# Classical theory fails to explain quantization of motions?

Because the degrees of freedom look like they're 'frozen' at low T. Statistically, we know there's going to be an average energy of $RT/2$ per mole for each degree of freedom. If you have translation, rotation and vibration that makes for a total of 7 degrees of freedom. By classical mechanics, there should be no lower limit to how much energy goes into them, so it should be $C_v = 7R/2$ from the very beginning. Instead, because of QM, there's an energy gap between ground and first excited state for each of these motions, and that means they don't contribute up to the point when $kT \sim \hbar \omega$ for each of them. That's what causes the 'steps' to appear in the heat capacity, and the very existence of those steps is only possible because of quantum effects. That the steps are smoothed out is merely a statistical effect due to the fact that not all modes will activate instantly across the gas.

The "wrong" thing in this picture is an illusion of "horizonality" of some parts of this curve. According to the Maxwell distribution $\propto \text{exp}(-mv^2/2kT)$, in the volume there are always high velocity molecules capable to get rotational and vibrational excitations, thus the curve has always a slope as a function of $T$.

For one molecule you have clear thresholds (a step-wise curve), but for a volume of molecules the thresholds are smeared due to statistics applied to calculate/measure the heat capacity of the volume.

Still, one can see a "thershold-like" behavior of the gas heat capacity indicating quantization of rotational and vibrational energies. Without quantization the curve would not be a step-wise at all.

By the way, with $T$ growing, excitations of electron levels come into play. Finally one can finish with fully ionized plasma ;-).