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Temperature of gas and solid in thermal equilibrium

  1. Nov 21, 2014 #1
    Hello

    I got to thinking about temperature when i prepared a lecture for my high school class. For a monoatomic ideal gas the temperature is proportional with the average kinetic energy of the atoms. For a diatomc gas we also have rotational energy (vibrations assumed to be freezed out), and for a polyatomic gas we have kinetic, rotaional and vibrational energy. For a solid we have vibrational energy.

    Lets assume we take a solid and a monoatomic gas and put into contact, and they reach thermal equilibrium. The monoatomic gas will have an average energy per atom of [tex] \frac{3}{2}k_b T [/tex] and the solid would have an average energy per atom of [tex] \frac{6}{2}k_b T =3 k_b T[/tex] (in the classical limit).

    The gas and solid being in thermal equilibrium will have the same temperature. If we take the simple definition of temperature, that two objects have the same temperature if there is no net energy transfer between them, there is something I can't rap my head around:

    Why will the solid not transfer its energy to the gas, when it has a higher average energy? I know I'm probably just forgetting some simple fact, but I just can't seem to figure it out somehow. It has never troubled me before, but when you teach, you suddenly stumble across things you thought was logical, but when you really look at it, it may not be.

    Anybody that can help me with a simple picture, that can help me see this i logical (preferable a explanation high school student would appriciate too, but any help is appreciated)

    Thanks in advance

    Anders
     
  2. jcsd
  3. Nov 21, 2014 #2

    Bystander

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    "Vibrational" energy includes what two components?
     
  4. Nov 21, 2014 #3
    Not sure I understand what you mean. There are 3 vibrational direction. Because of a kinetic and potential term, vibrational energy in a solid will contribute 3kbT
     
  5. Nov 21, 2014 #4
    Before they are brought into contact, they each had thermal energy to begin with. So when they equilibrate with one another, it is only the changes in thermal energy that matter, not the absolute amounts. They don't have to have equal amounts of thermal energy in the end. At equilibrium, the energies don't have to match, only the temperatures. Also, you left out the number of atoms of each one.

    Chet
     
  6. Nov 21, 2014 #5

    Bystander

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    For each degree of freedom, and only the kinetic term for a monatomic ideal gas and its three degrees, or the rotational degrees of polyatomic molecules. And for the vibrational degrees, again, kinetic plus potential terms.
     
  7. Nov 21, 2014 #6
    This is my point. I know it is this way, but why is it so? Normally energies have to be equal, in this case it is the temperature. Why is it temperatures that have to match, and not the energies?
     
  8. Nov 21, 2014 #7

    Bystander

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    You get, per mole (or per molecular unit), R or R/2 (k or k/2) per degree of freedom; in a monatomic gas you have 3 kinetic degrees of freedom, and energy of the gas is proportional to T. For more complex systems, energy is taken up in kinetic and/or vibrational degrees of freedom, and total system energy is a sum of T times "fudge factor" (it really ain't fudging, but for HS students ....) that expresses how "active" a particular degree of freedom is in storing/taking up energy, and that activity is temperature dependent.
     
  9. Nov 21, 2014 #8
    This is not correct. Normally, at equilibrium, the energies do not have to be equal. The changes in the energies have to be equal from the initial to the final states, not the absolute energies in the final state. If you think that the final energies normally have to be equal, please site some examples of this (other than degenerate cases in which you have equal amounts of exactly the same substance initially at two different temperatures). Here is a counter example: If you drop a marble at one temperature into the ocean (at a different temperature), is the amount of energy in the marble after the system equilibrates equal to the amount of energy in the ocean at the end.

    When two substances come to thermal equilibrium with one another, it is their temperatures that are always the same in the end.

    Why don't you work out the problem you posed in your original post, and show us what you get for the final temperature? Let the gas have an initial temperature T1, a mass of M1, and a heat capacity C1, and let the solid have an initial temperature T2, a mass of M2, and a heat capacity C2. What is the final temperature and how much energy do the gas and the solid each have at thermal equilibrium?

    Chet
     
  10. Nov 21, 2014 #9
    The energy per atom for the ocean and the marble would be the same, assuming the same amount of degrees of freedom for the two.

    I reformulate my question: When the energy per atom is not is not the same for a mono and diatomic gas, at the same temperature, why will the diatomic not transfer their excess energy to the less energetic monoatoms?

    PS. I can do all the calculations you ask about, but they does not help me with my question. Because I know how it is and how to calculate temperatures, degrees of freedom, and so on, but I can't understand why it is that way.
     
  11. Nov 21, 2014 #10
    If you have a hot body and a cold body in contact, do you know of any physical situation in nature where heat stops flowing from the hot body to the cold body before they attain the same temperature? Can you offer any evidence or references from the literature that indicate that a system reaches thermal equilibrium when the energy per atom of a hot body is equal to the energy per atom of the cold body, even if the hot body still has a higher temperature than the cold body? These things are simply not observed experimentally in the real world.

    Observational evidence always shows that heat continues to flow from a hot body to a cold body until the two equilibrate at the same temperature. Observational evidence also shows that if two bodies are at the same temperature, heat will not flow from one to the other to cause their temperatures to become unequal (irrespective of the amount of energy per atom each body contains).

    Chet
     
  12. Nov 22, 2014 #11
    Chet i appreciate your trying to explain, but I think you misunderstand my problem.

    I'm not questioning that equilibrium is reached when objects have the same temperature, I know that's what happens. I just don't understand why.

