Question

Calculate the ratio of average K.E of one molecule of diatomic (rigid) and diatomic (elastic)at temperature ${150}^{o}C$.

Answer 1

Average K.E is a result of transitional and rotational energies.

For rigid molecule, $K.E_{avg}=\frac{3}{2}{K}_{b}T$

For elsatic molecule, $K.E_{avg}=5K_{b}T$

$\therefore$ ratio=$\frac{3}{10}=0.3$ .