of molecules at a height, in addition to the total kinetic energy at a height, can be used to calculate the average kinetic energy at that height.

Yes. More precisely, the average kinetic energy of molecules in a small increment of height from ##h## to ##h + dh## is the total kinetic energy of molecules in that small volume, divided by the number of molecules in the volume. If we take the limit as ##dh \rightarrow 0##, we have ##k T = K / n## where ##T## is the temperature (average kinetic energy), ##K## is the kinetic energy density, ##n## is the number density of molecules, and ##k## is Boltzmann's constant (which is there only to allow ##T## to have conventional units of degrees Kelvin instead of energy units).

Now, let's suppose the kinetic energy density ##K## varies with height according to your assumption--i.e., that it varies the same way the kinetic energy of a free particle does. That is just the negative of the variation in potential energy with height, i.e., for a single free particle it is ##dK / dh = - m g##, where ##m## is the mass of the particle. Here, though, we want the variation in kinetic energy density of a gas of particles, which is just ##dK / dh = - n m g##. Furthermore, the number density ##n## is just ##\rho / m##, where ##\rho## is the mass density. So we have ##dK / dh = - \rho g##.

Now, if we observe that ##K = k n T##, we can write:

$$
\frac{dK}{dh} = k \left( T \frac{dn}{dh} + n \frac{dT}{dh} \right) = - \rho g
$$

which is actually just a rewrite of the hydrostatic equilibrium equation, as we can see by observing that the gas constant ##R = k / m##, so the number density ##n## is equal to ##\rho R / k##. This let's us rewrite the above (rearranging some factors) as:

$$
\frac{dK}{dh} = RT \frac{d \rho}{dh} + R \rho \frac{dT}{dh} = - \rho g
$$

where the middle expression is obviously just ##d / dh \left( \rho R T \right)##, as we would expect by using the ideal gas equation to eliminate the pressure ##P## in the hydrostatic equilibrium equation.

So what all this is telling us is that there are two ways for the kinetic energy density to vary with height as you assume it will (i.e., as the negative of the change in gravitational potential energy with height): either the temperature ##T## can vary, or the mass density ##\rho## can vary. Of course this in itself does not prove that it is the mass density that varies, not the temperature; but all along you have been implicitly assuming that the temperature must vary if the kinetic energy density varies. The above shows that that is not the case; it is perfectly possible for the kinetic energy density to vary with height at constant temperature, by having the mass density vary with height. That is why it is not inconsistent for the tube to be at constant temperature even though it is in a gravity field, meaning that the kinetic energy density of its molecules varies with height.

In other words, if you really want to know how the temperature varies with height in the tube, looking at the kinetic energy density (and the effect of the gravitational field on it) is a red herring. The temperature is determined by other constraints. In this scenario, you have imposed another constraint by assuming that the tube of gas is an isolated system: that means there is no heat exchange with the outside. And that means there cannot be any heat flow anywhere in the tube (my assumption #1 many posts ago), because such a heat flow could only be sustained in equilibrium by coupling the tube to some external heat source and sink. And "no heat flow" means "no temperature difference".

So the reason the temperature of the tube is constant in your scenario is that you assumed it was, by assuming an isolated system. It has nothing to do with the gravitational field at all; if you took the same tube and put it way out in space, far from all gravitating bodies, and just let it float there, isolated from everything else, the temperature would still be constant--what would change is the density and pressure profiles (they would be constant instead of changing with height).
 
  • Like
Likes russ_watters, mfb and gracy
  • #42
Sorry but I just must bring black holes into this discussion.:smile:

It's not a problem that in an universe with microwave background radiation with temperature 3 K, and a black hole that is radiating Hawking radiation with temperature 3 K, you freeze far from the black hole, but you burn near the black hole. Constant difference of temperature is allowed in a gravity field.

Temperature difference = difference of gravitational time dilation

Note that difference of gravitational time dilation is very small between two ends of a tube on earth.
 
  • #43
PeterDonis said:
The molecules that escape from the water and form the steam have higher than average kinetic energy, in order to break the inter-molecular bonds. If there is no heat source to keep the water at a constant temperature, the water that remains behind will cool down due to evaporation (since the molecules that remain will have lower than average kinetic energy).

Similar remarks apply to molecules escaping from a gravity well; they are preferentially the ones with higher than average kinetic energy, in order to achieve escape velocity. And again, if there is no heat source present, the gas that remains behind will be cooler.

Neither of these cases are really good analogies for the present discussion, because they involve open systems, and the OP specifically was talking about a closed, isolated system.

Ok I'll consider closed systems then:

A closed container with liquid water and steam tends towards uniform temperature for some reason.

A closed container with molecules at low potential energy and molecules with high potential energy tends towards uniform temperature for some reason.

And I bet it's the same reason in both cases.
 
  • #44
PaperProphet said:
Klimatos, I definitely believe you're the closest. It's something along those lines. It comes down to a Maxwell-Boltzmann distribution...if gravity were included as a variable. The delta-v definitely occurs...and the average energy should certainly be the same...but those molecules which are at a different Z somehow aren't at a different average energy.

I like to consider gas molecules that only collide with the container, not with each other.

In the container there will sometimes happen that a molecule has a low speed for a long time, like for a whole minute. But that can never happen at the top of the container, only at the bottom.
 
  • #45
@PaperProphet - you were given the answer in post #2 :

Bystander said:
Think about densities.

If there were to be a difference in temperature then there would be a difference in density as well . Simple buoyancy effects would just move hot gas up and cold gas down to maintain equilibrium .
 
  • #46
In post #32, PaperProphet said "..but those molecules which are at a different Z somehow aren't at a different average energy."
Why would they be? Gas temperatures are independent of gas densities at anything approaching NTP. Moreover, each molecule is undergoing some 5.11 x 109 inter-molecular collisions per second at 1000 hectopascals and 25° C. Half of these collisions are with upward-moving molecules and half with downward-moving molecules. The net resultant speed change along any single axis is zero over any significant period of time.

Finally, it is my understanding that every closed system tends toward thermal equilibrium over time.
 
  • #47
PeterDonis said:
And "no heat flow" means "no temperature difference".

Not necessarily. Suppose the gas is dry air. A temperature gradient equal to the dry lapse rate (1 °C per 100 meter descent for our atmosphere) will be stable. This temperature gradient does not result in heat flow, the vertical motion of air parcels is adiabatic, not isothermal.
 
  • #48
spareine said:
This temperature gradient does not result in heat flow, the vertical motion of air parcels is adiabatic, not isothermal.
The vertical motion itself is adiabatic, but it is still caused by heat flow: air on the ground is heated by the ground and air above is cooled by radiation. If there were no heat flow, (per the case in the OP) the bulk motion and thus the temperature gradient would disappear.
 
  • #49
russ_watters said:
The vertical motion itself is adiabatic, but it is still caused by heat flow: air on the ground is heated by the ground and air above is cooled by radiation. If there were no heat flow, (per the case in the OP) the bulk motion and thus the temperature gradient would disappear.

The model of adiabatic motion of air says the dry lapse rate is equal to g/cp, so it does not depend on any heat flow. It is a stable temperature gradient without heat flow.
 
  • #50
jartsa said:
It's not a problem that in an universe with microwave background radiation with temperature 3 K, and a black hole that is radiating Hawking radiation with temperature 3 K, you freeze far from the black hole, but you burn near the black hole.

Once again this is an open system, so it's not relevant to this discussion.
 

Similar threads

Back
Top