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Another Thermodynamics Question

  1. Oct 31, 2004 #1
    Why does U = (3/2)nRT?
  2. jcsd
  3. Oct 31, 2004 #2
    Thats not true in all cases. Its onlytrue for , if I remember correctly, monatomic ideal gases. The reason why its true has to do with the "degrees of freedom" of the gas. Monoatomic gaese have only 3 degrees of freedom(they can only move in the x, y and z directions and cannot rotate) For diatomic gases, it would be U = 5/2 nRT since we added two more degrees of freedom(two planes of rotation)
  4. Oct 31, 2004 #3
    Yes i understand. I should have asked why

    where f = degrees of freedom
  5. Nov 2, 2004 #4


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    This is done by the equi partition theorem.
    It states that we add 1/2KT per degree of freedom and 1KT per degree of Vibrational freedom.
    An interesting case is when we consider a diatomic gas like Hyrdogen gas.
    We expect U to be
    [tex] N_a[ \frac{3}{2}KT + \frac{2}{2}KT + 1KT] [/tex]

    from velocities in x,y,z directions, the rotation about x,y and vibrational (1/2mv^2 and 1/2kx^2) respectivley
    [tex] = \frac{7}{2}RT [/tex]

    But experimentally we find that
    [tex] U= \frac{5}{2}RT [/tex]

    This is because at room temperature vibration does not seem to contribute

    Therefore [tex] \gamma = \frac{7}{5} [/tex] at room temperature
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