    If the best explanation is that that's what we always observe that's also ok. I just hopped someone had a more intuitive understanding of what is happening.

    I just find it remarkable that we have systems in contact where the average energy per atom is different, yet there is no transfer of energy. Yet again I know it is so, not questioning that, and not looking for a counterexample to thermodynamics, I trust that too.

    Just wanted to understand why the higher energy atoms don't transfer their energy, to the lower energy atoms. If the only explanation there that is what is observed, then fine, that's how physics work sometimes. However a more fundamental understanding of this would be nice in my appinion.
     
  13. Nov 22, 2014 #12

    Bystander

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    We're not at the classical limit at ordinary temperatures; equipartition does not hold for the vibrational modes. Jog the memory?

    Appendix added: It's been fifty years, so the memories a little slow. Recall heat capacities of solids at low T, Debye is one of several models, none of which are terribly accurate, for which heat capacity is zero at absolute zero T, increases through some characteristic temperature, theta, and eventually reaches 3kT/molecule; other thing to remember is that energy levels for translational and rotational degrees of freedom for gases are very closely spaced, and equipartition holds under almost all conditions, while vibrational energy levels have larger spacing, and are not all accessible. Electronic energy levels (additional degrees of freedom for each electron in a molecule) are not even active until high temperatures are reached, and then contribute further to heat capacity (and to integral of CxdT from 0 to T of interest, or the total energy of the molecule).
     
    Last edited: Nov 22, 2014
  14. Nov 22, 2014 #13

    Philip Wood

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    The original question is (if I may say so) very clearly posed. It is also non-trivial and is bound to crop up when teaching Physics thoughtfully even at introductory level.

    We've steered clear in this discussion of examining the idea of temperature, except that the important point has been made that if bodies are in thermal equilibrium (no microscopic transfer of energy on average) their temperatures are the same. One way to proceed would be to consider molecules of different types (say a diatomic and a monatomic) colliding, and working out the condition for there to be no transfer on average. This would be the kinetic theory approach. 'Only' dynamics would be involved (and some probability theory) if we treat the system classically, but I wouldn't like to tackle it. I've no doubt that the condition would turn out to be that the molecules have equal mean energies per degree of freedom. In The Kinetic Theory of Gases, Jeans considers spherical molecules of different masses colliding. One is tethered by a spring to a fixed point and represents a molecule of the container wall. He does show by simple dynamics that the condition for no mean energy transfer between the two is that their mean kinetic energies are equal. But this doesn't go far enough to give you what you need.

    The modern (i.e. post Boltzmann and Gibbs!) way to reach the result (equal energies per degree of freedom in the classical régime) uses statistical mechanics: counting microstates and so on. I doubt if this would be much help to your students.
     
    Last edited: Nov 22, 2014
  15. Nov 22, 2014 #14
    Hello Philip

    Thanks for your reply. You state my problem very clear. I will try to have a look at the reference you state, maybe his approach can inspire me to see a solution.

    It just puzzles me that the fact that equal energy per degree of freedom, is the condition for equilibrium, has not puzzled more people, and a beautiful simple calculation or argument cant be found anywhere (maybe i just dont know where to look). I think the other posters in this thread cant even understand my problem, because they have used these facts so many times they have become natural to them. Before i starter to teach i would have been in the same boat as them, so i completely understand why they answer as they do.

    The reason i think this discussion is so important is because it explains why we need to have a thing called temperature. In most situation equal energy (or average energy) is the condition for equilibrium, and there would be no need to define temperature, but not in thermodynamics. Hence we need to define temperature , to state the equilibrium condition, this fact is fascinating to me.

    I hope someone else can help me with an argument, calculation or reference to a book.

    PS. Youre right that an approach with microstates would not help my students. Id rather not try to define temperature through microstates and entropy
     
  16. Nov 22, 2014 #15

    Philip Wood

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    I hope so, too. I'm keen to learn.

    It appears that Maxwell carried out the kinetic theory calculations that I was droning on about in my previous post, and arrived at equal mean energies for each translational and rotational degree of freedom as the condition for no mean transfer of energy between molecules, i.e. same temperature:
    http://books.google.co.uk/books?id=4HJVAgAAQBAJ&pg=PA149&lpg=PA149&dq=justifying equipartition of energy&source=bl&ots=8Rap8D14ZB&sig=TEjsnjdbz5LI1tzy888PGpG5Jy8&hl=en&sa=X&ei=0JhwVMjdM8H5PPDlgOgH&redir_esc=y#v=onepage&q=justifying equipartition of energy&f=false
    This kinetic theory approach is, I believe, very difficult, and each new case (e.g. vibrational degrees of freedom) would need a new treatment. For this reason, I don't think Maxwell's approach was developed further. It was superseded by the statistical mechanics approach.
     
    Last edited: Nov 22, 2014
  17. Nov 22, 2014 #16

    RonL

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    Perhaps you can compare atoms of Gold, Wood, and water, show the nucleus and electrons of each,
    At room temperature there is a vast difference of energy at work between the nucleus and electrons, just to maintain their composition.
    Then show the temperature and pressure conditions that allow water to become a vapor, then the condition that allows Gold to become vapor deposited on a surface in a semiconductor process.


    Hope I haven't completely missed an explanation above or the objective of your thread.
     
    Last edited: Nov 22, 2014
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