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 ?
Answer
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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