Why isn't rotation included when calculating the average energy of a monatomic gas?

[sanbyakuman]

Greetings, this is my first post, though I have been reading these forums for a while.

I understand that the average energy of each degree of freedom in a thermodynamic system in equilibrium is kT/2. My textbook says that for a monatomic gas particle, the only degrees of freedom that count are movement in three dimensional space, so the average energy of such a particle is 3kT/2.

My question is, why don't rotations contribute toward the average energy? My textbook suggests that this is because the moment of inertia is vanishingly small. However, my thought is that if the moment of inertia is very small, it just means that the particle would be spinning extremely fast in order to reach an average energy of kT/2 for each rotational axis.

sanbyakuman

 

[Naty1]

Your description is one of classical physics. You are correct: rotational energy may be significant.

Try these sections in this link:

http://en.wikipedia.org/wiki/Heat_capacity#The_storage_of_energy_into_degrees_of_freedom


The effect of quantum energy levels in storing energy in degrees of freedom
Energy storage mode "freeze-out" temperatures

It should be noted that it has been assumed that atoms have no rotational or internal degrees of freedom. This is in fact untrue. For example, atomic electrons can exist in excited states and even the atomic nucleus can have excited states as well. Each of these internal degrees of freedom are assumed to be frozen out due to their relatively high excitation energy. Nevertheless, for sufficiently high temperatures, these degrees of freedom cannot be ignored. In a few exceptional cases, such molecular electronic transitions are of sufficiently low energy that they contribute to heat capacity at room temperature, or even at cryogenic temperatures. One example of an electronic transition degree of freedom which contributes heat capacity at standard temperature is that of nitric oxide (NO), in which the single electron in an anti-bonding molecular orbital has energy transitions which contribute to the heat capacity of the gas even at room temperature.

 

[AlephZero]

However, my thought is that if the moment of inertia is very small, it just means that the particle would be spinning extremely fast in order to reach an average energy of kT/2 for each rotational axis.

If that was correct, you need a way to apply forces to a particle to make it spin very fast. It's not very obvious how that would work.

Making an unsymmetrical object like a diatomic molecule spin isn't a problem. All you have to do is hit one end of it in a collusion, for example.

 

[Rap]

Greetings, this is my first post, though I have been reading these forums for a while.

I understand that the average energy of each degree of freedom in a thermodynamic system in equilibrium is kT/2. My textbook says that for a monatomic gas particle, the only degrees of freedom that count are movement in three dimensional space, so the average energy of such a particle is 3kT/2.

My question is, why don't rotations contribute toward the average energy? My textbook suggests that this is because the moment of inertia is vanishingly small. However, my thought is that if the moment of inertia is very small, it just means that the particle would be spinning extremely fast in order to reach an average energy of kT/2 for each rotational axis.

sanbyakuman

Its a quantum mechanics effect. The rotational energy of a monatomic gas particle is quantized, it has a ground value of zero, and then it jumps to a non zero energy for the first (rotationally) excited state, and then a higher number for the second, etc. If the energy of the first excited state is much, much larger than kT/2, then there won't be many particles in that state, or any higher state. For all practical purposes you can say that all particles are in the ground state - that is, they are not rotating. For a monatomic gas at room temperature, this is the case. If you raise the temperature of a monatomic gas high enough, there will be a significant number of particles in the first and possibly higher excited states, and then you cannot assume that they are not rotating, and the average energy per particle will be higher than 3kT/2.

 

[chill_factor]

single atoms are considered to not be able to rotate. this is taught in basic statistical mechanics. there are electronic degrees of freedom but they require such high temperature to excite for most molecules that they don't play a major role.

 

[sanbyakuman]

Thanks everyone for your replies. I understand now.