Can Altitude Hypothesis Challenge the Second Law of Thermodynamics?

[striphe]

Post number 71

D H hasn't given full calculations but has provided the results of calculations.

 

[klimatos]

I didn't read the whole thing, but I believe they resolve the paradox by showing that the coldest atoms in the maxwell-boltzmann distribution don't have enough energy to travel far up the tube, so that while all the atoms lose energy as they travel upwards, by shedding the coldest atoms, the temperature in fact doesn't change as we move upwards in the tube.

There seems to be quite a bit of misunderstanding of kinetic gas theory and statistical mechanics throughout this thread. A molecule that is at the bottom of the tube now is still likely to be at the bottom of the tube a minute from now. At NTP a molecule changes its translational speed and direction more than five billion times a second. Its path over that second can be described as a random walk. It is no more likely to go up than it is to go down, and its direction of movement is independent of its translational velocity.

If you want to give a molecule a "temperature" based on this translational velocity, then consider that in that single second that molecule will have changed its temperature some five billion times--more or less. The Maxwell-Boltzmann distribution of molecular velocities not only applies to a population of molecules; it also applies to the distribution of speeds of a single molecule over any significant portion of time.

Therefore, the idea of a "cold" molecule being less likely to go up the tube is meaningless. It is not the physical molecules themselves that conduct thermal energy, it is the impulses that they generate at collisions. These impulses travel up and down the tube at roughly the speed of sound, while the actual migration of molecules is a matter of self-diffusion.

 

[Smacal1072]

Hi klimatos,

I absolutely agree with you - However, the paper I mentioned doesn't deal with gases at NTP, it only discusses an apparent paradox involving ideal gases.

 

[JaredJames]

Hi klimatos,

I absolutely agree with you - However, the paper I mentioned doesn't deal with gases at NTP, it only discusses an apparent paradox involving ideal gases.

Gonna stick my neck out here, but I was under the impression ideal gases don't exist? So any paradox doesn't either?

 

[klimatos]

Gonna stick my neck out here, but I was under the impression ideal gases don't exist? So any paradox doesn't either?

Of course you're right, Jared. Ideal gases don't exist. They are usually either teaching devices for teachers or speculative devices for scholars. They are useful in both roles. The atmospheric sciences, for one, would be lost without the concept of the ideal gas.

On the other hand, a large number of the threads in this section seem to deal with idealized situations and hypothetical devices.

Jared, I think you're just having fun!

 

[JaredJames]

Jared, I think you're just having fun!

But am I wrong?

If you use an 'ideal' situation for analysis, it could certainly violate the laws of thermodynamics (PMM's when ignoring friction losses for example).

So just because under perfect conditions you can violate a law, if those conditions can never exist then there's no problem.

You can ignore friction to help make a problem easier, but that doesn't mean you can apply the same thinking all the time and extend it to other areas - which is why you end up with perpetual motion devices.

 

[striphe]

One can understand that under ideal conditions a wheel that experiences no friction can spin forever or that entropy won't increase in a number of hypothetical occurrences.

This is different, we have something reducing entropy; which I would say is impossible in ideal circumstances according to the second law of thermodynamics.

In a closed static body of gas, under the force of gravity there is going to be a pressure difference. As one is dealing with a gas, there are less particles per volume at the lower pressure end then the higher pressure end.

If there is less particles at the lower pressure end, each gas particle has to be moving faster for the temperature be the same throughout the body. Can someone explain how this would be so?

 

[D H]

As a retired professor of atmospheric sciences, I am familiar with lapse rates. I referred to a closed system at equilibrium. Adiabatic lapse rates are phenomena of moving air--not an equilibrium situation. The normal atmospheric lapse rate is an artifact of the Earth's heat budget. It would not exist without an outside source of energy--the Sun.

You have this exactly backwards. The adiabatic lapse rate is a direct consequence of the second law of thermodynamics. A lapse rate that deviates from the adiabatic lapse rate is a phenomena of moving air.

 

[klimatos]

If there is less particles at the lower pressure end, each gas particle has to be moving faster for the temperature be the same throughout the body. Can someone explain how this would be so?

