Find the average translational kinetic energy of nitrogen molecules

[mit_hacker]

[SOLVED] Translational Kinetic Energy

Homework Statement

(Q) Find the average translational kinetic energy of nitrogen molecules at 1600K.

Homework Equations

Translational KE per degree of freedom = 1/2kT.

The Attempt at a Solution

Since Nitrogen molecules are diatomic, it has 5 degrees of freedom so KE = 5/2kT.

The problem is that this yields the wrong answer and the answer at the back of the book uses the formula 3/2kT. Can someone please explain to me why this is so?

Thank-you very much in advance.

 

[G01]

Yes, the nitrogen molecule has 5 degrees of freedom, but translational motion can only happen in three of them.

(i.e. The rotational and vibrational motions are not translational motion, so the kinetic energy for these types of motion is not translational kinetic energy.)

Thus, we only consider the 3 degrees of freedom for which the molecule can undergo translational motion. So, we get:

$$<KE> = 3/2kT$$

Does this make sense?

 

[mit_hacker]

Thanks a ton!

I understand. Thanks a lot for your extremely quick help!

 

[G01]

Anytime.