Does a Thermometer Truly Reflect Air Temperature in Kinetic Theory?

[klimatos]

Let’s take an ordinary household thermometer and expose it to the air. The molecules of the air impact on the surface of the thermometer and transfer kinetic impulses. The thermometer transfers kinetic impulses to the impacting molecules as well. Eventually, thermal equilibrium is reached and we say that the thermometer and the surrounding air are at the same temperature.

Now the thermometer can only measure the impulses transferred to it by the select sub-population of molecules that impact upon its surface. The kinetic energies of those molecules in the general population that do not impact upon the thermal surface are not measured.

Statistical mechanics and kinetic gas theory tell us that the mean kinetic energies of translation normal to the thermal surface of those molecules that impact upon the surface is exactly twice the mean kinetic energy of translation of the general population. This is because a molecule traveling at the speed of 5v is five times more likely to strike a given surface in a given period of time than one traveling at speed v. The sub-population has a different distribution of velocities and a different mean velocity that those of the general population.

Since the thermometer is at thermal equilibrium with the impacting sub-population and not with the general population, it would seem that the thermometer would be hotter than the air that it is supposed to be measuring!

Explanation, anyone?

 

[soothsayer]

"Statistical mechanics and kinetic gas theory tell us that the mean kinetic energies of translation normal to the thermal surface of those molecules that impact upon the surface is exactly twice the mean kinetic energy of translation of the general population.

Could you elaborate on this? I don't understand what you mean. Perhaps an equation?

 

[Andrew Mason]

Since the thermometer is at thermal equilibrium with the impacting sub-population and not with the general population, it would seem that the thermometer would be hotter than the air that it is supposed to be measuring!

Explanation, anyone?

The thermometer is in thermal equilibrium with all of the molecules in the air, not the faster subpopulation of molecules.

First of all, if the thermometer was in thermal equilibrium with a faster subpopulation of molecules of air, this would violate the second law of thermodynamics - heat would be transferred from the cooler air to the hotter thermometer.

You are making some odd assumptions about the transfer of molecular kinetic energy from the air to the thermometer. The distribution of energies of the molecules follows a Maxwell-Boltzmann curve. While it is true that a molecule traveling 5 x faster than the average will, on average, strike more frequently that slower molecules, the size of population of such molecules (determined by the Maxwell Boltzmann distribution) is relatively small.

AM

 

[klimatos]

Could you elaborate on this? I don't understand what you mean. Perhaps an equation?

The distribution functions for the general population and the distribution functions for the select subset are different--as are their means.

I am not yet fluent in LATEX and cannot do them in simple text. I use MathType in my own writings. The derivation of the two distribution functions and their means is given in:

edit: removed personal website

I apologize for not being able to post the work here for all to see.

 

[klimatos]

1. The thermometer is in thermal equilibrium with all of the molecules in the air, not the faster subpopulation of molecules.

2. First of all, if the thermometer was in thermal equilibrium with a faster sub-population of molecules of air, this would violate the second law of thermodynamics - heat would be transferred from the cooler air to the hotter thermometer.

3. You are making some odd assumptions about the transfer of molecular kinetic energy from the air to the thermometer. The distribution of energies of the molecules follows a Maxwell-Boltzmann curve. While it is true that a molecule traveling 5 x faster than the average will, on average, strike more frequently that slower molecules, the size of population of such molecules (determined by the Maxwell Boltzmann distribution) is relatively small. AM

1. How can a thermometer be in thermal equilibrium with molecules with which it has absolutely no contact of any kind?

2. There is no violation of the Second Law since heat can only be transferred to and from the sub-population. The only molecules that contact the thermometer are those of the sub-population. And they, of course, are in thermal equilibrium with the thermometer. It is even in radiative equilibrium with this sub-population. A photon emitted by the thermometer is more likely to be absorbed by a fast moving molecule that a slow one, and a fast moving molecule is more likely to occupy a space where it will have a "clear shot" at the thermometer than a slow one. Keep in mind that molecules are only part of this sub-population in the free path immediately prior to and the free path immediately subsequent to impact with the thermometer. At all other times they are an ordinary part of the general population.

