Can Altitude Hypothesis Challenge the Second Law of Thermodynamics?

[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! :tongue2:

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! :tongue2:

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.

 

[D H]

Apologies to all.

I've done the nasty math and programming, and the constant temperature wins.A system in hydrostatic equilibrium that follows an adiabatic lapse rate does represent a local max in entropy, but a local max is not the same as a global max. A system in hydrostatic equilibrium with a constant temperature profile also represents a local max in entropy. For a vertical gas system with a fixed total energy and mass that is in hydrostatic equilibrium, the constant temperature configuration has a higher entropy than the adiabatic lapse rate configuration.

So why do we see a lapse rate in the atmosphere? First off, this state is a local max in entropy, and a rather strong one at that. That the troposphere is heated from the bottom / radiates out into space (real gases radiate; ideal gases don't) means that this local max is in many cases the favored state. Getting to the global max from this local max requires going through states that are disfavored entropically.

 

[striphe]

I've had a bit of a rethink about this and I'm not getting something.

T= (mv^2)/k

Naturally a particle will be affected by gravity, so as its velocity moves against gravity, it will slow and if ones velocity has it moving with gravity it will speed up. T= (mv^2)/k so changes in velocity result in changes of temperature.

How is the velocity even throughout the vessel if the particles are affected by gravity?

 

[klimatos]

I've had a bit of a rethink about this and I'm not getting something.

T= (mv^2)/k

Naturally a particle will be affected by gravity, so as its velocity moves against gravity, it will slow and if ones velocity has it moving with gravity it will speed up. T= (mv^2)/k so changes in velocity result in changes of temperature.

How is the velocity even throughout the vessel if the particles are affected by gravity?

Gas temperatures measure mean molecular kinetic energies of translation and not individual molecular energies. They are proportional to the mean of the squares of the individual axial velocities. These individual molecular energies follow a Maxwell-Boltzmann distribution. As long as conditions are isothermal, the mean values in any portion of the vessel will be the same as in any other portion. This isothermal condition requires that the incessant downward pull of gravity be matched by the average of the highly intermittent upward forces of thermal agitation.

By the way, in your formula m is the mean molecular impulse mass (not quite the same as the mean molecular mass for atmospheric air) and v is the root-mean-square of the axial molecular velocity, not the true-path velocity.

Since, under conditions of equilibrium, the mean axial velocity is zero, this makes v the standard distribution of the axial velocity distribution.

 

[striphe]

What are these thermal agitations?

 

[klimatos]

What are these thermal agitations?

The various molecular motions (translation, rotation, vibration, libration) usually subsumed under the concept of molecular kinetic energies.

Generally speaking, however, gas temperatures are usually considered to be functions only of the molecular kinetic energies of translation; i. e., T = mv^2/k. Here, Boltzmann's Constant (k) is the constant of proportionality that relates Kelvins to the KMS system of units.

 

[striphe]

Now to help me understand this, i will need a time based description of how this occurs in a particular instance.

Imagine a planet covered in an atmosphere of ideal gas. On this planet we have our super-long insulated upright tube, but at this moment, its a vacuum; nothing is inside of it.

To fill it with gas, the top cap is opened and gas flows into it. Once the pressure inside equalizes, it is re-sealed.

Now the fact that we have had gas move with gravity, the particles that are lower down, have gained more heat energy due to the conversion of potential energy, into kinetic energy and are hotter 'at the moment at least'.

now that it is left to become static how does temperature even out?

I find it hard to understand how, when a particle from a lower position, will loose energy to gravity when it changes location, to collide with a particle at a higher position.

 

[klimatos]

Now to help me understand this, i will need a time based description of how this occurs in a particular instance.

Imagine a planet covered in an atmosphere of ideal gas. On this planet we have our super-long insulated upright tube, but at this moment, its a vacuum; nothing is inside of it.

To fill it with gas, the top cap is opened and gas flows into it. Once the pressure inside equalizes, it is re-sealed.

Now the fact that we have had gas move with gravity, the particles that are lower down, have gained more heat energy due to the conversion of potential energy, into kinetic energy and are hotter 'at the moment at least'.

now that it is left to become static how does temperature even out?

I find it hard to understand how, when a particle from a lower position, will loose energy to gravity when it changes location, to collide with a particle at a higher position.

Through simple conduction of enthalpy (kinetic energy) and through thermal radiation exchange (if your ideal gas allows for it). Put two masses of air at different temperatures in physical contact and heat will flow from the warmer to the cooler. It will continue to do so until thermal equilibrium is reached.

Forget about individual molecular movements. It is the means that are significant.

 

[striphe]

every particle that moves up loses heat energy to gravity; as the mean velocity of a particular level has been determined by its fall, the average velocity of particles entering a particular level are going to be the same as the particles in that level.

em radiation would allow heat energy to move from one level to another without being affected by gravity in a classical sense, but the general theory of relativity says otherwise. Even light has gravitational potential energy.

 

[Delta Kilo]

I find it hard to understand how, when a particle from a lower position, will loose energy to gravity when it changes location, to collide with a particle at a higher position.

At equilibrium, the molecule will collide with more molecules on average when it is lower than when it is higher due to density gradient but the energy exchanged per collision will be the same on average.

Imagine a column of gas where thin layers are separated by weightless membranes. The molecules in each layer hit harder on the lower membrane than on the upper, but it is compensated by more molecules in the layer below so the net force on the membrane is zero.

