Physics Question re. Temperature of Gases

as soon as you find "perfect insulation" let me know, not only will you be rich but you'll go down as a guy who rewrote some of the laws of physics.

Hmm, you've never studied thermodynamics, have you? "Perfectly insulating walls" are a convenient fiction for analyzing processes, from a theoretical point of view.

Perfectly insulating walls is a valid assumption. If you consider a column of air within the atmosphere at a given altitude, the air is the same temperature as the adjacent air, so no heat flows horizontally as if the "walls" were perfectly insulating.

The "atmosphere" is not contained within a cylinder.

Where is your cylinder, within the gravitational field?

In your examples, if the gravitational field is that of the surface of the earth, #2, if not, wherever the gravitational field is strongest.

#1 is correct.

Heat transfer will eliminate any thermal gradient as the system approaches equilibrium.

Remember hotter air is less dense than cooler air, meaning it is buoyant, so it rises. The cool air falls. Heat transfer also occurs from hotter gasses to cooler gasses even if the hotter gasses are already above the cooler gasses.

As no energy is being added or removed from the system, everything will settle toward an equilibrium. Gravity will create a density gradient but no temperature gradient.

PV=nRT=NkT

That's what I think as well, but it doesn't explain why the second answer is wrong.

If a molecule travels a vertical distance (delta)h between collisions, it will lose or gain kinetic energy of m*g*(delta)h, where m is the mass of the molecule and g is the acceleration of gravity. So how does that not lead to a warmer temperature at the bottom of the cylinder than at the top?

OK, just for fun, let's say the warm air does move to the bottom. In order for that to be equilibrium, you have to make two huge assumptions:

1. Heat will not transfer from warmer to colder.

2. That buoyancy is a flawed concept.

I'm not ready to throw out heat transfer and buoyancy just to not hurt #2's feelings.

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The simple answer is that the system will not be in equilibrium until the temperature gradient is eliminated.

Nothing's wrong with the concept of buoyancy, if properly applied.

I'm not saying anything is wrong with buoyancy or heat transfer. In fact my argument is based on the idea that these concepts are dependably valid.

But Lyn has a point, in that in a gravity field, as long as the actual lapse rate in the gas is less than its adiabatic lapse rate, the gas column will be stable. If a parcel of gas rises, it will expand and cool to a temperature lower than that of the surrounding gas. So "warmer at the bottom" does not necessarily mean unstable.

The question I was originally trying to answer was precisely the question of conductive heat flow in a gas under the influence of gravity. It could be restated as asking whether the normal adiabatic lapse rate in the atmosphere is an equilibrium condition, or does it necessarily imply conductive heat flow from lower to higher altitudes?

'....So "warmer at the bottom" does not necessarily mean unstable.....'

Warmer at the bottom does indicate it has not reached equilibrium. Your argument would deny the readily observable phenomena of natural convection.

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Think of it this way, if running up the center of the hypothetical tube was a rod of copper which was perfectly insulated except for the portion close to the top of the chamber and a portion close to the bottom of the chamber. Plenty of room remains for the gas in the tube....or if you like you can imagine the copper conductor exiting the tube near the bottom and then reentering the tube again near the top.

Would you agree that heat transfer would occur and heat would move from the bottom of the rod where it is warmer to the top where it is cooler?

And if your proposed solution were to be valid with the warmer air sinking to the bottom of the tube, wouldn't this create a form of perpetual motion, with never ending flow of gas in the tube?

The key here is that heat transfer occurs in the tube moving from warmer to colder, and this occurs whether or not you have a copper heat conductor in the channel or not.

"..your argument would deny the readily observable phenomena of natural convection."

No, convection depends on the difference between the actual vs. adiabatic lapse rates. The adiabatic lapse rate for air is pretty small -- 10 K per km, or 0.01 K per meter -- and you're probably not used to thinking about it as non-zero. If you're thinking on the scale of rooms or lab space, then it's usual to take the pressure as constant throughout the room, and ignore the very small difference in pressure of air at the ceiling vs. air at the floor. But the difference is there. And air at a higher pressure can be warmer than air at a lower pressure and still be denser, as long as it's not too much warmer.

That's actually a very important consideration for meteorology. The air at lower elevations is almost always warmer than the air at higher elevations, but it is not unstable if the difference is less than the adiabatic lapse rate would have it. Convection only starts when the input of energy at ground level raises the temperature gradient above the adiabatic lapse rate.