3. For 5x the population is quite small, but this was just a convenient example. Almost every textbook on the kinetic theory of gases that I have read has a section on the difference between the general population and the sub-population that either passes through a plane or impacts on a surface. It's not something I invented. I'm just pointing out some of the peculiar consequences of this notion.

Klimatos

 

[Ken G]

1. How can a thermometer be in thermal equilibrium with molecules with which it has absolutely no contact of any kind?

If I may, thermodynamic equilibrium is not a relationship between subsets of systems, it is a relationship between entire systems. It includes, for example, the requirement that every subprocess be balanced by its inverse subprocess, when summed over the entire interacting systems. So it already accounts for what all the various subpopulations are doing. In other words, if the thermometer is mostly getting heat from the faster air molecules, then the faster air molecules are also mostly getting the heat that is coming back from the thermometer. It doesn't matter whether the fast or slow air molecules are getting most of the action, what matters is the balance or imbalance they are experiencing, and the concept of temperature already assumes the full distribution is the most statistically likely one, so it already includes what the more "inert" subpopulations are doing too.

2. There is no violation of the Second Law since heat can only be transferred to and from the sub-population. The only molecules that contact the thermometer are those of the sub-population. And they, of course, are in thermal equilibrium with the thermometer.

Again, isolated subpopulations cannot be in thermal equilibrium-- that equilibrium has to include every subpopulation or it is not thermal equilibrium.

3. For 5x the population is quite small, but this was just a convenient example.

Yes, it doesn't matter if the subpopulation is small or not, it can still be the dominant actor in whatever equilibrium is going on. For example, if you have a high density gas in ionization equilibrium at some temperature, it is not unusual for the electrons responsible for collisional ionizaton to be in the tail of the Maxwellian only. All the same, the ionization temperature is the temperature of the entire population, not the tail population-- the tail is necessarily part of the rest of the Maxwellian, or else it does not have a meaningful temperature.

 

[LeonhardEuler]

Let’s take an ordinary household thermometer and expose it to the air. The molecules of the air impact on the surface of the thermometer and transfer kinetic impulses. The thermometer transfers kinetic impulses to the impacting molecules as well. Eventually, thermal equilibrium is reached and we say that the thermometer and the surrounding air are at the same temperature.

Now the thermometer can only measure the impulses transferred to it by the select sub-population of molecules that impact upon its surface. The kinetic energies of those molecules in the general population that do not impact upon the thermal surface are not measured.

Statistical mechanics and kinetic gas theory tell us that the mean kinetic energies of translation normal to the thermal surface of those molecules that impact upon the surface is exactly twice the mean kinetic energy of translation of the general population. This is because a molecule traveling at the speed of 5v is five times more likely to strike a given surface in a given period of time than one traveling at speed v. The sub-population has a different distribution of velocities and a different mean velocity that those of the general population.

Since the thermometer is at thermal equilibrium with the impacting sub-population and not with the general population, it would seem that the thermometer would be hotter than the air that it is supposed to be measuring!

Explanation, anyone?

It's worth noting that whatever thermal state of the metal on the thermometer in equilibrium with the gas at, say 20C, it would be have to correspond to a temperature of 20C by definition. The temperature is not defined by the average velocity of the molecules that collide with the material.

 

[lalbatros]

klimatos,

You did not mention that:

• the thermometric fluid is in equilibrium with "a sub population" of glass
• the glass is in equilbrium with "a sub population" of thermometric fluid
• the glass is in equilibrium with "a sub population" of ambiant air
• the ambiant air is in equilibrium with "a sub population" of glass
• "the sub population" of thermometric fluid is in equlibrium with the main population of thermometric fluid
• "the sub population" of glass is in equilibrium with the main population of glass
• "the sub population" of ambiant air is in equilibrium with the main population of ambiant air

When taking all that into account, statistical mechanics fits by design with thermodynamics.

