Why wouldn't the temperatures be different in this scenario

  • Context: Undergrad 
  • Thread starter Thread starter PaperProphet
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on the thermodynamic behavior of gas molecules in a vertical tube under the influence of gravity. Participants explore why a temperature difference does not occur between the top and bottom of the tube, despite the kinetic energy changes of the gas molecules. Key principles include hydrostatic equilibrium and the ideal gas law, which dictate that temperature remains constant throughout the gas when in equilibrium, thus preventing any spontaneous temperature gradient that would violate the second law of thermodynamics.

PREREQUISITES
  • Understanding of hydrostatic equilibrium in gases
  • Familiarity with the ideal gas law (PV=nRT)
  • Knowledge of thermodynamic principles, particularly the second law of thermodynamics
  • Basic physics background, especially regarding kinetic energy and gravity's effects on gas behavior
NEXT STEPS
  • Study the principles of hydrostatic equilibrium in gases
  • Learn about the implications of the ideal gas law on temperature and pressure relationships
  • Explore the second law of thermodynamics and its applications in closed systems
  • Investigate the effects of thermal conductivity in gas-filled environments
USEFUL FOR

Physics students, thermodynamics enthusiasts, and anyone interested in understanding gas behavior under gravitational forces will benefit from this discussion.

  • #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'.
 
Science news on Phys.org
  • #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   Reactions: 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.
 
  • #51
spareine said:
This temperature gradient does not result in heat flow, the vertical motion of air parcels is adiabatic, not isothermal.

But the vertical motion of air parcels itself transports heat, because the air parcels contain heat. The heat transport is convection, not conduction, but it's still heat transport. Also, as russ_watters pointed out, the heat transport is ultimately driven by heat exchange with the ground at one end and radiative emission at the other.

In the scenario under discussion in this thread, there is, by hypothesis, no convection; the gas inside the tube is stationary. Also, by hypothesis, there is no heat exchange between the tube and the outside. So the mechanism you describe for maintaining a temperature gradient in the presence of gravity is not available.
 
  • Like
Likes   Reactions: russ_watters
  • #52
PeterDonis said:
Once again this is an open system, so it's not relevant to this discussion.

Gravitational time dilation is not limited to open systems. As jartsa already stated the effect is very small in a closed tube on Earth but it exists and it should lead to a vertical temperature gradient. Without such a gradient the bottom of the tube could be heated up with the blue-shifted thermal radiation from the top.
 
  • #53
PeterDonis said:
But the vertical motion of air parcels itself transports heat, because the air parcels contain heat. The heat transport is convection, not conduction, but it's still heat transport.

For each parcel moving up another one is moving down, the parcel doesn't leave a vacuum behind it. The net heat transport is zero. That is the dry lapse rate model.

In the scenario under discussion in this thread, there is, by hypothesis, no convection; the gas inside the tube is stationary. Also, by hypothesis, there is no heat exchange between the tube and the outside. So the mechanism you describe for maintaining a temperature gradient in the presence of gravity is not available.

For each parcel moving up another one is moving down, the parcel doesn't leave a vacuum behind it. The net gas transport is zero.
The temperature gradient is stable. That is the dry lapse rate model.
 
  • #54
spareine said:
The net heat transport is zero.

Is it? If so, what drives the gas flow? A convection cell doesn't just magically come into existence. A particular parcel of air near the ground only starts rising because it is warmer than other parcels of air near the ground. How did it get warmer? Conversely, a particular parcel of air at altitude only starts falling because it is cooler than other parcels of air at altitude. How did it get cooler?

In other words, the dry lapse rate model implicitly assumes that there is heat transfer at the ground and at altitude, in order to drive convection. Without that convection, the flow of air parcels will stop. And if there is heat being added at the ground and taken away at altitude, there must be net heat flow within the convection cell itself; the rising air must be transporting more heat upward than the falling air is transporting downward.

spareine said:
The net gas transport is zero.

But there is still gas moving. In the scenario under discussion in this thread, there is no gas moving at all, no bulk flow, not even a circulation in which upflow exactly balances downflow. And, based on the reasoning I gave above, that is the equilibrium state in the absence of heat exchange with the environment.
 
  • #55
PeterDonis said:
Is it?

Yes it is. Convection starts when the temperature gradient exceeds the adiabatic lapse rate. Without convection there is heat conduction and radiative transport only.
 
  • #56
DrStupid said:
Convection starts when the temperature gradient exceeds the adiabatic lapse rate. Without convection there is heat conduction and radiative transport only.

Conduction and radiative transport are still heat transport. If they are taking place, the net heat transport is not zero. I agree they are not the same as convection; but it seemed to me that spareine (not you) was arguing that the adiabatic lapse rate is determined by convection.
 
  • #57
spareine said:
the vertical motion of air parcels is adiabatic

It's also worth noting that this requires the air to be unconfined. In the OP's scenario, the gas is confined inside a tube, so it can't expand adiabatically.
 
  • #58
PeterDonis said:
but it seemed to me that spareine (not you) was arguing that the adiabatic lapse rate is determined by convection.

That's correct. Due to adiabatic expansion in upward direction and adiabatic compression the other way around convection results in the adiabatic lapse rate. Heat conduction and radiation alone would result in other temperature profiles.

PeterDonis said:
the gas is confined inside a tube, so it can't expand adiabatically.

Did you forgot the hydrostatic equilibrium?
 
  • #59
Back to the original post:
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 effect of gravity and the gas laws is: any initial temperature gradient will be stable if it is equal to or less than the dry lapse rate (g/cp), if you don't interfere. Alternatively, you could harvest an amount of energy from the temperature difference between the cold top and warm bottom, but it is not replenished for free. After harvesting, it would take energy to recreate the initial temperature profile.
 
  • #60
DrStupid said:
Did you forgot the hydrostatic equilibrium?

No. Hydrostatic equilibrium exists even in the absence of convection, i.e., even in the absence of adiabatic expansion.
 

Similar threads

  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 43 ·
2
Replies
43
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K