 

[striphe]

Does anyone agree with the above statement?

 

[klimatos]

every particle that moves up loses heat energy to gravity; as the mean velocity of a particular level has been determined by its fall, the average velocity of particles entering a particular level are going to be the same as the particles in that level.

Since you seem to be hung up on gravitational changes to molecular velocities, let us approach the problem from that aspect. I am going to use examples from our own atmosphere rather than a hypothetical one because I am more comfortable there and that is where I have data at my fingertips.

At a temperature of 25° C and a pressure of 1000 hPa, the average air molecule will undergo some 3.18 billion collisions per second. The mean molecular speed along the vertical axis will be around 234 meters per second. A temperature differential of only 1° between two adjacent layers means an axial speed difference of some 40 cm per second. The gravitational acceleration or deceleration in one billionth of a second is infinitesimal.

Thus, thermal changes in molecular speeds of ascents and descents will far, far outpace gravitational ones.

 

[klimatos]

Does anyone agree with the above statement?

Delta Kilo properly stated the conditions for an ideal gas under conditions of equilibrium (except for gravity, of course). Keep in mind that both ideal gases and conditions of true equilibrium only exist in computer simulations and in the imaginations of scientists. That is what makes the study of atmospheric phenomena so interesting.

 

[RonL]

Does anyone agree with the above statement?

http://en.wikipedia.org/wiki/Non-equilibrium_thermodynamics

http://en.wikipedia.org/wiki/Equilibrium_(thermodynamics )

There seems to be a big divide between these two categories ? in how people look at system evaluation.

I think I see where you are in this quest for understanding, I am still trying to get things clear in my mind and just in the recent past found this wiki page on non-equilibrium, I have often used Tornado's, hurricane's, and different temperature air masses to try and make some point, only to have some rebuttal based on an equilibrium condition ?

I'm still trying to absorb so much, with still a long way to go.

Ron

 

[striphe]

At a temperature of 25° C and a pressure of 1000 hPa, the average air molecule will undergo some 3.18 billion collisions per second. The mean molecular speed along the vertical axis will be around 234 meters per second. A temperature differential of only 1° between two adjacent layers means an axial speed difference of some 40 cm per second. The gravitational acceleration or deceleration in one billionth of a second is infinitesimal.
.

Now I would absolutely agree that, if two bodies of gas come into contact with each other (one on top of the other) and at their point of contact, that a 1C temperature gradient existed, that heat would transfer from the hotter to the colder.

But when it it comes to the example I there is going to be some distance between a level that is a particular temperature and a level that is one degree less. The dry adiabatic lapse rate according to wiki for air is 9.8C/Km.

 

[Delta Kilo]

You just have to choose whether you talk about real Earth atmosphere or some idealized model, and if it is the latter, then which one in particular.

My remark was for system in static equilibrium which is obviously quite different from adiabatic lapse rate condition. Real Earth athmosphere is yet another different thing.

 

[stevmg]

Simple explanation -

Air particles (N2, O2) move at whatever speed they do on the ground. That is reflected in the air temperature on the surface.. In order for them to get to a higher altitude, their kinetic energy is converted (somewhat) into potential energy. By reducing their kinetic energy the temperature of the gas mixture would drop somewhat (5 degrees Fahrenheit for every 1000 feet elevation.) Analogous to throwing a ball into the air. The kinetic energy which is its speed will drop as the ball goes up and will pick up as the ball descends.

Some of our idiot friends who shoot off their guns on New Year's Eve so-called harmlessly into the air forget "whatever goes up, must come down" and it will come down at almost the same speed it went up at (less air drag) and that bullet can still kill. We have numerous deaths in Miami every New Year's Eve because of that.

 

[stevmg]

Oh, by the way, what do they call the inter-molecular gravitational forces which would theoretically slow down the expansion of a gas into a pure vacuum? With an ideal gas there would be no loss in kinetic energy of the expanding molecules (because they would do no work) and the temperature would remain the same.

I do understand that radiant ("electromagnetic" energy) would deplete as to the square of the radius of expansion (like light does) but not the knietic energy of motion. Although, Einstein did relate kinetic energy to mass m = m0/[sq rt (1 - v^2/c^2)] which would alter things a bit.

Are these called "Newtonian forces?"

 

[striphe]

Stevmg, I can not see how the velocity of particles could be the same high up as they are down low within this closed system. Like a ball or bullet, a gas particle is affected by gravity.

When a gas particle moves by its own velocity from a lower position to a higher position within the closed system, kinetic energy is converted to gravitational potential energy and hence the velocity of the particle is reduced.

When a gas particle moves by its own velocity from a higher position to a lower position within the closed system, gravitational potential energy is converted to kinetic energy and hence the velocity of the particle has increased.

As I have said the only possibility of the closed system having no temperature gradient lies with em radiation.

As the the kinetic/gravity interactions would create a temperature gradient, a greater quantity of em radiation will be emitted from the lower particles than the higher particles due to their temperature difference, this will work against the highlighted effects of gravity.

I see it as being unlikely that the em radiation could neutralise the effects of gravity.

When we add upon a modern understanding of physics, photons that move from a lower position to a higher position, in a gravity field undergo red shift, which reduces their energy. Photons that move from a higher position to a lower position, in a gravity field undergo blue shift, which increases their energy.