A gas can have the same temperature at a wide variety of pressures. Temperature measures the mean kinetic energy of translation of the gas molecules. It is an average. T= (mv^2)/k, where T is the temperature, m is the molecular mass, v is the root-mean-square molecular velocity along a single axis of movement, and k is Boltzmann's Constant.

Pressures, on the other hand, are usually a function of both temperature and molecular number density: P = nkT, where n is the number density. Pressure is a total. It is the number of molecular impacts times the mean impulse transferred per impact.

 

[striphe]

This doesn't answer the question of how the molecules have a higher average velocity at the top than the bottom.

 

[D H]

If there is less particles at the lower pressure end, each gas particle has to be moving faster for the temperature be the same throughout the body. Can someone explain how this would be so?

The mean velocity of the molecules that comprise an ideal gas is a function of temperature only, striphe. For example, the root mean square velocity of Helium gas at 300 K is 1368 meters per second. Pressure and density are irrelevant.

 

[striphe]

If I had one mole of gas in a metre cubed container and then expanded this container so it was 2m^3. The gas particles would not change in mean velocity, but the temperature will drop.

The gas maintains the same amount of heat energy, but the temperature drops.

When there is less gas particles per volume, you require faster molecules for it to be the same temperature.

 

[Smacal1072]

If I had one mole of gas in a metre cubed container and then expanded this container so it was 2m^3. The gas particles would not change in mean velocity, but the temperature will drop.

The gas maintains the same amount of heat energy, but the temperature drops.

When there is less gas particles per volume, you require faster molecules for it to be the same temperature.

Striphe, that's not really true. For an ideal gas, temperature is related to the average kinetic energy of a gas particle. Yes, if you allow a gas to expand, it will do work and decrease it's temperature. However, less dense gases are not automatically cooler that more dense gases just by being less dense.

D H, you have explained that a tall vertical tube of gas at hydrostatic equilibrium will be cooler at the top, and warmer at the bottom.

The question Striphe is posing is: If we disturb this equilibrium (say, by momentarily enforcing the top and bottom to be the same temperature), and then allow the system to relax to equilibrium again, will the temperature gradient re-emerge?

 

[D H]

If I had one mole of gas in a metre cubed container and then expanded this container so it was 2m^3. The gas particles would not change in mean velocity, but the temperature will drop.

Not true. Google the term "free expansion". The gas has to do work in that expansion to lose energy. If it doesn't do any work, as is the case in a free expansion, there is no change in energy and hence no change in temperature.

When there is less gas particles per volume, you require faster molecules for it to be the same temperature.

No, you don't. Here is a fairly simple overview of the kinetic theory of gases: http://Galileo.phys.Virginia.EDU/classes/252/kinetic_theory.html.

The question Striphe is posing is: If we disturb this equilibrium (say, by momentarily enforcing the top and bottom to be the same temperature), and then allow the system to relax to equilibrium again, will the temperature gradient re-emerge?

Sure. It might take a long time, however. You have just choked off convection, so about all that is left is diffusion. Diffusion is a very slow process. This might be a part of klimatos' issue. As an atmospheric scientist, he views a situation in which the lapse rate is smaller than adiabatic as indicative of a stable atmosphere. There is little convection in such situations, almost none in the case of a temperature inversion. (That's why Los Angeles has such a problem with smog.)

Why am I so sure? For a fixed amount of total energy, entropy will reach a maximum under isentropic conditions. The second law of thermodynamics dictates that this is the equilibrium condition of this isolated system.

 

[striphe]

Thanks for the reading, i guess you can miss even the most fundamental of things when your using the internet alone to learn.

I assume that at the top of the gas column, there is going to be less of a concentration of particles as well as a lower pressure.

If they are in proportion, a static column of gas in equilibrium will not have a heat gradient. P/n = mv^2 so if the pressure is half at the top compared to the bottom and the concentration of molecules is half that of the bottom, both the top and the bottom have the same temperature.

This doesn't really agree with my visualisations of a column with very few particles in it. If you think of them as super elastic bouncy balls in the column. One particle bounces of the bottom and heads straight to the top. Even thought it has enough energy to hit the top, it hits with less velocity than it hits the bottom (it has done work to reach the top and has cooled). So any reading of temperature, which is based on the velocity of the molecules will be less as you head up. This would be rather easy to quantify also.

