Question

Match the followung two columns for 2 moles of an ideal diatomic gas at room temperature T.

Answer 1

For a diatomic molecule, there are three degrees of freedom for translation.

Thus translational kinetic energy=$n\frac{f}{2}RT$

$=2\times \frac{3}{2}RT=3RT$

For a diatomic molecule, there are 2 degress of freedom for rotation.

Thus rotational kinetic energy=$n\frac{f}{2}RT$

$=2\times \frac{2}{2}RT=2RT$

Total degrees of freedom of a diatomic molecule=$6$
Thus total internal energy of the molecule=$n\frac{f}{2}RT$

$=2\times \frac{6}{2}RT=6RT$