Note that I'm not saying that the gas in my hypothetical closed isolated cylinder could have a non-zero temperature gradient in a gravitational field and still be at equilibrium. I don't think it can, but that's the point in question. I still don't see how to get from the low density limit of a single-molecule gas -- where the kinetic energy of the molecule is unquestionably a function of its elevation in the cylinder -- to the many-molecule gas where the kinetic energy is (presumably) independent of elevation.

I guess we need to back up one step:

In my modification of the hypothetical tube, do you agree or disagree that heat would be transferred by the copper bar from the area of higher temperature to the area of lower temperture?

About the copper rod business, I don't actually know with certainty that the temperature at equilibrium would be independent of position along the rod in the presence of a gravitational field. In quantum mechanical terms, that's equivalent to asking whether phonons have mass. I don't think they do, and that would suggest that the temperature distribution is not affected by the gravitational field.

But I hate quantum mechanics and ignore it when I can, as a matter of principle.

One other thing, you refer to warmer air sinking to the bottom of the tube; of course a warmer parcel of air would not sink to the bottom. That's not at all the model I'm asking about. I'm asking about how the presence of a gravitational potential gradient affects the diffusion of gas molecules in a closed, isolated system.

The warmer air sinking to the bottom of the tube is not what I am suggesting actually happens with the copper conductor introduced.

The sinking warm molecules are a consequence of the temperature gradient you are proposing in combination with the other restrictions you have presented.

In restricting the interaction between molecules, you have restricted significant energy transfer between molecules. If the molecules aren't interacting with other gas molecules and the copper bar has transferred some heat up the tube, what other choice do the warm molecules at the top have when trying to conform to what you propose?

With no interaction with other molecules, if the hypothetical gradient is to be reestablished, the energetic molecules must migrate to the proper areas to maintain the gradient you propose.

'...About the copper rod business, I don't actually know with certainty that the temperature at equilibrium would be independent of position along the rod in the presence of a gravitational field. ...'

This is definitely within the realm of things that could be supported or refuted via the results of inexpensive experimentation.

Well, since we're already committed to using perfect thermal insulation, why not go for broke and make our 'copper wire/rod' a thermal superconductor? (such things exist, actually, but inconveniently at He2 temperatures) and dispense with the clutter of such (irrelevant) detail altogether? Perfect insulation needs a perfect conductor! Really, copper's so mundane when you could've picked silver or, better yet, diamond. :)

You are right, but I wanted to make the example as uncomplicated as possible. A heat pipe would work even better than a diamond, but I thought it would confuse the issue by having anything but a solid involved.

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It actually doesn't make a difference as long as the heat conducting rod either always remains inside the cylinder, or if the insulation encapsulating it is perfect.

Perfect insulation was not something I introduced, it was part of the original though exercise, so it seemed reasonable to use.

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So what are your feelings on the original question, Europium? How does the heat conducting rod change the system?

"Introduce a length of perfect thermal conductor (composition unspecified) running from the top of the column to the bottom, perfectly insulated along its entire length except for its ends, which are exposed."

There. That should do it! :)

The basic approach is to start with the Ideal and add Reality as we go along. There's nothing wrong with that approach and a great deal right about it. For one thing it strips away all the extraneous clutter that only confuses things. Then, once we(ve mastered the basic problem, we add a bit of Reality and deal with that variable, then add a bit more Reality and so forth and so on.

ok, but what do you think happens? Do you think the thermal conductor changes the system at equilibrium, or do you think either way the temperature will be the same everywhere at equilibrium.

When real columns of atmosphere are at equilibrium, they are colder the top and warmer at the bottom so, yes, it will disturb the system because thermal superconductors are absent from real columns, forcing heat to flow the old-fashioned way.

However, gases constrained within the confines of a cylinder are prevented from undergoing adiabatic expansion as they rise by convection, and so what works in free air does not necessarily apply in this situation. In free air, the gas expands as it rises. Not so here.

So you allow for the system being in continual motion if a super (heat) conducting rod is running up the middle.

What if it has just a little bit of resistance?

A pièce de résistance, you mean?

OK that was pretty funny.

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But, yeah, just a little resistance to the flow of heat.

Well, then - assuming our formerly perfect thermal superconductor conducts heat energy instantaneously and still does - the system will take longer to settle into a new equilibrium condition which is also not representative of the system in the (now ordinary) conductor's absence, as it provides a path for energy to flow which bypasses the mechanism which predominates in the Original problem. By introducing this second path we have created in effect a new problem, not identical to the first and leaving the Original problem unsolved. Best to nix the conductor altogether.

