Law of Equipartition of Energy and Specific Heats of Gases

The law of equipartition of energy -- taken here as a STATED result, without proof, exactly as WB's own syllabus specifies -- asserts that when a gas is in thermal equilibrium at absolute temperature $T$, its total energy is distributed EQUALLY among all of its available degrees of freedom, with EACH degree of freedom contributing, on average, exactly $\tfrac{1}{2}k_BT$ of energy PER MOLECULE (equivalently $\tfrac{1}{2}RT$ per mole). A molecule with $f$ total degrees of freedom therefore has average total energy

This single rule, combined with the degree-of-freedom counts of Section 9.7, correctly predicts how much energy -- and so how much heat -- each type of gas needs to raise its temperature by a given amount, i.e. its molar specific heats.

Molar specific heat at constant volume, $C_v$. At constant volume, no work is done by the gas as it is heated (no expansion), so ALL the heat supplied goes directly into raising the gas's internal energy $U$. Since $U = \tfrac{f}{2}nRT$, the heat needed to raise the temperature of $n$ moles by $dT$ is $dU = \tfrac{f}{2}nR\,dT$, so

Molar specific heat at constant pressure, $C_p$. At constant pressure, the gas also does work $P\,dV$ as it expands on heating, so more heat is needed for the same temperature rise than at constant volume. Using the first law of thermodynamics and the ideal gas equation, this extra work per mole per degree works out to exactly $R$ more than $C_v$ (Mayer's relation, derived formally in the Exercises):

The ratio $\gamma = C_p/C_v$. This ratio, which appears throughout thermodynamics (e.g. in the adiabatic relation $PV^\gamma=\text{const.}$), works out to $\gamma = (f+2)/f$, so it depends ONLY on the degrees of freedom $f$, and therefore only on the STRUCTURE of the gas's molecule, not on the particular gas's chemical identity or its molar mass. Applying this with the degree-of-freedom counts from Section 9.7 (see the table for this section):

• Monatomic ($f=3$): $C_v=\tfrac{3}{2}R$, $C_p=\tfrac{5}{2}R$, $\gamma = 5/3 \approx 1.67$.
• Rigid diatomic ($f=5$): $C_v=\tfrac{5}{2}R$, $C_p=\tfrac{7}{2}R$, $\gamma = 7/5 = 1.4$. …