 

[Smacal1072]

The question Striphe is posing is: If we disturb this equilibrium (say, by momentarily enforcing the top and bottom to be the same temperature), and then allow the system to relax to equilibrium again, will the temperature gradient re-emerge?

Sure. It might take a long time, however. You have just choked off convection, so about all that is left is diffusion. Diffusion is a very slow process. This might be a part of klimatos' issue. As an atmospheric scientist, he views a situation in which the lapse rate is smaller than adiabatic as indicative of a stable atmosphere. There is little convection in such situations, almost none in the case of a temperature inversion. (That's why Los Angeles has such a problem with smog.)

Why am I so sure? For a fixed amount of total energy, entropy will reach a maximum under isentropic conditions. The second law of thermodynamics dictates that this is the equilibrium condition of this isolated system.

I guess that's the point of confusion then. I did a little digging, and found an excerpt from Maxwell's "Theory of Heat" (pg 300-301):

...If two vertical columns of different substances stand on the same perfectly conducting horizontal plate, the temperature of the bottom of each column will be the same; and if each column is in thermal equilibrium of itself, the temperatures at equal heights must be the same. In fact, if the temperatures of the tops of the two columns were different, we might drive an engine with this difference of temperature, and the refuse heat would pass down the colder column, through the conducting plate, and up the warmer column; and this would go on till all the heat was converting into work, contrary to the second law of thermodynamics.

This sort of summarizes one of the apparently paradoxical situations we thought up earlier. (I honestly just stumbled upon this today). Since we know that 2 columns of different substances "at equilibrium of itself" do have temperature gradients, it falls into Maxwell's paradox above.

 

[D H]

This sort of summarizes one of the apparently paradoxical situations we thought up earlier. (I honestly just stumbled upon this today). Since we know that 2 columns of different substances "at equilibrium of itself" do have temperature gradients, it falls into Maxwell's paradox above.

So maybe we have found a way to defeat the second law of thermodynamics!

Or maybe not. Something is going to happen to kill this idea. There are a bunch of things that can work against this. The second law of thermodynamics will eventually remain the unbeaten champ.

One thing that can happen is that the gases in the columns will simply stop following an adiabatic lapse rate curve. Look at the Earth's atmosphere. Temperature nominally drops with increasing altitude from the ground up to the tropopause, where temperature starts climbing with increased altitude. The reason is the ozone layer. Absorbing all that ultraviolet light heats up the stratosphere. The column that is cooler at the top (with my suggested hydrogen and xenon columns, it would be the xenon column) will set up such a temperature profile. The heat engine will stop.

Suppose we use this temperature difference very sparingly. We don't need a practical way to defeat the second law here; we only need a theoretical way to defeat it. So, we'll draw heat from the system so slowly that this temperature inversion doesn't appear. I don't know if this would work, and even if it does, it still won't work to defeat the second law. Eventually one of the columns will stop acting like an ideal gas. The temperature at the top of one of the columns will get too close to the boiling point.

 

[willem2]

So maybe we have found a way to defeat the second law of thermodynamics!

Or maybe not. Something is going to happen to kill this idea. There are a bunch of things that can work against this. The second law of thermodynamics will eventually remain the unbeaten champ.

A spontaneously forming temperature gradient really breaks the second law.
You need just a single column that has a gradient, and is supposedly at termal equilibrium,
and that can't exchange heat with the environment. Just put an object on the ground. Since the entire column except for the bottom is colder, the object will radiate heat and become colder than its surrounding.
Now we can use this temperature difference to produce some current with a thermocouple, we lead this current out of the column, and use it to heat something that's hotter than the entire column. We have now heated something hot, with something cold and have used no work.

It's true that this can't go on forever, because we're only breaking the second law and not the first, so the air will get colder and colder.

The simplest explanation is that the temperature gradient doesn't form.

see "On a paradox concerning the temperature distribution of an ideal gas in a gravitational field" paper or this:

http://www.vttoth.com/barometric.htm"

 

[striphe]

so what gets your backing?

(1) a temperature gradient will not appear in a body of gas.

(2) a temperature gradient will appear in a body of gas, but the same kind of gradient will appear in anybody of gas.

(3) a differing temperature gradient will appear in different bodies of gas of the same hight. but for other reasons it can not lower entropy.