Then the heat and gas would go around a little slower.

There is no problem with a perpetual motion machine: it's perpetual work machines which are impossible.

OK, so the major difference in using a material that doesn't conduct heat as well, will be heat is transferred more slowly.

This still means that equilibrium will not be reached, right?

Depends on your definition of equilibrium. Is a ball spinning in space (billions of light years from any other object) in equilibrium?

If it can be considered a closed system, and no forces will act on it to slow the 'spin' (which would be converted to heat) and no processes are converting other types of energy to heat, then if the temperature is uniform throughout then it is in thermal equilibrium.

But generally, no.

We can sidestep the term equilibrium if it make you uncomfortable.

Let me make sure I understand your position:

You believe as things settle to a stead state over time, the gas at the bottom of the container will be at a higher temperature than the gas at the top.

You accept the idea that a rod made of a material that is a decent conductor of heat placed in the center and extending from top to bottom insulated everywhere except for the potion at the top and bottom, and leaving room for the gas to travel between it and the outer wall of the container; would transfer heat from higher temperature to lower temperature, so heat the gas at the top of the container and cool the gas at the bottom. Correct?

Further you believe that this warm gas, by the same process that created the original condition of the warm gas collecting at the bottom of the container, should sink back to the bottom. Is that right?

Yes.

It does seem very counterintuitive when you put it that way, but, I think you have to try to think about individual atoms/molecules rather than pockets or clouds of gas.

Provided the ball is perfectly symmetric, has no magnetic field and carries no charge, ideally it might spin forever, but it will still slow down. In the first case, were it not symmetric, it will (very) slowly lose angular momentum due to gravitational radiation - gravitational waves. In the second and third cases magnetic fields and electric charges - both in motion relative to the 'fixed' stars - will cause the ball to lose angular momentum through electromagnetic radiation. Even when the ideal conditions are satisfied, the ball will still lose energy via frame-dragging. Kind of like a mixer blade in a bowl of infinitely-stretchable Jell-O. :)

Maybe. But, I think the question in the original post can be answered to an acceptable level of accuracy without resorting to anything more exotic than Newtonian mechanics.

There are two plausible answers: (1) uniform temperature independent of height within the column; and (2) cooler with increasing height, consistent with the adiabatic lapse rate for the gas in the column. There are reasonable sounding arguments supporting either answer, but they can't both be correct.

Well, if neither of your plausible answers is convincing, it may be because neither is correct. Neither describes the actual temperature gradient of the atmosphere. You might examine why the temperature gradient of the 'real' atmosphere is neither uniformly the same throughout, nor does it decrease uniformly with height.

Actually, the problem is that both answers are convincing, but they contradict. Obvious falsehoods I can deal with; it's the convincing ones that cause trouble.

Of course the temperature gradient of the "real" atmosphere is all over the map. The atmosphere is a dynamic system with huge energy inputs and losses. It never comes close to equilibrium. What's surprising is that a constant lapse rate in unsaturated air, in the absence of a frontal system, is so often a very good approximation.

If it's fully sealed and insulated, neither gaining nor losing heat from any source (the Earth, the Sun, the 3K blackness of space) then you're only dealing with internal convection and radiation.

So the question, I think, comes down to the opacity of the gas within the tube. (Much like the early conditions of the universe; are matter and energy coupled or decoupled?) For a blackbody, the temperature is given by the signature (spectral distribution) of the radiation field. If the gas is opaque to the radiation, then you will get a temperature gradient, since one end of the tube will have a different radiation spectrum compared to the other end of the tube. If the gas is transparent to the radiation then both ends of the tube will have the same radiation spectrum and therefore will have the same temperature.

There is a difference in pressure and also density between the bottom and the top. This difference manifests also as a difference in temperature. None of the gas will remain stationary, but the system as a whole will settle into a state of constancy and stability. The temperature at the top must be lower than the temperature at the bottom, simply because the lower pressure and density can only account for the average velocity of the gas present, and that velocity cannot be greater as you measure higher and higher up the cylinder. You only have gas as fast as you have when you start, and it does not get faster except in the downward direction. It cools as it rises and slows down.

Also, the taller the cylinder, and the lower the pressure, the further the velocity of the gas will allow it to go away from the gravitational field resisting it's rise. The temperature will fall as the gas loses velocity and energy is expended in the movement away from the gravitational field. With no further input of energy and an infinitely tall cylinder, eventually, the gas will settle near the source of gravity.

