Question

Given below are two statements:
Statement I: In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution.
Statement II: In a diatomic molecule, the rotational energy at a given temperature equals the translational kinetic energy for each molecule.

• A. Both Statement I and Statement II are true.
• B. Both Statement I and Statement II are false.
• C. Statement I is true but Statement II is false.
• D. Statement I is false but Statement II is true.

Hint:

Remember the law of equipartition of energy: the average energy per degree of freedom is $\frac{1}{2}kT$. For a diatomic gas, degrees of freedom are 3 (translational) + 2 (rotational) = 5 at normal temperatures. This helps compare translational and rotational energies.

Answer 1

The Correct Option is C

Let's analyze each statement.
Statement I: The Maxwell-Boltzmann distribution describes the distribution of speeds or energies of particles in a system at thermal equilibrium. Rotational energy is a form of kinetic energy, and at a given temperature, the distribution of rotational energies among the molecules of a gas will follow a Maxwell-Boltzmann-like distribution. Thus, Statement I is true.
Statement II: According to the equipartition of energy theorem, the average energy associated with each quadratic degree of freedom is $\frac{1}{2}kT$.
A diatomic molecule has 3 translational degrees of freedom. The average translational kinetic energy is $3 \times \frac{1}{2}kT = \frac{3}{2}kT$.
A diatomic molecule also has 2 rotational degrees of freedom (at ordinary temperatures). The average rotational energy is $2 \times \frac{1}{2}kT = kT$.
Since $\frac{3}{2}kT \neq kT$, the average rotational energy is not equal to the average translational kinetic energy. Thus, Statement II is false.
Therefore, Statement I is true and Statement II is false.