Specific Heats And Gamma

Question

Molar specific heat at constant volume for an ideal diatomic gas at room temperature, using the equipartition principle, takes which of the following values?

RankGuru · Accepted Answer

Deriving this capacity combines equipartition with the definition of internal energy from NCERT Class 11, Chapter 13 (Kinetic Theory). A diatomic gas at room temperature has five degrees of freedom, three translational and two rotational, so by equipartition each mole stores internal energy $U = \frac{5}{2}RT$. Since the molar heat capacity at constant volume is $C_V = \frac{dU}{dT}$, differentiating gives $C_V = \frac{5}{2}R$. The option $\frac{3}{2}R$ applies to a monatomic gas with only three translational degrees. The option $\frac{7}{2}R$ would apply only if vibrational modes were excited, which does not occur at ordinary room temperature. The option equal to $R$ is far too small and matches no physical gas. The reason vibrational modes stay dormant is that they require far larger energy quanta to activate, so at room temperature only translation and rotation contribute to the heat capacity. A consistency check ties this to the specific heat ratio: with $C_V = \frac{5}{2}R$ and $C_P = C_V + R = \frac{7}{2}R$, the ratio $\gamma = \frac{7}{5} = 1.4$, precisely the measured value for common diatomic gases such as nitrogen and oxygen, confirming the result.

Specific Heats And Gamma

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