If we have an isolated pentane molecule at room temperature (RT) the Boltzmann energy $(E=k_\mathrm{B}T)$ is approxately $0.59\ \mathrm{kcal/mol} \overset{\wedge}{=} 207\ \mathrm{cm^{−1}}$. There is not enough energy for an electronic or vibrational excitation, so the molecule is in its electronic and vibratory ground state. However usually 10–20 rotational states are populated at RT. Can we attribute mathematically (to a first approximation) this $207\ \mathrm{cm^{−1}}$ amount of energy to $x\:\%$ translational energy ($E_\mathrm{kin} = \frac{m v^2}{2}$) and to $(100-x)\:\%$ rotational energy?
This question is related to the question about the internal degrees of freedom for a molecule: Rotational degrees of freedom (3N-5 and 3N-6)
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