 

[soothsayer]

The distribution functions for the general population and the distribution functions for the select subset are different--as are their means.

I am not yet fluent in LATEX and cannot do them in simple text. I use MathType in my own writings. The derivation of the two distribution functions and their means is given in:


I apologize for not being able to post the work here for all to see.

Ok, got it, that statement wasn't clicking with me at first. If you're interested, there are some sites which will allow you to enter LATEX code via a graphic interface, much easier, and pretty fast. If you google "Latex code editor" you should find a few.

 

[Andrew Mason]

1. How can a thermometer be in thermal equilibrium with molecules with which it has absolutely no contact of any kind?

That is the nature of thermal equilibrium. You need to review some basic thermodynamics. The paper you cited is full of obvious errors.

2. There is no violation of the Second Law since heat can only be transferred to and from the sub-population.

The sub-population of molecules that strikes the thermometer consists of molecules with the same distribution of speeds that is found in the general population. Otherwise, the air would not have a temperature (the air would not be in thermal equilibrium).

3. For 5x the population is quite small, but this was just a convenient example. Almost every textbook on the kinetic theory of gases that I have read has a section on the difference between the general population and the sub-population that either passes through a plane or impacts on a surface. It's not something I invented. I'm just pointing out some of the peculiar consequences of this notion.

Then give us a cite. If the sub-population is in thermal equilibrium with the surrounding air, it doesn't matter. The sub-population has the same distribution of speed as the rest of the air.

Your premise that the thermometer will get hotter than the surrounding air is a very testable theory. It is odd that no one has actually found it to be true. It has been unequivocally shown that heat does not flow from cold to hot by itself.

AM

 

[klimatos]

Explanation, anyone?

Dear fellow posters,

Time’s up! Here’s the solution to the Ideal Gas Temperature Paradox.

Firstly, I would like to thank all of you who responded—even those who thought me an ignoramus who never should be allowed to post. I am an ignoramus. There are whole worlds of knowledge of which I know little or nothing. I just keep chipping away as best I can.

Secondly, much of the confusion stems from my posting the problem from the micro-scale perspective—the perspective of kinetic gas theory and statistical mechanics. This perspective is one that I thought most physical scientists were familiar with, but it appears I was mistaken. Many of you chose to approach it from the macro-scale perspective of thermodynamics. The two perspectives are compatible; but we get into trouble when we try to mix them.

Most of you saw the logical end result: the thermometric surface has to be in thermal equilibrium with the entire population of gas molecules. However, you could not resolve that principle with the fact (and it is a fact) that the average speed with which those molecules strike the thermometric surface is higher than the average speed of the general population at other times. The resolution of this seeming paradox is simple; but it has to be couched in micro-scale terms.

Thirdly, here is the solution. The sub-population that impacts on the thermometric surface is the same population that does not impact on the thermometric surface—just at different times! An individual molecule belongs to the sub-population only during its free paths prior to and subsequent to impact. At all other times, it is a part of the general population. Hence, the thermometric surface is in thermal equilibrium with the entire population of gas molecules.

Let me illustrate. Let us put our ideal gas inside of a closed cubic container, one of whose sides is thermometric. Let us postulate an infinite series of equally-spaced imaginary planes inside that cube and parallel to the thermometric surface.

Now let us take a single molecule, Moe. We can simplify things further by considering only a single arm of a single axis of movement: normal to and towards the thermometric plane. Over the course of a second under equilibrium conditions Moe will change his speed along that axis billions of times. Each one of those changes results in a specific speed. Graph those speeds against the frequency with which those speeds occur, and you will get a single-dimensional Maxwell-Boltzmann curve. That curve will have a mean speed (v). That speed is the mean speed of all molecules having a component of motion toward the thermometric plane.

