Can Altitude Hypothesis Challenge the Second Law of Thermodynamics?

[D H]

So I guess we are back at part B of the hypothesis.

Which hypothesis? This one?

If I break the hypothesis into two parts:
(a) A contained body of gas that is within a field of gravity will have differing temperature differing locations within the body.
(b)These differing temperatures can be utilised to by heat engine, to convert heat energy into other forms.

You haven't specified how you are going to utilize this temperature difference. No matter how you do it, you will not be violating the laws of thermodynamics (which includes conservation of energy). Let's look at your two column system (posts 23 & 23). Suppose you have two isolated columns containing gases, each a couple of kilometers high. Fill one with hydrogen, the other with xenon, such that at the bottom of each column the pressure is 1 atmosphere and the temperature is 300 K. Note: The column of xenon can't be all that tall because xenon has a very low specific heat and therefore the temperature gradient will be phenomenally steep 61.96 K/km with g=9.81 m/s² throughout. (The temperature gradient in the hydrogen column will only be 0.6858 K/km).

With this, the temperatures at the top of the 2 km towers will differ by 122.55 K. Not a huge difference, but any difference will suffice for a heat engine. So, let's "break the seal" at the top of the columns to take advantage of this difference. We'll be transferring heat from the top of the hydrogen column to the top of the xenon column, stealing some of that transferred heat in the form of useful energy. What's going to happen in the columns? Simple: The lapse rates will no longer be adiabatic. The hydrogen column will have a super-adiabatic lapse rate while the xenon column will have a sub-adiabatic lapse rate. Eventually the two columns will stabilize with equal temperatures at the tops of the columns. Our heat engine of course will become worthless at this point.

Before this happens, let's see if we can take advantage of what is happening at the bottoms of the columns. The xenon column will be warmer at the bottom than will the hydrogen column. So, let's break the seal there as well and install another heat engine. Have we got a perpetual motion machine? Nope. Eventually we'll get equal temperatures at the top and the bottom as well. There ain't no such thing as a free lunch in thermodynamics.

Another what if game: Let's force the bottoms to have a common temperature of 300 K, forever. Now we can draw energy at the top of the column, forever. Is this a perpetual motion device? Nope. That forcing at the bottom requires an energy input, and this energy input will be greater than the amount of energy we can draw out at the top. So once again, no free lunch.

 

[striphe]

It was my consideration that if the bottom of the tubes were the same temperature than the top would have a different temperature and vise versa.

The environment at the bottom has a temperature of 300 K and we have a tube of hydrogen and one of xenon 2km high. Let's consider that these heat engines when inactive insulate the tubes perfectly and that the engines are activated at choice.

Initial conditions, bottom of tubes the same temperature and the tops vary. The heat engines connecting the tube extracts energy as they converge to the same temperature as heat moves from the hydrogen to the xenon. Once the temperatures have converged the heat engine at the top is deactivated. My consideration holds that the temperature of the hydrogen column at the bottom is colder than the environment and the xenon column has a temperature at the bottom that is hotter than the environment. The two heat engines at the bottom then are activated, with heat moving from the environment into the hydrogen column and heat moving from the xenon column into the environment.

The bottom of the columns reach the same temperature as the environment. The heat engines are deactivated at the bottom returning the devise to its initial conditions.

So when you consider that the devise operates in bursts, both mine and your considerations defy the second law of thermodynamics. Unless I've misunderstood.

 

[D H]

Still no violation of the second law. Here:

The bottom of the columns reach the same temperature as the environment.

What environment? By saying that you are making this a non-isolated system. Once you do that you have to account for energy and entropy transfer with the external environment.

While there are more detailed versions of the second law that take into account interactions with the external environment, the simplistic version of second law of thermodynamics embodied by $dS/dt \ge 0$, and Clausius' statement in particular, pertains to isolated systems only.

 

[striphe]

So the device would operate as I've depicted it, but wouldn't defy the second law of thermodynamics because it isn't an isolated system?

 

[D H]

I didn't say that. You have a column that is made of some substance that is a perfect insulator and reflector. Unobtainium, in other words.

It appears you are actively trying to find a way to violate the laws of conservation of energy and the laws of thermodynamics. If that is the case, this is not the site to carry on such discussions.

 

[striphe]

It's all theoretical. Giving things perfect properties just makes things less complex, but none the less relevant.

The heat gradient formed by gravity and the second law of thermodynamics, weren't compatible in my understanding and this post is an attempt to rectify my understanding.

No one has explained how these are compatible as of yet.

