Question

According to law of equipartition of energy the molar specific heat of a diatomic gas at constant volume where the molecule has one additional vibrational mode is:-

• A. $\frac{3}{2} R$
• B. $\frac{7}{2} R$
• C. $\frac{9}{2} R$
• D. $\frac{5}{2} R$

Answer 1

The Correct Option is B

The problem asks us to determine the molar specific heat at constant volume for a diatomic gas with one additional vibrational mode, using the law of equipartition of energy.

According to the law of equipartition of energy, each degree of freedom contributes \(\frac{1}{2} R\) (where \(R\) is the universal gas constant) to the molar specific heat.

For a diatomic gas, consider the following degrees of freedom:

• Translational motion: 3 degrees of freedom (x, y, z).
• Rotational motion: 2 degrees of freedom (around two axes, as rotation around the bond axis is negligible).
• Vibrational motion: 1 degree of freedom, as given in the problem.

In total, the degrees of freedom are 3 (translational) + 2 (rotational) + 1 (vibrational) = 6.

Therefore, the molar specific heat at constant volume \(C_v\) is given by:

\(C_v = \frac{1}{2}R \times \text{(Number of degrees of freedom)}\)

Substituting the degrees of freedom:

\(C_v = \frac{1}{2}R \times 6 = \frac{6}{2}R = 3R\)

However, this total includes one vibrational mode's energy, which involves both kinetic and potential energy contributions. Each vibrational mode effectively contributes \(R\) (i.e., \(\frac{1}{2}R\) from kinetic + \(\frac{1}{2}R\) from potential), thus we need to account for \(2R\) for the additional vibrational mode, not just \(1R\).

As a diatomic molecule in general without vibration contributes \(\frac{5}{2}R\), adding the vibration contribution, the correct molar specific heat becomes:

\(C_v = \frac{5}{2}R + R = \frac{7}{2}R\)

Thus, the molar specific heat of a diatomic gas at constant volume, where the molecule has one additional vibrational mode, is \(\frac{7}{2}R\). Therefore, the correct answer is:

Correct Answer: \(\frac{7}{2} R\)