A moment’s thought will show that when Moe was traveling at speed 2v he passed through twice as many planes as when he sped along at speed v. Similarly, at speed 3v he passed through three times as many—and so on and so forth. Hence, the probability of Moe’s passing through (or impacting upon) any given single plane is proportional to Moe’s speed at the time.

Consequently, the mean speed with which a population of molecules passes through a given plane or impacts upon a given surface is greater by some factor than the mean speed (v). Statistical mechanics gives that factor the value of π/2.

Average all of Moe’s travels toward the thermometric plane and you get the mean speed (v). Average only those travels in which he impacts on the thermometric plane and you get (π/2v).

Paradox resolved.

 

[JaredJames]

Paradox resolved.

There was no paradox and you were given the answer. Including the above post telling you the paper you cited had errors.

 

[klimatos]

1. The sub-population of molecules that strikes the thermometer consists of molecules with the same distribution of speeds that is found in the general population.

2. Your premise that the thermometer will get hotter than the surrounding air is a very testable theory. AM

1. Not true. See my post #11 or check any good textbook on kinetic gas theory.

2. I never said the thermometer would get warmer. It can't, it is in thermal equilibrium with those molecules that impact on its surface. It is that kinetic energy, not the kinetic energy of the general population that the thermometer actually measures. The rest is just a matter of calibration.

 

[Andrew Mason]

1. Not true. See my post #11 or check any good textbook on kinetic gas theory.

You cite this paper which at page 5 says:

"That is, the thermal energy of a system is equal to the kinetic energy of translation
of that system. This is because ideal gases have no internal energies."

I am not sure how you defend that statement. An ideal gas at any temperature above absolute 0 certainly has internal energy.

The thermal energy of a system is the internal energy of the system. The internal energy takes into account all of the kinds of motion of molecules (translational, rotational, vibrational). The temperature depends upon the translational kinetic energy of the molecules in the gas. The temperature is determined by the Maxwell-Boltzmann distribution of those translational kinetic energies. Temperature is not the same as thermal energy.

2. I never said the thermometer would get warmer. It can't, it is in thermal equilibrium with those molecules that impact on its surface. It is that kinetic energy, not the kinetic energy of the general population that the thermometer actually measures. The rest is just a matter of calibration.

Perhaps you could support your "theory" with some actual data and perhaps an authoritative source.

AM

 

[Ken G]

Most of you saw the logical end result: the thermometric surface has to be in thermal equilibrium with the entire population of gas molecules. However, you could not resolve that principle with the fact (and it is a fact) that the average speed with which those molecules strike the thermometric surface is higher than the average speed of the general population at other times.

No, that presented no difficulty. Yes it's true that the average speed of the interacting particles is higher than the average speed of the full population, that is quite a routine state of affairs and presents no difficulties.

Thirdly, here is the solution. The sub-population that impacts on the thermometric surface is the same population that does not impact on the thermometric surface—just at different times!

No, that is not the solution, because it does not assert a required element of the situation. It is not required that Moe, or any other particle, sample all the speeds of the full distribution. You would get exactly the same answer if Moe always had the speed he has, and every other particle always have the speed they have, as long as the full distribution is Maxwellian. In thermodynamic equilibrium, every process is balanced by its own inverse process, so if Moe has speed 3v, then the rate of processes that make Moe lose a little energy, say down to 2.9v, is balanced by the rate of processes that make the particles at 2.9v jump up to 3v. So on balance, we can just treat Moe as if he never changed from 3v, and everything is exactly the same as you get in thermodynamic equilibrium.

 

[Ken G]

"That is, the thermal energy of a system is equal to the kinetic energy of translation
of that system. This is because ideal gases have no internal energies."

I believe what that quote is intended to mean is that "translational energy" is just the kinetic energy of the particles, not bulk translation of the gas, and "internal energy" is internal to the particles, not to the gas as a whole. It's awkward language, but it's not completely wrong because it applies to monoatomic ideal gases with fixed excitation and ionization. However, it is certainly sloppy language because a gas can be ideal and still have internal degrees of freedom, like ideal diatomic molecules. The real definition of an ideal gas is no interactions between the molecules other than elastic collisions once equilibrium is achieved, so the molecules can harbor internal energy but they thermalize to the same temperature as the translational modes.

