# Do rotational degrees of freedom contribute to temperature?

Recently I have come across a mathematical problem where I was said to calculate the temperature increase of certain mol of N2 gas confined in a room.
However, I found that there was only consideration of kinetic energy in the issue of temeperature.
My question is why don't we include rotational DOF in calculating temeperature increase?

If you start with a monatomic gas then the only degrees of freedom available are the three translational degrees of freedom. Each of them absorbs $\tfrac{1}{2}kT$ of energy, so the specific heat (at constant volume) is $\tfrac{3}{2}k$ per atom or $\tfrac{3}{2}R$ per mole.

If you move to a diatomic molecule there are two rotational modes as well - only two extra modes because rotation about the axis of the molecule has energy levels too widely spaced to be excited at normal temperatures. Each of those two rotational degrees of freedom will soak up another $\tfrac{1}{2}kT$, giving a specific heat of $\tfrac{5}{2}k$ per molecule or $\tfrac{5}{2}R$ per mole.

But the rotational energy levels are quantised with an energy spacing of $E = 2B, 6B, 12B$ and so on, where $B$ is the rotational constant for the molecule:

$$B = \frac{\hbar^2}{2\mu d^2}$$

where $\mu$ is the reduced mass and $d$ is the bond length. So these rotational energy levels will only be populated when $kT$ is a lot greater than $B$ - say 10 to 100 times greater. You can look up the rotational constant of nitrogen, or it's easy enough to calculate, and the result is:

$$B \approx 3.97 \times 10^{-23} \text{J}$$

which is about $3k$. So as long as the temperature is above say $30K$ the rotational modes will be excited and nitrogen will have a specific heat of $\tfrac{5}{2}R$. If you go down to temperatures of $3K$ and below then the specific heat will fall to $\tfrac{3}{2}R$ just like a monatomic gas.

The specific heat of nitrogen at constant volume is 0.743 kJ/(kg.K), and converting this to J/mole.K we get 20.8 J/(mole.K) and this is indeed 2.50R (to three significant figures).

The conformist mentions that the vibrations of the nitrogen molecule will contribute to the specific heat, and indeed they will. However the energy of the first vibrational mode is 2359 cm$^{-1}$, which converted to non-spectrogeek units is $4.7 \times 10^{-20}$ J or about $3400k$. So the vibrational mode isn't going to contribute to the specific heat until the temperature gets above 3400K.

• This looks like the solution to a sample problem in a good physics reference book. :) – Gaurav Aug 6 '15 at 16:17
• @JohnRennie, thanks for the conversion to "non-spectrogeek" units ... it was very helpful. – David White Aug 6 '15 at 17:21
• Nothing here mentions the density. I know the OP mentions gas, but out of curiosity, if the pressure is great enough such that the system is in a condensed phase, how does the specific heat change wrt rotational DOF? – gogators Aug 6 '15 at 20:17

N$_2$ is a diatomic molecule and therefore you must and do consider the degree of freedom from rotation. This is why for a diatomic gas, the mean energy is $$\frac{5}{2} K_B T$$ if the molecule is rigid. These extra degrees of freedom come from rotation. It is also worth pointing out that if the molecule can vibrate along the line joining the two centres, its mean energy is $$\frac{7}{2} K_B T$$. On the other hand, for a monatomic gas like He or Ne, the mean energy is $$\frac{3}{2} K_B T.$$This is an experimentally verified result. The result will vary depending on whether the molecule is made up of the same kind of atom, how many atoms of different kinds and the relative strength of the bonds because all these factors will change the centre of mass of the molecule. The wikipedia page gives a clear explanation to this: https://en.wikipedia.org/wiki/Degrees_of_freedom_(physics_and_chemistry)

• Shouldn't that be 5/2, not 7/2? – Jon Custer Aug 6 '15 at 15:16
• Didn't you mean $\tfrac{5}{2}kT$? $\tfrac{1}{2}kT$ for each of the three degrees of translational freedom and another $\tfrac{1}{2}kT$ for each of the two rotational modes. – John Rennie Aug 6 '15 at 15:18
• Well, yes if the molecule is ridid. It's 7/2 if the molecules can vibrate along the imaginary axis joining the two centres. Thanks for the spot, I will edit my answer. – user58089 Aug 6 '15 at 15:19
• You need to do the kind of analysis that John did for the rotational mode on-set temperature for the vibrational modes. $\mathrm{N}_2$ is tightly bound and the ground state energy for it's rotational modes is surprisingly high resulting in their not being excited until quite high temperatures indeed. – dmckee Aug 6 '15 at 16:25
• Of course, John's answer was much more precise. But I think there is a slight miscommunication on my part in what I wanted to say. I'm not claiming that nitrogen has significant vibrational dof's. Just that if molecules also have significant vibrational dof's, then you also consider. The point was supposed to be that no dof is left behind. – user58089 Aug 6 '15 at 16:43