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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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laminin
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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).

Jan Jensen
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  • 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
  • If there is only a single molecule (very very low gas pressure), which cannot collide, will its translational energy change during the time coordinate? If there is no collision I expect no Brown'sche molecular motion, so it will fly along a one-dimensional coordinate (if we neglect grativational forces) with the same energy as it rotates. I guess the molecule tumbles along the different rotational axis a,b,c. Or will the molecule rotate only along one axis? – laminin Jun 20 '15 at 13:49
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    The usual approximation is that translational, rotational, and vibrational motion is uncoupled. So if the molecule doesn't collide with anything then the translational kinetic energy (and the velocity) will be constant. However, for an individual molecule that never collides the translational and rotational energy will not be equal. That's only true on average and after many, many collisions. Most likely, the rotational motion will reflect rotation along 3 distinct axes. – Jan Jensen Jun 21 '15 at 07:33
  • What does $f(v)$ tell me? – laminin Jun 21 '15 at 10:07
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    $f(v)$ 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 08:47
  • As you said the translational and rotational energies are different if there is no collision. Is there any formulae which relates the two energy values. I mean (with no collision), if I indice a dipole transition with light (and there is not enough energy for vibration or electronic transition) how many % will roughly go into rotational motion and how many % into translational motion in the case of isolated pentane? – laminin Jun 22 '15 at 09:43
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    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
  • Ok, that's a very important point for me what you mentioned (that there is no internal to external energy transfer)! I guess (for our case) only the total energy and the total angular momentum is conserved. If we heat an isolated molecule it's thermal energy = kinetic translational energy will increase. Is it possible for the molecule to lower its translational energy by going to some higher rotational levels (= increasing the internal energy)? – laminin Jun 22 '15 at 14:27
  • In the case of a supersonic expansion the expanded gas is cooled down and gets to the ground state of rotational and vibrational levels. I'm thinking if the translational motion is also decreased (not only the internal degrees of freedom) because the internal degrees of freedom might not have enough energy for this adiabtic expansion. – laminin Jun 28 '15 at 00:51