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Homework Help: Heat capacity ratio for diatomic gases

  1. Jan 5, 2010 #1
    Calculate the heat capacity ratio for a diatomic ideal gas.


    Using the equipartition theorem, I calculate:

    Cp / Cv = (Cv + R) / Cv = (7/2 R + R) / (7/2 R) = (9/2 R) / (7/2 R) = 9/7 ~ 1.286

    According to the equipartition theorem, I assign each vibrational degree of freedom 1 R and the translational and rotational degrees of freedom are 1/2 R. A diatomic ideal gas has 3(2) = 6 d.o.f. with 3 translational, 2 rotational, 1 vibrational.

    However, here the vibrational d.o.f. are not considered separately - and in the first link it shows that the experimental results are actually closer to 1.4 (which you would predict if the vibrational d.o.f. were assigned the same values as the other two). What's going on?

  2. jcsd
  3. Jan 5, 2010 #2


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    Homework Helper

    Counting the degrees of freedom for a diatomic ideal gas is actually a little bit tricky. In principle you assign 3 translational, 2 rotational (for a symmetrical molecule) and 2 vibrational (one for potential and 1 for kinetic); however, it's not as simple as that, because it turns out that at regular temperatures like room temperatures the vibrational degrees of freedom are "frozen out" - there isn't enough thermal energy at room temperature to excite the vibrational degrees of freedom, and so the diatomic molecule behaves like a dumbell instead of two balls connected by a spring, and so the total count for the degrees of free (near room-temperatures) is 5.

    At very low temperatures even the rotational degrees of freedom freeze out, but since we live near room temperature, we typically take the room temp count of 5 degrees of freedom (vibrations ignored).
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