Why does it get your backing?

 

[willem2]

so what gets your backing?

(1) a temperature gradient will not appear in a body of gas.

(2) a temperature gradient will appear in a body of gas, but the same kind of gradient will appear in anybody of gas.

(3) a differing temperature gradient will appear in different bodies of gas of the same hight. but for other reasons it can not lower entropy.

Why does it get your backing?

(1) - because it violates the 2nd law
- Because of the two articles I mentioned

 

[D H]

A spontaneously forming temperature gradient really breaks the second law.

No, it doesn't, no more than does the existence pressure gradient. The same reasoning can be used to argue that a pressure gradient can't exist, either. Both arguments are fallacious.

see "On a paradox concerning the temperature distribution of an ideal gas in a gravitational field" paper or this:

http://www.vttoth.com/barometric.htm"

The key phrase is in that page is this (emphasis mine):

The condition for equilibrium is that the integral ∫ dS be maximal, which is a problem from the calculus of variations. Unfortunately, I was not able to obtain a solution this way.

There is a way to obtain a solution this way, for an ideal gas at least. Rather than relying on fallacious reasoning, why not use the relationship between energy and entropy for an ideal gas?

 

[striphe]

I really have difficulty seeing how a heat gradient can be compatible with the Clausius statement. You could add heat energy at the cooler top and watch the heat energy passively spread to the hotter bottom. Unless someone can explain why that wouldn't happen or how I am misinterpreting the Clausius statement.

 

[willem2]

No, it doesn't, no more than does the existence pressure gradient. The same reasoning can be used to argue that a pressure gradient can't exist, either. Both arguments are fallacious.

I'm sorry, but I can't find no reason why that is in any of your posts

The key phrase is in that page is this (emphasis mine):

The condition for equilibrium is that the integral ∫ dS be maximal, which is a problem from the calculus of variations. Unfortunately, I was not able to obtain a solution this way.
There is a way to obtain a solution this way, for an ideal gas at least. Rather than relying on fallacious reasoning, why not use the relationship between energy and entropy for an ideal gas?

I'm sorry, but I've been unable to do that, and I've not seen you do it.
Not that he was able to find the solution in another way.


Your reasoning with regards to the argument that a temperature differential breaks the 2nd law seems to go as follows.

propostion A: a temperature difference with altitude develops in an isolated column of gas

proposition B: this can be exploited to extract energy from the gas

proposition C: this would mean that the 2nd law is false.

now obviously, if A and B are true C must be true, and it's actually wrong, but for
some reason that I don't understand you reject B instead of A.

 

[D H]

I really have difficulty seeing how a heat gradient can be compatible with the Clausius statement. You could add heat energy at the cooler top and watch the heat energy passively spread to the hotter bottom. Unless someone can explain why that wouldn't happen or how I am misinterpreting the Clausius statement.

I assume you are talking about this statement, striphe, emphasis mine:

Heat generally cannot flow spontaneously from a material at lower temperature to a material at higher temperature.​

The way to read the word "spontaneously" is "without work". This makes for a loophole in the second law. Here is another form of the Clausius statement:

No process is possible whose sole result is the transfer of heat from a cooler to a hotter body.​

When you heat the vertical column from the top you are adding energy to the system. After reaching equilibrium, the center of mass of the gas will be a bit higher than is was before you added heat to the top of column. Some of the energy that you added went into raising the center of mass. The transfer of heat from the cooler body (the top of the column) to the hotter body (the bottom of the column) was not the sole result of the process. Some of that added energy went into the form of an increase in potential energy. Another way to look at it: By raising the center of mass, work has been done on the system.

Both of the above statements of the second law of thermodynamics were qualitative. Here is a quantitative expression of the second law:

The entropy of an isolated system is maximized at equilibrium.​

The meaning of "isolated system" here is one that has constant mass and constant total energy. Our vertical tube is isolated in the this sense because gravitation is a conservative force. The equilibrium condition is that given by the adiabatic lapse rate, not a constant temperature.

 

[D H]

I'm sorry, but I've been unable to do that, and I've not seen you do it.

1. Ergodicity. Distributing the entropy evenly over all of the particles of the gas will result in a minimum or maximum of the entropy. In this case it is a maximum. The derivation of the adiabatic lapse rate assumes entropy is evenly distributed over all of the gas particles.