#2 is correct. The reason your heat engine won't work is that the relationship between velocity and temperature is also related to the mass of the molecules. The difference between the two cylinders with different gasses is not in the temperature, but in the relative pressure at varying heights. In a gravitational field and when considering energy transfer via motion, mass is everything. Your engine will run until the heavier gas is cooled off, then it will stop and both columns will settle. You also have to account for the difference in density. A lower density gas cannot transfer as much heat as a high density gas, so your engine will either cool off on the hot side and/or decrease the density on the hot cool side as the cooler gas expands with the heat you deliver to it. The two columns will end up at differing heights, but the same temperature at the bottom. Gas weighs less as it rises away from the gravitational source. Every molecule at the bottom weighs more than every identical molecule at the top. There is a big difference between weight and mass.

That's my 2 cents.

'.... A lower density gas cannot transfer as much heat as a high density gas....'

If by 'lower density', you mean something like Helium or H2; and if by 'high density' you mean something like Argon or SF6, you've got it exactly backwards.

Helium and Hydrogen are far more effective at heat transfer than a similar number of Argon or SF6 molecules.

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I also disagree with your suggestion that velocities cannot be the same or greater at increased elevation. Imagine you have a tank half filled with liquid water right at the boiling point. Just above there is saturated steam. Which molecules move faster? Steam or liquid water?

If I then begin to very slowly draw liquid water from the bottom of the tank, lowering the pressure so that some of the water flashes to steam and cools the remaining water. As I continue to do this, the later steam that flashed is cooler, won't it have a tendency to remain lower in the tank?

Eventually the higher temperature upper steam will lose heat to the lower temperature steam and temperature will equalize. There is nothing that prevents this temperature driven heat flow, regardless if hot temperature is positioned above, below, or even beside the colder temperature.

No temperature gradient exists in a perfectly insulated closed system at thermal equilibrium

You left out the gravity. There is NOT going to be a single temperature for bottom and top. Gravity itself is a gradient field. The upper regions of the column are under less gravitational influence than at the bottom. The velocities are effected by this change because gravity affects the rate of time and for a given energy and mass, affects the velocity of that mass.

A gradient is "forced" upon the gas column because the space-time gradient of the gravity field cannot be escaped, all other things being equal, and no other source of energy or drain of energy is present.

So then are you suggesting that heat is transferred slower in the direction away from the predominant gravitational source than heat transfer toward the predominant gravitational source?

That is a pretty remarkable claim, if that is what you are saying. Do you suggest it holds true only for gasses?

It would have to be a measurable effect if gravity can keep higher temperature gas from losing heat to lower temperature gas it contact and above it. But that would be required in your 'temperature gradient at equilibrium in a closed system' assertion.

Is whatever you were taking when you wrote this available legally?

As a parcel of air rises in a column of air, it cools at the adiabatic lapse rate. If the temperature gradient in the column is greater than this rate, this parcel will continue to rise. In other words, the air is unstable and convection will ensue. (The adiabatic lapse rate is a function of the amount of water vapor in the air due to the latent heat of condensation.)

I think the answer is that the temperature can decrease with altitude at any rate less than the adiabatic lapse rate and is determined by other factors such as radiation from the ground.

It is an extremely rare thread here, that is answered in one or two definitive responses.

Even those which only have one possible answer.

When I was working with High vacuum systems I learned that even one molecule of gas can maintain an even pressure within the confines of a chamber of constant temperature. Heating the chamber results in a higher pressure even if there is no addition of gas to the system. What happens is the velocity of the molecule increases with the increase in temperature.

Gravity is due to the presence of mass which decreases the rate of time flow in the space around it. The further away one gets from the mass, the faster the time flows. This affects the velocity of another smaller mass. Total energy is preserved, but velocity is not preserved, it decreases as the smaller mass moves into a faster and faster time potential and away from the larger mass. The energy is maintained in the distance as potential energy.

The total energy of the gasses in the column will remain the same, but the gravitational gradient will maintain a temperature gradient in the column because it will maintain a pressure and density gradient in the column. The gravitational field could be characterized by measuring the pressure and temperature at varying points between the bottom and the top of the column.

Air cools at it rises because of the change in density from expansion as well as the conversion of velocity to potential energy (slowing against gravity).

I maintain the #2 is the correct answer and that the bottom will be warmer than the top.

In a closed system the average kinetic energy of the molecules will remain constant despite variation in gravitation along the height.

