Question

We know that, for an ideal monoatomic gas, the total internal energy is a result of the translational kinetic energy only. For the oxygen molecule $O_2$ (assume it doesn't oscillate along the bond axis), which rotates as well as translates, choose the correct statement(s) from the following

• A. Change in translational kinetic energy of a sample of n moles = $n{C}_V\Delta T$
• B. Translational kinetic energy < rotational kinetic energy
• C. Translational kinetic energy > rotational kinetic energy
• D. Translational kinetic energy = rotational kinetic energy.

Answer 1

The correct option is C Translational kinetic energy > rotational kinetic energy

The equipartition theorem postulates - each degree of freedom contributes an energy equal to $\left(\frac{1}{2}\right)kT$ to the total internal energy of a molecule, depending on the temperature T. Let's count the number of degrees of freedom for an $O_2$ molecule.

(1) Possible directions of translational motion are the three axes, x, y and z. Thus, number of translational degrees of motion = 3.

(2) Given the negligible size of the atoms, rotation about the bond-axis will not be significant. It can only rotate about the two axes shown (perpendicular to the bond-axis). Thus number of rotational degrees of motion = 2.

Therefore, according the equipartition theorem,

(i) Translational Kinetic energy, $K_{Tr}=\frac{3}{2}kT$,

(ii) Rotational Kinetic energy, $K_{Rot}=kT$.

Clearly, $K_{Tr}$ > $K_{Rot}$.