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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?

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rafi ur rashid
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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.

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John Rennie
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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)