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?
[Physics] Do rotational degrees of freedom contribute to temperature
degrees of freedomrotational-kinematicsthermodynamics
Related Solutions
WetSavanaAnimal aka Rod Vance has given a good introduction to the issues involved.
When I originally wrote an answer, I though that you were right: if you had a perfectly ideal system of perfectly hard spheres, with perfectly elastic collisions, in a container with perfectly rigid walls, and if all the particles started out with exactly the same speed, then the velocities could not evolve into a Maxwell-Boltzmann distribution, because I thought there was no process that could make the velocities become non-equal. However, I've realised I was wrong about that. For example, consider this collision:
The total $x$-momentum is zero but the total $y$-momentum is positive. This must be the case after the collision as well, so the motion must end up looking like this
with the top particle having gained some kinetic energy while the bottom particle loses some. Through this type of mechanism the initially equal velocities will rapidly become unequal, and the system will converge to a Maxwell-Boltzmann distribution just by transferring energy between particles, with no need for energy to enter or leave the system.
However, it's still possible to imagine special initial conditions where this won't happen. For example, we can imagine that all the particles are moving in the same direction, exactly perpendicular to the chamber walls, and are positioned such that they will never collide. A system in this configuration will remain in this special state forever, and not go into a Maxwell-Boltzmann distribution.
However, such a special state is unstable, in that if you start out with even one particle moving in a slightly different direction than all the others, it will eventually collide with another particle, creating two particles out of line, which can collide with others, and so on. Soon all particles will be affected and the system will converge to the Maxwell-Boltzmann distribution.
In reality, as Rod Vance said, in practice the walls will not be perfectly rigid and will be in thermal motion, which would prevent any such precise initial state from persisting for very long.
Even so, this seems to imply that the hard sphere gas system has at least one special initial state from which it will never reach a thermal state. But this isn't necessarily a problem for statistical mechanics. In this case (if my intuition is correct) the states with this special property form such a small proportion of the overall phase space that they can essentially be ignored, since the probability of the system being in such an initial state by chance is technically zero.
There can be cases where every initial state has a property like this. This means that the system will always remain in some restricted portion of the phase space and never explore the whole thing. But these are just the cases where there is some other conservation law, in addition to the energy, and we know how to deal with that in statistical mechanics.
People used to worry a lot about proving that systems were "ergodic", which essentially means that every possible initial state will explore every other state eventually. But nowadays a lot less emphasis is put on this. As Edwin Jaynes said, the way we do physics in practice is that we use statistical mechanics to make predictions and then test them experimentally. If those predictions are broken then that's good, because we've found new physics, often in the form of a conservation law. When the new law is taken into account, the new distribution will be seen to be a thermal one after all. So we don't need to prove that systems are ergodic in order to justify statistical mechanics, we just need to assume they are "ergodic enough", until Nature tells us differently.
For example, rotational kinetic energy of gas molecules stores heat energy in a way that increases heat capacity, since this energy does not contribute to temperature.
This description is misguiding in two ways.
First, the statement that
rotational energy does not contribute to temperature
makes an impression that temperature is a quantity that is closely connected with the translational kinetic energy, but not rotational kinetic energy. But that is false; according to classical theory (applicable when temperatures are high) in thermodynamic equilibrium, all quadratic degrees of freedom, translational and rotational, correspond to kinetic energy $k_BT/2$ on average.
It is only true that rotational energy does not contribute to translational kinetic energy of molecules, since the two energies are exclusive contributions to total kinetic energy.
Second, heat capacity when molecules are allowed to rotate is not higher because rotational energy does not contribute to translational kinetic energy of molecules.
It is higher because for the same temperature, such system has higher energy than system without rotation. This is because there are additional degrees of freedom, to which corresponds additional average kinetic energy.
Equilibrium implies temperature implies average energies of molecules. Value of average kinetic energies of molecules neither implies temperature exists nor implies temperature is only connected to translational kinetic energy.
Best Answer
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.