How to distribute the Boltzmann energy to internal and external motion?

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)

The translational energy of a mole of molecules is $\tfrac{3}{2}RT$ which corresponds to an average kinetic energy of $\tfrac{1}{2}M\langle v^2\rangle$. The average reflects a Boltzmann distribution of speeds $$f(v)=4\pi\left(\frac{M}{2\pi RT} \right)^{3/2}v^2\operatorname{e}^{-\tfrac{1}{2}Mv^2/RT}$$ i.e. some molecules have very low velocity and some have very high velocities.
Kinetic energy is constantly being transferred, so if you could monitor its velocity over time its average kinetic energy would be $\tfrac{1}{2}M\langle v^2\rangle=\tfrac{3}{2}RT$
In addition one mole of gas has $\tfrac{3}{2}RT$ of rotational energy ($RT$ for linear molecules), which also reflects an average rotational kinetic energy. For individual molecules at any given time, the amount of translational and rotational energy is independent. For example, a molecule can move very slowly while rotating very rapidly – until it collides with another molecule and things change. However, on average there is an equal amount of translational and rotational energy (for non-linear molecules).
• Is $f(v)$ equal to $\langle v^2\rangle$? I'd like to obtain numbers for pentane for the rotational energy and for the translational energy. Can we do this with formulas or do we really have to measure this values in experiments? – laminin Jun 20 '15 at 12:03
• $f(v)$ is a function, $\langle v^2 \rangle$ is a number, so no. The rotational and translational energies for pentane and any other non-linear molecule are $\tfrac{3}{2}RT$ – Jan Jensen Jun 20 '15 at 12:44
• $f(v)$ tell you the probability of a molecule having velocity $v$. Also $\langle v^2 \rangle=\int_0^\infty v^2 f(v) dv$ – Jan Jensen Jun 22 '15 at 8:47
• The absorption of light depends on the wavelength. Sub-IR radiation typically means microwave, which excites rotational modes. None of this internal energy can be transferred to translational motion. The photon also has momentum ($p=h/c$), which can be transferred to the molecule. Everything we've talked about previously assumes thermal energy transfer and does not apply to light-induced excitations. – Jan Jensen Jun 22 '15 at 13:21