 

[Andrew Mason]

I believe what that quote is intended to mean is that "translational energy" is just the kinetic energy of the particles, not bulk translation of the gas, and "internal energy" is internal to the particles, not to the gas as a whole.

You are being very charitable. If that was the case, why does he reference to "thermal energy" ?

AM

 

[Ken G]

The thermal energy is the sum of all the translational energies of the particles, it's not translation of the whole system. That would indeed be the thermal energy of an ideal gas of particles with no internal degrees of freedom (though of course that is not the actual definition of an ideal gas).

 

[Andrew Mason]

The thermal energy is the sum of all the translational energies of the particles, it's not translation of the whole system.

It is the sum of all translational, rotational and vibrational energies of the molecules in thermal equilibrium (ie. energies with respect to all degrees of freedom) as measured in the frame of the centre of mass of the system. The translational energy of the centre of mass of the system is mechanical energy, not thermal energy.

That would indeed be the thermal energy of an ideal gas of particles with no internal degrees of freedom (though of course that is not the actual definition of an ideal gas).

I don't understand what that means. Thermal energy means energy associated with molecular motions of a substance that is in thermodynamic equilibrium.

AM

 

[Ken G]

It is the sum of all translational, rotational and vibrational energies of the molecules in thermal equilibrium (ie. energies with respect to all degrees of freedom) as measured in the frame of the centre of mass of the system.

Yes, that's why I specified that the quote was only true for monatomic ideal gases, which don't have rotational or vibrational modes. That's what is meant by internal degrees of freedom-- internal to the particles, not internal to the gas.

The translational energy of the centre of mass of the system is mechanical energy, not thermal energy.

Yes, the source of the quote appears to be aware of that, which is why they were never talking about translational energy of the center of mass. The only error, beyond being quite unclear in their meaning, was in overlooking multi-atomic ideal gases.

 

[klimatos]

I believe what that quote is intended to mean is that "translational energy" is just the kinetic energy of the particles, not bulk translation of the gas, and "internal energy" is internal to the particles, not to the gas as a whole. It's awkward language, but it's not completely wrong because it applies to monoatomic ideal gases with fixed excitation and ionization. However, it is certainly sloppy language because a gas can be ideal and still have internal degrees of freedom, like ideal diatomic molecules. The real definition of an ideal gas is no interactions between the molecules other than elastic collisions once equilibrium is achieved, so the molecules can harbor internal energy but they thermalize to the same temperature as the translational modes.

Earlier in the paper, the ideal gas under consideration was defined to consist of rigid spheres with no internal energies. There are many other kinds of ideal gases, of course.

 

[Ken G]

Earlier in the paper, the ideal gas under consideration was defined to consist of rigid spheres with no internal energies. There are many other kinds of ideal gases, of course.

It was clear from context that is what the author intended, but all the same, careful writers simply do not make false statements like "This is because ideal gases have no internal energies." But I grant you that none of this is relevant to the issue at hand, we can certainly choose to deal with ideal gases with no internal energies if we want, and thermodynamic equilibrium can still involve a subpopulation that is active in heat exchange while the larger population is not, with no difficulties presented to any statistical mechanical theorems. Indeed, we can consider a very simple situation, where two identical gases with Maxwellian velocity distributions at the same temperature and density are separated by a thin film. We might stipulate that only particles with speeds 100 times the thermal speed can penetrate the film, while the rest bounce off. These two gases will be in thermal equilibrium, it makes no difference that only the fastest particles communicate heat between the gases, and it makes no difference if the particles can ever change speed. Replacing one of the gases with a thermometer also makes no fundamental difference once equilibrium is reached-- issues like the speed of the particles that are active in transferring heat are only relevant to the approach to equilibrium, where such details will indeed matter a lot.