May be we can consider the Hilsch Tube.

http://www.me.berkeley.edu/~gtdevera/notes/vortextube.pdf

In this, centrifugal force separates the high energy molecules from the low similar to what gravitation might do. However, this is an open system allowing separation of the two, which is not the case with a closed system.

OK, I've been thinking about the problem while the discussion has been going on, and I think I've got a sufficient handle on it to answer my own question. I want to thank everyone for the responses they've given and the stimulus they've provided for my thoughts.

Before I get to my answer, I think a little more background on how this question came up might be in order. It was in correspondence with a climate change "skeptic", regarding the nature of the greenhouse effect and its role -- or lack of role -- in the very high surface temperatures on Venus. This person maintains that the surface temperatures on Venus have nothing to do with any greenhouse effect, but are simply a consequence of gravity and the very thick atmosphere of Venus. His reasoning is that the thermal infrared radiation needed to balance Venus' radiation budget originates from the top of the sulfuric acid cloud layer in the upper reaches of Venus' atmosphere. That layer is at the approximate blackbody temperature needed to produce the requisite Stefan-Boltzmann flux of outgoing radiation. At lower altitudes, the atmospheric temperature simply rises, under the influence of gravity, in accordance with the adiabatic lapse rate. Nothing to do with greenhouse effect.

I maintain that he has the premise right, but the conclusion wrong. He's right about the temperature at the top of the high cloud layer, and right about the temperature profile in the atmosphere. But the greenhouse effect is essential in maintaining that profile. If the atmosphere were equally transparent or equally opaque to all frequencies of the spectrum, then the profile would be quite different. If it were transparent to all frequencies, then the blackbody equilibrium temperature would be found at the surface of the planet, not at the cloud tops. And if the atmosphere were opaque to all frequencies then there would be no heating anywhere below the cloud tops, and conduction and radiative transfer within the body of the atmosphere would create an isothermal, strongly stratified atmosphere.

At that point he said no, and brought up the kinetic vs. gravitational potential energy of the atmospheric gas molecules as a function of altitude. I was pretty sure he was wrong, and that in the absence of heat input at the bottom, an opaque atmosphere would be quasi-isothermal. Any other answer seemed to run afoul of the 2nd law. But I couldn't refute his kinetic vs. potential energy argument.

Well, I think I've finally got it figured out. But I'm probably approaching the word limit for postings, so I'll postpone the answer itself. Since I haven't been able to reduce it to an obviously correct, "oh, of course!" answer yet, I'll give you other gurus a chance to see if this background has triggered any insights toward a better answer.

Here's my take on how the second answer is wrong. (BTW, the question for me was never which answer was right; it had to be the first (uniform temperature) because the 2nd law violation otherwise was compelling. The problem was to understand what was wrong with the argument for the second answer.)

There's no question that the trajectory of individual gas molecules between collisions is affected by the gravity field. Molecules on ascending trajectories lose kinetic energy, and their paths bend toward the horizontal. Molecules on descending trajectories gain kinetic energy, and their paths bend toward the vertical. So the distribution of trajectories in the presence of a gravity field is not isotropic.

Since the gas is stipulated to be at equilibrium, both the energy and momentum of molecular trajectories into and out of a differential volume of the gas must be equal. If the temperature were a function of elevation within the field, such that the average kinetic energy of ascending and descending trajectories were equal at the volume element, then the balance condition would be violated. The distribution of trajectories for molecules leaving the volume element after undergoing collision within the element would would carry a net flux of ascending thermal energy.

That's not intuitively obvious, and I don't know a good way to prove it, short of a Monte Carlo computer simulation. If I steel myself to work through the math in detail, something simple and elegant might emerge. Unfortunately, my math abilities have severely atrophied over the years. One of the hazards of working with computers is that the power of simulation and numerical methods tends to extinguish advanced math analysis skills. Sigh!

Oh, stop complaining, geezer. It's called getting old; get used to it.

because the 2nd law violation otherwise was compelling.

But you haven't answered my post 42 yet. You said yourself in post 33

" and there would be a net loss of thermal energy equal to the work taken out of the system by the heat engine."

How is anything going to keep doing work under those circumstances?

"How is anything going to keep doing work under those circumstances?"

It can't. That's the point. You can't extract any work from a closed system at equilibrium. That's the second law. If you can extract any work, then the system is, ipso facto, not at equilibrium. If the temperature at the top of the column is lower than the temperature at the bottom, then you can extract work. That follows from the lapse rate being dependent on the mass of the gas molicules. Therefore, colder at the top cannot be the equilibrium condition.

