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Why isn't rotation included when calculating the average energy of a monatomic gas?

  1. Sep 14, 2012 #1
    Greetings, this is my first post, though I have been reading these forums for a while.

    I understand that the average energy of each degree of freedom in a thermodynamic system in equilibrium is kT/2. My textbook says that for a monatomic gas particle, the only degrees of freedom that count are movement in three dimensional space, so the average energy of such a particle is 3kT/2.

    My question is, why don't rotations contribute toward the average energy? My textbook suggests that this is because the moment of inertia is vanishingly small. However, my thought is that if the moment of inertia is very small, it just means that the particle would be spinning extremely fast in order to reach an average energy of kT/2 for each rotational axis.

    sanbyakuman
     
  2. jcsd
  3. Sep 14, 2012 #2
    Re: Why isn't rotation included when calculating the average energy of a monatomic ga

    Your description is one of classical physics. You are correct: rotational energy may be significant.

    Try these sections in this link:

    http://en.wikipedia.org/wiki/Heat_capacity#The_storage_of_energy_into_degrees_of_freedom


    The effect of quantum energy levels in storing energy in degrees of freedom
    Energy storage mode "freeze-out" temperatures



     
  4. Sep 14, 2012 #3

    AlephZero

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    Re: Why isn't rotation included when calculating the average energy of a monatomic ga

    If that was correct, you need a way to apply forces to a particle to make it spin very fast. It's not very obvious how that would work.

    Making an unsymmetrical object like a diatomic molecule spin isn't a problem. All you have to do is hit one end of it in a collusion, for example.
     
  5. Sep 14, 2012 #4

    Rap

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    Re: Why isn't rotation included when calculating the average energy of a monatomic ga

    Its a quantum mechanics effect. The rotational energy of a monatomic gas particle is quantized, it has a ground value of zero, and then it jumps to a non zero energy for the first (rotationally) excited state, and then a higher number for the second, etc. If the energy of the first excited state is much, much larger than kT/2, then there won't be many particles in that state, or any higher state. For all practical purposes you can say that all particles are in the ground state - that is, they are not rotating. For a monatomic gas at room temperature, this is the case. If you raise the temperature of a monatomic gas high enough, there will be a significant number of particles in the first and possibly higher excited states, and then you cannot assume that they are not rotating, and the average energy per particle will be higher than 3kT/2.
     
  6. Sep 14, 2012 #5
    Re: Why isn't rotation included when calculating the average energy of a monatomic ga

    single atoms are considered to not be able to rotate. this is taught in basic statistical mechanics. there are electronic degrees of freedom but they require such high temperature to excite for most molecules that they don't play a major role.
     
  7. Sep 15, 2012 #6
    Re: Why isn't rotation included when calculating the average energy of a monatomic ga

    Thanks everyone for your replies. I understand now.
     
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