My understanding doesn't highlight a conservation of energy defiance, maybe this is key.
Can you elaborate on how this is defied in the thought experiment?

 

[D H]

But you are making things overly complex, striphe. Rube Goldberg devices do not in general help understanding. All they are good for is annoying students (some profs revel in coming up with overly complex problems) and befuddling patent examiners (the over unity devices that erroneously manage to receive a patent are almost inevitably overly complex devices in which it is hard to see the flaw).

 

[striphe]

So how would you display the issue to make it less complex and more approachable?

I've honestly tried to make it as simple as I could, but i understand that doesn't mean that it is the simplest it could be.

 

[striphe]

How is two long tubes and three heat engines overly complex?

 

[Smacal1072]

Which hypothesis? This one?



You haven't specified how you are going to utilize this temperature difference. No matter how you do it, you will not be violating the laws of thermodynamics (which includes conservation of energy). Let's look at your two column system (posts 23 & 23). Suppose you have two isolated columns containing gases, each a couple of kilometers high. Fill one with hydrogen, the other with xenon, such that at the bottom of each column the pressure is 1 atmosphere and the temperature is 300 K. Note: The column of xenon can't be all that tall because xenon has a very low specific heat and therefore the temperature gradient will be phenomenally steep 61.96 K/km with g=9.81 m/s² throughout. (The temperature gradient in the hydrogen column will only be 0.6858 K/km).

With this, the temperatures at the top of the 2 km towers will differ by 122.55 K. Not a huge difference, but any difference will suffice for a heat engine. So, let's "break the seal" at the top of the columns to take advantage of this difference. We'll be transferring heat from the top of the hydrogen column to the top of the xenon column, stealing some of that transferred heat in the form of useful energy. What's going to happen in the columns? Simple: The lapse rates will no longer be adiabatic. The hydrogen column will have a super-adiabatic lapse rate while the xenon column will have a sub-adiabatic lapse rate. Eventually the two columns will stabilize with equal temperatures at the tops of the columns. Our heat engine of course will become worthless at this point.

Before this happens, let's see if we can take advantage of what is happening at the bottoms of the columns. The xenon column will be warmer at the bottom than will the hydrogen column. So, let's break the seal there as well and install another heat engine. Have we got a perpetual motion machine? Nope. Eventually we'll get equal temperatures at the top and the bottom as well. There ain't no such thing as a free lunch in thermodynamics.

Another what if game: Let's force the bottoms to have a common temperature of 300 K, forever. Now we can draw energy at the top of the column, forever. Is this a perpetual motion device? Nope. That forcing at the bottom requires an energy input, and this energy input will be greater than the amount of energy we can draw out at the top. So once again, no free lunch.

In your last thought experiment, suppose instead of forcing the bottoms to have a common temperature of 300 K, we just connect the bottoms with a thermally conductive metal, like copper.

If we "run" the heat engine at the top at a rate slower than that at which the system comes to hydrostatic equilibrium, do we get a free lunch then?

 

[D H]

Whether a device is simple Carnot heat engine or a complex Rube Goldberg contraption, there is no free lunch in thermodynamics. Conservation of energy says ignoring losses, the best you can possibly do is break even. Entropy concerns say you cannot ignore losses.

There ain't no such thing as a free lunch -- particularly in thermodynamics.

 

[striphe]

Although its easy enough to say we won't get a free lunch, as the second law doesn't allow for free lunches. It seems excessively difficult to explain how such is true in with these hypothetical devises.

The principle of all these hypothetical devises are based on the seeming overlook (at least what I've seen) by thermodynamics to consider that a temperature differences will arise in a body of gas under the force of gravity or simulated gravity (centrifugal force).

Any number of devises that attempt to take advantage of this particular temperature difference could be conceived. I expect what ever argument against one will wipe them all out. But the inability of users to cope with either the 'complexity' of these devises and the fact that there is a temperature difference induced within a body of gas, has gotten physics no closer to dismissing these devises, with what I would consider a required explanation.

 

[klimatos]

The atmosphere's heat budget shows that it receives some 102 watts per square meter of solar insolation, some 268 watts of terrestrial radiation, roughly 17 watts by conduction from the warmer surface, and a final 3 watts by hydrologic cycling (water evaporates from the warm surface higher temperatures than it condenses at in the atmosphere). Thus, an average of 74% of its heat comes from the surface of the Earth.