Technically, that's only 99.9999..% correct. There is a general relativity effect to which Deefburger alluded earlier. Clocks run at different speeds as a function of gravitational potential energy. Even photons lose energy ascending in a gravitational field. Since photons don't interact, the radiation temperature at the top of a black body cavity is minutely lower than the radiation temperature at the bottom. And I do mean "minutely". One would need a thermometer accurate to 18 digits to measure the effect in a laboratory. It's many, many orders of magnitude smaller than the temperature difference produced by the adiabatic lapse rate in a circulating gas.

The reason that the minute temperature difference from relativistic effects does not give rise to any violation of the second law is that it's a function of the space-time configuration, and is totally independent of the properties of materials. It applies equally to the conductive copper pipe, a column of light gas, or a column of heavy gas. No differential temperature that could be tapped by a heat engine will arise.

(2) is correct.

The error in the argument for (1) is as follows:-

So you could have two cylinders side by side, holding different gases. If their bottom temperatures were equal, their equilibrium temperatures at the top would differ. One could then run a heat engine off of the temperature difference. The ensemble would be a perpetual motion machine of the second type.

The average temperature of both columns would start to decrease, and, the heat engine would gradually generate less power.

Sorry, doesn't wash. The cylinder with the light gas (and higher temperature at the top) would lose temperature, while the one with the heavy gas (lower temperature at the top) would gain temperature, and there would be a net loss of thermal energy equal to the work taken out of the system by the heat engine. But as long as the two cylinders were thermally connected at the bottom, the flow wouldn't stop. If there's no mechanism operating to maintain the temperature differentials between top and bottom, then that's not the equilibrium state. That takes us back to answer #1.

and there would be a net loss of thermal energy equal to the work taken out of the system

At what temperatures would both columns be at after say 1 year; 100 years; 1000 years?

After how much time would no gas be reaching the tops of the columns?

I am surprised at the number of sensible, well versed CR4 members who have expressed a strong conviction that #2 (non-uniform temperature) will be the ultimate result.

Anything except #1 (uniform temperature) would be a violation of the 2nd law of thermo, in a truly closed system left long enough.

Although I am convinced that my analysis is correct, there are several CR4 members for whom I have respect coming up on the other side of this argument. This gives me pause, but even after running through everything again, I come to the same conclusion.

I would appreciate it if some of you would comment on how #2 can be supported.

Looking at the following heat transfer equations, none of the denominators in the following equations could reasonably be considered to be infinite, and as long as a temperature difference exists, the numerator will be non-zero. This suggests that heat transfer will occur until uniform temperature is reached, i.e. thermal equilibrium.

Transfer Mode	Rate of Heat Transfer	Thermal Resistance
Conduction
Convection
Radiation		, where

In light of the equations above in combination with the Second Law of Thermo, particularly as described in the Clausius Statement, "Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time."; Support for the idea that the ultimate state of an isolated system to which no heat is being added, to which no work is being done, in which no reactions are occurring, could contain a thermal gradient seems untenable.

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If I am missing something please enlighten me.

OK lets try to look at this in another extreme way.

Suppose you only have one molecule or atom of a gas in the cylinder. And that the cylinder is tall enough so that when you release the particle from a stationary position at the top it reaches a speed equivalent 25°C at the bottom (where it bounces and just returns to the top)?

What about 2 atoms; 1 million; 1billion......... 1kg per m³.

At what density does the physics change from one "kind of physics" to "the other", and, why?

If you have more than one molecule then there will be heat transfer, and eventually the molecules will average over time roughly the same kinetic energy.

That the molecules are more often found in the bottom is seen as density not temperature.

The increased force against the bottom of the container as opposed to the top of the container is a difference in pressure not temperature.

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By what mechanism are you suggesting makes this specific set up different from every other set up anywhere, in that a difference in temperature fails to drive heat transfer?

Your hypothesis suggests the ultimate steady state of the system contains a measurable temperature gradient that is miraculously devoid of heat flux.

Please show me some law of thermodynamics that allows for exclusions from heat transfer when a temperature difference exists. If you are going to state the assumption of perfect insulation, remember there is none of that blocking heat transfer between the top and bottom of the container.

I don't think that I am excluding any form of heat transfer.

I think I am simply including the potential height energy of the particles, which can normally be ignored in most other thermal calculations.

Then why doesn't the difference in temperature cause heat transfer leading to uniform temperature?

Certainly looks like a violation of the 2nd Law of Thermo, to me

The second law of thermodynamics states that the entropy of an isolated system never decreases, because isolated systems spontaneously evolve towards thermodynamic equilibrium.

