# Translational Kinetic Energy

1. Nov 30, 2007

### mit_hacker

[SOLVED] Translational Kinetic Energy

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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?

2. Nov 30, 2007

### 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?

3. Nov 30, 2007

### mit_hacker

Thanks a ton!!

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

4. Nov 30, 2007

Anytime.