When a parcel of air sinks, more molecules have a downward component of motion than have an upward component. The opposite is true when a parcel of air is forced up (nothing moves against the force of gravity unless pushed by a stronger force). Gravity accelerates downward molecular motions and decelerates upward motions. Changes in mean molecular velocities are measured as changes in temperature. Thus, sinking air warms and rising air cools.

 

[klimatos]

The principle of all these hypothetical devises are based on the seeming overlook (at least what I've seen) by thermodynamics to consider that a temperature differences will arise in a body of gas under the force of gravity or simulated gravity (centrifugal force).

Could you cite a source for this development of temperature differences in a parcel of gas at equilibrium under the influence of gravity as it sole outside force. Differences in density: yes. Differences in temperature: no way. At NTP, each gas molecule collides and exchanges kinetic energy more than five billion times a second. We measure this mean kinetic energy of translation as temperature. The mathematical probability that some significant potion of a gas would develop a temperature difference from the rest without some outside force other than gravity is incredibly small. Statistical mechanics tells us that for all intents and purposes it just isn't going to happen.

 

[striphe]

klimatos, my support of the temperature difference is based on the understanding that, if molecular collisions were the only way heat was being transferred within a body of gas, then particles moving down speed up and particles moving up slow down; resulting in a temperature gradient. But heat energy can be transferred through em radiation which wouldn't be affected by such a force in classical physics.

I consider it plausible that the em radiation may even out the temperature in a closed body of gas.

As D H has given quantitative calculations as to the temperature difference within a closed and static body of gas and that he is the most senior member of the forum posting on this thread; it gives me a lot of confidence that the temperature difference does arise.

When it comes down to it, I will sway to the side with the most evidence and rationale. Science and stubbornness are incompatible.

 

[D H]

Could you cite a source for this development of temperature differences in a parcel of gas at equilibrium under the influence of gravity as it sole outside force. Differences in density: yes. Differences in temperature: no way.

Yes way. It's called "lapse rate" (google that term). Here is a plot of the dry and moist adiabatic lapse rates:

http://www.fas.org/irp/imint/docs/rst/Sect14/moist_dry.jpg

Here is a largish (17.1 MB) reference:
http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19770009539_1977009539.pdf

Although its easy enough to say we won't get a free lunch, as the second law doesn't allow for free lunches. It seems excessively difficult to explain how such is true in with these hypothetical devises.

Since you are the one making the extraordinary claim, striphe, the burden of proof falls upon you to prove that such a device would allow a violation of the second law of thermodynamics. You are assuming that the gases in these columns will follow adiabatic conditions. They won't. You are forcing conditions to be other than adiabatic, so the temperature profile will be something other than adiabatic.

 

[Smacal1072]

Striphe, you might want to simplify your analysis by considering an ideal gas.

If you consider a monatomic, ideal gas (like Xenon), then you can assume that when a gas atom travels upward in the tube, it loses energy due to increasing it's gravitational potential energy. You can make your idea more concrete by calculating how much kinetic energy a particle loses by traveling upwards against gravity using something like 1/2kT - mgh (for some reason latex won't work...)

HOWEVER, you should read this paper:

"On a paradox concerning the temperature distribution of an ideal gas in a gravitational field" by S Velasco, F L Román and J A White.

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.

 

[klimatos]

Yes way. It's called "lapse rate" (google that term). Here is a plot of the dry and moist adiabatic lapse rates:

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.

 

[klimatos]

klimatos, my support of the temperature difference is based on the understanding that, if molecular collisions were the only way heat was being transferred within a body of gas, then particles moving down speed up and particles moving up slow down; resulting in a temperature gradient. But heat energy can be transferred through em radiation which wouldn't be affected by such a force in classical physics.

I consider it plausible that the em radiation may even out the temperature in a closed body of gas.

Under conditions of equilibrium, the number of molecules having a downward component of motion is essentially the same as the number of molecules having an upward component of motion. Gravitational energy gains match gravitational energy losses, and no temperature gradient emerges.

Heat (enthalpy) transfer within a gas may be accomplished by any or all of the three classical methods: molecule-to-molecule electromagnetic radiation, molecule-to-molecule conduction (collisions) and mass transfer (fluid flow). Under conditions of equilibrium, fluid flow is ruled out but the other two remain. Most of the heat transfer in the Earth's atmosphere is brought about by electromagnetic radiation rather than conduction.

 

[klimatos]

As D H has given quantitative calculations as to the temperature difference within a closed and static body of gas and that he is the most senior member of the forum posting on this thread; it gives me a lot of confidence that the temperature difference does arise.

Can you give me a posting number for these calculations? I would like to read them.

 

[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.