Understanding the Equipartition Theorem for Ideal Gases

[GravityX]

Homework Statement

At what temperature does the degrees of freedom freeze (estimate)

Relevant Equations

none

Hi,

I am unfortunately stuck with the following task

I started once with the hint that at very low temperatures the diatomic ideal gas behaves like monatomic gas and has only three degrees of freedom of translation ##f=3##. If you then excite the gas by increasing the temperature, you add two degrees of freedom of rotation, ##f=5## and if you then excite the gas even further, you add two more degrees of freedom of vibration ##f=7##.

The Equipartition theorem states that the internal energy is distributed equally among the degrees of freedom. The calculation of the internal energy for the ideal gas is ##U=\frac{3}{2}RT## for the translation, ##U=\frac{5}{2}RT## for the rotation and ##U=\frac{7}{2}RT## for the oscillation.

Unfortunately, I don't know either ##U## or ##T##, but I can't think of any other way to estimate the temperature.

 

[vela]

Did you use the provided hint for each type of motion?

 

[GravityX]

Thanks vela for your help

I would now proceed as follows

Translation:##\frac{3}{2}RT=\frac{\pi^2 \hbar^2}{2ML^2}(n_x^2+n_y^2+n_z^2)##

Rotation: ##\frac{5}{2}RT=\frac{\pi^2 \hbar^2}{2ML^2}(n_x^2+n_y^2+n_z^2)+\frac{\hbar^2l(l+1)}{2\theta}##

Oscillation: ##\frac{7}{2}RT=\frac{\pi^2 \hbar^2}{2ML^2}(n_x^2+n_y^2+n_z^2)+\frac{\hbar^2l(l+1)}{2\theta}+\hbar\omega(n+\frac{1}{2})##

Now I can solve the individual equations according to the temperature with

For translation, ##n_x^2,n_y^2,n_z^2=1##
For rotation ##n_x^2,n_y^2,n_z^2=2## and ##l=1##
During oscillation ##n_x^2,n_y^2,n_z^2=2## , ##l=1## and ##n=1##

 

[vela]

The Hamiltonian is for a single particle, so you want to use the Boltzmann constant, not the gas constant.

 

[GravityX]

Thanks vela for your help

So ##\frac{3}{2}k_bT## instead of ##\frac{3}{2}RT##.