2. Principle of minimum potential energy. This yields the potential temperature being constant throughout the column, and that is once again equivalent to the adiabatic lapse rate conditions.

Not that he was able to find the solution in another way.

He didn't. What he did was flat out wrong.

Reason 1: This diagram.

[PLAIN]http://www.vttoth.com/images/area.gif

What's wrong with it? He already knows that the pressure varies with altitude. That diagram implicitly assumes constant density. Constant density, constant temperature, and non-constant pressure violate the ideal gas law (which he also assumed).


Reason 2: The mass of the atmosphere is finite.
Extend this column out to infinity. A constant temperature combined with an inverse square gravity field and a gas in hydrostratic equilibrium will result in an infinite mass.

Your reasoning with regards to the argument that a temperature differential breaks the 2nd law seems to go as follows.

I never said anything of the kind. I said that a temperature differential is a direct consequence of the second law of thermodynamics.

Stop putting words in my mouth.

 

[willem2]

Suppose you start with an column, at constant temperature. If a temperature differential form, the gas at the bottom should expand some, while the gas at the top contracts. You actually need workt, because the gas will rise on average. You'll have to do a better job of explaining why this is against the second law.

 

[D H]

Suppose you start with an column, at constant temperature. If a temperature differential form, the gas at the bottom should expand some, while the gas at the top contracts. You actually need workt, because the gas will rise on average. You'll have to do a better job of explaining why this is against the second law.

That's exactly backwards, willem.

If you start with a column at a uniform temperature with a pressure gradient given by the hydrostatic equilibrium equation, the column will go to equilibrium with the temperature at the bottom of the column higher than the initial temperature. The temperature at the top of the column (assuming the gas now reaches the top of the column) will be lower than the initial temperature.

The end result is that mass-weighted mean temperature of the column will be higher than the initial temperature while the center of mass of the gas will be lower than the initial center of mass. This is precisely in line with the principle of minimum potential energy.

 

[willem2]

Suppose you start with an column, at constant temperature. If a temperature differential form, the gas at the bottom should expand some, while the gas at the top contracts. You actually need workt, because the gas will rise on average. You'll have to do a better job of explaining why this is against the second law.

It's interesting to compare this with the case where a temperature difference is created by forced convection. Suppose you have two air tanks with an altitude difference between them and two tubes between them so you can have the air cycle around.
If you put a pump in one of the tubes and start pumping the air around, a temperature difference DOES develop, because the rising air expands and cools (it's not expanding in a vacuum), and the descending air heats up.
This also takes energy, because the air in the rising tube will be colder, and the air in the descending tube will be warmer, until the lapse rate is equal to the adiabatic lapse rate everywhere.
If we now try to exploit the temperature difference to extract energy, this will make the air on top warmer, and the air at the bottom colder, and you get warmer and lighter air in the descending tube and colder and heavier air in the rising tube again, so the pump has to do work.

 

[D H]

willem: The standard explanation of how stars form and then ignite is that the gravitationally-induced temperature gradient inside a protostar is a direct consequence of the second law of thermodynamics. The temperature gradient inside the atmosphere, or inside an isolated vertical column of gas in equilibrium, arises from the same phenomena.

How do you explain star formation and ignition without this gravitationally-induced temperature gradient?

 

[willem2]

willem: The standard explanation of how stars form and then ignite is that the gravitationally-induced temperature gradient inside a protostar is a direct consequence of the second law of thermodynamics. The temperature gradient inside the atmosphere, or inside an isolated vertical column of gas in equilibrium, arises from the same phenomena.

How do you explain star formation and ignition without this gravitationally-induced temperature gradient?

Easy, If you start out with a diffuse globe of gas and let it contract, it will heat up most
in the center, because you have the highest pressure there, and also because the heat
will radiate away from the outside. There is gravitational energy to create this temperature difference.

If you would use an isolated static column of air that starts out at the same temperature and
pressure, you would indeed get a temperature difference as the air sinks to the bottom
and the pressure gradient is established. After this, diffusion and radiation will equalize
the temperature, and some energy will get released during this process, because the gas at the bottom will cool and shrink somewhat and the gas at the bottom will expand, so the center of gravity of the gas will be lower.