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I understand that the average energy of each degree of freedom in a thermodynamic system is $\frac12kT$. And so, for an ideal monatomic gas, there are three degrees of freedom associated with the translational components of each atom, which gives $E$ = $\frac32kT$ for each atom in this system.

For an ideal diatomic gas, there are the three degrees of freedom from the translational components plus two more degrees of freedom associated with the rotational components of each molecule. So we end up with $E$ = $\frac52kT$ for each molecule.

My question here is why did we ignore the rotational degree(s) of freedom in case of a monatomic gas ? Like why isn't there another degree of freedom for the rotational component of a single atom ? Also, if that is neglected for some reason, then when does it become significant and can no longer be ignored ? For example, shouldn't it be included in calculations under extreme conditions like temperature of millions of degrees ?

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    $\begingroup$ Just curious, are you reposting an old question from physicsforums.com/threads/… ? $\endgroup$
    – CuriousOne
    Commented Jul 6, 2015 at 2:25
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    $\begingroup$ General assumption for ideal gas is that the particles are point-like, thus don't have any rotational degrees of freedom. This degree of freedom exist in molecules with more than 1 atom (thus have some spatial extension), and there exist some statistical models for this kind of molecules (which include also vibrational degrees of freedom). $\endgroup$
    – Alexander
    Commented Jul 6, 2015 at 2:28
  • $\begingroup$ Hi Abanob. The link I've suggested isn't an obvious duplicate, but it does answer your question. The rotational energy levels of a single atom are far too widely spaced to be excited under most circumstances. $\endgroup$
    – John Rennie
    Commented Jul 6, 2015 at 5:33

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Intuitively, the moment of inertia of a single atom is far smaller than a diatomic molecule because the nucleus is at the origin, while in a diatomic molecule the nuclei are half the bond length from the origin. The minimum excitation energy for rotation is then much higher, well above room temperature, so it doesn't contribute, because $E=\frac 12I\omega^2$ and the angular momentum is quantized as $N=n\hbar =nI\omega=nE^2/I$. To formalize this, you need to compare the angular momentum of a nucleus with the angular momentum of the electron cloud and the energy spacing of the rotational modes with room temperature. There is clearly a temperature where it will become important, but maybe the atoms are ionized before that.

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  • $\begingroup$ So in case of an ion, can the rotational energy be completely neglected ? $\endgroup$
    – Abanob Ebrahim
    Commented Jul 6, 2015 at 3:19
  • $\begingroup$ Aside from hydrogen, an ion has almost the same moment of inertia as an atom. I think (but didn't do the calculation) that the electron cloud has most of the MOI of an atom in the classical picture. The electrons are $2E3$ times less massive than the protons, but the radius (which gets squared) is $1E5$ times greater or more. $\endgroup$
    – Ross Millikan
    Commented Jul 6, 2015 at 3:25
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    $\begingroup$ You said there is a temperature where it will become important, but maybe atoms are ionized before that. So does that mean when the atom is ionized, we can ignore the rotational energy ? Assume Hydrogen atom $\endgroup$
    – Abanob Ebrahim
    Commented Jul 6, 2015 at 3:29
  • $\begingroup$ Then the MOI is decreased because now you have a bare proton. That means the temperature where the degree of freedom becomes important is higher-I suspect much higher for the reasons in my last comment. Other things may be going on by then-increasing number of electrons contributing the $\frac 32kT$ would be my first guess. $\endgroup$
    – Ross Millikan
    Commented Jul 6, 2015 at 3:43
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    $\begingroup$ @AbanobEbrahim - no: the average energy of a single atom cannot exceed 3kT. There is no limit on the energy of an individual atom. Boltzmann distributions and all that... $\endgroup$
    – Floris
    Commented Jul 6, 2015 at 4:15

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