I think we're back with trying to define more rigorously what thermodynamic equilibrium really means when you start considering ridiculously tall columns in a gravitational field.

Of course it may simply be that taking the potential energy of height into account resolves this difficulty.

There is definitely no reason to ignore gravitational potential energy.

Similarly there is no reason to ignore heat transfer via radiation and conduction.

Warmer gas is less dense than cooler gas, as such, if it is below a mass of cooler gas, the cooler gas typically moves down displacing the warmer less dense gas upwards.

Isn't that your experience?

Yes when you're talking about pockets of gas at different temperatures, but, apparently not when the gas reaches a steady stratified state.

How does that make sense? And if it is true up until stratification, how did stratification as a temperature inversion occur in the first place.

Stratification also does not present a barrier to radiative heat transfer.

how did stratification as a temperature inversion occur in the first place

The gas mixes up.

I don't think that radiative heat transfer is very significant.

Lets look at the numbers at typical temperatures around sea level.

The specific heat capacity of dry air is about 1000 J/Kg/°K
Potential height energy = mgh = about 10 J/Kg/m = 1000 J/Kg/100m

The adiabatic lapse rate for dry air is about 10°C per km

So it looks like the two calculations agree with that figure.

Another way to look at it is that a unit mass of air at 15°C at 1km up has the same energy (thermal plus potential height) as another unit mass of air at 25°C at ground level.

1. Stratification:

Stratification does not occur by 'mixing up'. 'Mixing up' destroys stratification, it does not create it.

2. The link you provided:

Please note in the second section (labeled Definition) of the linked page you provide:

'....Because the atmosphere is warmed by conduction from Earth's surface, this lapse or reduction in temperature is normal with increasing distance from the conductive source.....'

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meaning this is not a quality of an isolated column of gas in a gravitational field, but a dynamic condition created by the introduction of heat at the lower portion of the atmosphere.

Later on it says this:-

The adiabatic lapse rates - which refer to the change in temperature of a parcel of air as it moves upwards (or downwards) without exchanging heat with its surroundings. The temperature change that occurs within the air parcel reflects the adjusting balance between potential energy and kinetic energy of the molecules of gas that comprise the moving air mass. There are two adiabatic rates:

But it also says lots of other things which lead me to believe that this is not at all straight forward.

However, on the whole I'm still convinced that with the problem as stated in the OP there will be a temperature gradient.

I agree with you on ' this is not at all straight forward'

This blog is certainly evidence of that! I do have faith that we can resolve this.

I think inclusion in the definition of lapse rate, of conductive heating of the lower portion of the atmosphere, should allow for the possibility that the phenomena occurs in relation to that.

With that possibility (which disallows using lapse rate as an explanation), it seems like the following would bring you around to my way of thinking (which of course I think is correct, but the jury is still out):

-Is a gas at a higher temperature and the same pressure more or less dense than a gas at a lower temperature and the same pressure?

Hopefully we can agree that at the same pressure a warmer gas is less dense.

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-Will a less dense gas eventually be displaced upwards by a more dense gas with which it is in contact (assuming no energy is being added to maintain the unstable state)?

Hopefully we can agree, 'yes'.

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-As the warmer gas is displaced upward by the cooler more dense gas, the warmer gas expands and cools, so pressure decreases and so does temperature...I know we agree on this. Now as for where equilibrium will be reached.....as the gas expands and cools, if there is still cooler more dense gas above it, it will continue to be displaced upwards. If momentum carries it too high and it cools too much becoming more dense than the gas below, it will displace the gas below until it reaches equilibrium. I think this is the place where our thinking diverges.... can you tell me why?

-Is a gas at a higher temperature and the same pressure more or less dense than a gas at a lower temperature and the same pressure?

Yes

Again I think that trying to analyse this in terms of pockets of gas rather than individual molecules will lead to misleading results. However:-

Let's look at a 1km column again.

Although we are only looking at sealed and insulated cylinders with everything else "in balance" you could briefly remove the lid of such a cylinder without anything dramatic happening so the following argument is still valid. If this statement is not clear you could calculate the pressure differential from first principals.

At 15°C the atmospheric pressure for dry air is about 10% lower 1km up than at ground level:-

From http://en.wikipedia.org/wiki/File:Atmospheric_Pressure_vs._Altitude.png

Using the general gas equation:-

P₁V₁/T₁ = P₂V₂/T₂

Or V₂ = P₁V₁T₂/P₂T₁

Then: taking P₁/P₂ =10/9 (from above), and, T₂/T₁ = 278/288 (that's the 10° from an earlier post)

then V₂ = ~ 1.07 V₁

In other words the the colder air is less dense than the warmer air because your assumption of same pressure is not valid.

