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Relationship between Temperature and Mean Translational Kinetic Energy

  1. Oct 20, 2014 #1
    While I am familiar with the derivation of the relationship between mean translational kinetic energy, ##<E_k>##, and temperature, ##T##, for an ideal gas (namely ##<E_k> = \frac{3}{2}k_B T##); I fail to see how we can establish such a relationship for solids, liquids, and even real gases. Could someone please guide me through this?
     
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  3. Oct 20, 2014 #2

    stevendaryl

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    That relationship doesn't hold in general, for non-ideal gases. That relationship assumes that energy for a single particle is just [itex]E = p^2/(2m)[/itex]. If there are interactions between the particles, then the energy for each particle is more complicated, and the relationship between energy and temperature is more complicated.
     
  4. Oct 21, 2014 #3

    Andrew Mason

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    The concept of temperature is tied to heat flow. If heat flows spontaneously from body A to body B we say that body B has a lower temperature than body A. The heat flow stops when the molecules in both bodies reach a common thermal equilibrium. At the molecular level, the heat flow from one body to another (ie. the transfer of molecular kinetic energy) results from a difference in the translational kinetic energies of the molecules (i.e. molecular collisions). If the molecules were just vibrating or rotating about their centres of mass there would be no flow of kinetic energy from one molecule to another.

    When the bodies reach thermal equilibrium, the total kinetic energies of the molecules will be equally partitioned between all modes (ie. active vibrational, active rotational and translational motion). But the transfer of energy is due to the differences in translational kinetic energies. This applies whether you are dealing with ideal gases, real gases, solids or liquids.

    AM
     
  5. Dec 18, 2015 #4
    Hi Andrew ,

    Could you please explain the first line , more specifically "When the bodies reach thermal equilibrium" ?

    My understanding is that " the energy transferred would be equally partitioned between all modes until the average translational kinetic energy of the two bodies become equal . As a consequence of which the bodies reach thermal equilibrium ". But you have written that after the bodies reach thermal equilibrium ,then energy is partitioned .

    How is it that first thermal equilibrium is reached and then transferred energy is partitioned in different modes ?

    Doesn't thermal equilibrium mean equality of average translational kinetic energy of molecules of the two bodies ?

    Sorry if I have misunderstood you . I am having little difficulty in understanding relation between temperature and average translational kinetic energy .
     
    Last edited: Dec 18, 2015
  6. Dec 26, 2015 #5
    @Andrew Mason , could you please respond to the above post ?
     
  7. Dec 27, 2015 #6

    stevendaryl

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    I think you're just misinterpreting what Andrew is saying. He's using "then" in the logical sense: If__then. He's just saying that if a system is in thermal equilibrium, then the energy is equally distributed among all mode. He's not saying that the first happens and then later the second happens.
     
  8. Dec 28, 2015 #7

    Andrew Mason

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    As heat flow occurs from body A to body B (due to higher average translational kinetic energy of the molecules in body A compared to body B) the energy in body A decreases and that of body B increases. When thermal equilibrium is reached the internal energy in each of body A and B is equally partitioned between all modes. Before thermal equilibrium is reached, all modes will have energy but they will not necessarily be equally partitioned.

    AM
     
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