As long as we are not constraining the system to something like the column having negligible width, then we can consider the interaction between warmer and cooler portions (right down to one molecule) that are essentially beside each other....and therefor at the same pressure. This is where the interaction is most direct.

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So what happens when the column is mixed and in the same level, some warmer gas is directly adjacent to some cooler gas?

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You admit the cooler gas, at the same pressure is more dense, and the warmer is less dense.... so the warmer gas should, over time, move to be above the cooler gas.

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The fact that as the warmer gas cools as it expands and the fact that the cooler gas increases in temperature as it is compressed, does not change dynamic that gas at the same level that is warmer and hence less dense will be displaced above the cooler more dense gas that is at essentially the same height.

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If this column is so thin that molecules cannot effectively pass each other, then there won't be any mixing.....but even then, radiative heat transfer will eventually lead to uniform temperature. I don't think that is what the original question specified though.

The adiabatic lapse rate which is widely observed in the real world (but you have to leave the classroom to see it) is the result of convection. As a parcel of air moves upward, the pressure reduces and so does the temperature. When a parcel descends to higher pressure, it warms up. (Remember the experiment pumping up a tire in sixth grade?) If you put the copper rod in the cylinder, it will warm the upper reaches and cool the lower, resulting in what the weather girl would call a temperature inversion. The result is no convection and ultimately a temperature equilibrium. The convection is the result of heating the air at the surface, mostly by conduction. At night, the air near the surface cools, and an inversion results. When the sun comes up, convection recommences (about 2 or 3 pm in the Mohave Desert).

I assert that the "greenhouse effect" is largely fictional or irrelevant. The analysis of Venus (cloud tops set the temperature) was essentially accurate. The effects of water, clouds, (or H2SO4 if Venus) and convection overwhelm the greenhouse effect, if it exists. Consider if you measure and compare radiated energy at a point (an instrument) going up and radiated energy going down, in the shade or integrated over days. It's easy to do with a computer model, but it is nearly impossible to find a difference in the real world; which might explain why climate change takes centuries even though the greenhouse effect operates at the speed of light.

To get your head in the right mode, consider that you are swimming in a sea of radiating molecules. The pressure at the surface is low and is high at the bottom, but the pressure difference, upward and downward, at any given level is zero. There is no "pressure flux" up or down. Why would you expect a "photon flux"?

The effect of man made CO2 is clearly slight compared to water. As a bit of experimental evidence, consider that during the Great Depression, man made CO2 emissions from fuel were reduced by perhaps 30% as factories and steel mills shut down, yet the rate of "global warming" was the maximum so far measured. Driving a Prius (I do) isn't going to alter global warming. It would be expensive to repeat the experiment, and would you expect a different result?

So near and yet so far.

Your first paragraph is entirely correct. But then you jump the rails and assert that the greenhouse effect is "largely fictional or irrelevant".

The "greenhouse effect" is a name for what happens when sunlight impinges on a greenhouse medium. A greenhouse medium is one having a difference in aggregate transmission / absorbtion coefficients between short wave (visible and near infrared) and long wave (thermal infrared) electromagnetic radiation. The atmosphere is most definitely a greenhouse medium, and that fact is central to atmospheric physics.

Convection does not cause the atmospheric lapse rate, it is caused by it and serves to limit it. What drives the lapse rate is the greenhouse effect -- energy deposited at the bottom of the atmosphere in the form of visible and near-infrared radiation, and disproportionately re-radiated into space as thermal IR from the upper layers of the atmosphere. If there were no greenhouse effect, the atmosphere would be uniformly cold and stagnant.

You're right that the effect of CO2 in the atmosphere (man made or otherwise) is slight compared to that of water. That is the source of a great deal of confusion and misunderstanding about climate issues. I won't try to explain why CO2 levels are nonetheless a critical "signal" in determining global climate, but here's something to consider, by analogy: the amount of iron in your body is a trivial fraction of a percent, and its role in bodily metabolism, compared to everything else that goes on, is "slight". But without it, we'd all be dead.

1) uniform temperature independent of height within the column. The thermal idea has to do with energy being applied to change the pressure of the gas to produce heating and cooling through compression and expansion of the gas. The difference in height of the gas in the stable column would not effect the temperature.