Specific Heat Capacity from Kinetic Theory

From equipartition to heat

The internal energy of an ideal gas is purely the energy stored in the motion of its molecules. The law of equipartition of energy fixes how that store grows with temperature: in thermal equilibrium at absolute temperature $T$, each quadratic mode of energy holds an average $\tfrac12 k_B T$. A molecule with $f$ active degrees of freedom therefore carries average energy $\tfrac{f}{2}k_B T$, and one mole — $N_A$ molecules — carries

$$ U = \frac{f}{2}\,k_B T \cdot N_A = \frac{f}{2}\,RT \qquad\text{(per mole)} $$

where $R = N_A k_B = 8.314~\text{J mol}^{-1}\text{K}^{-1}$. Because $U$ depends only on $T$, the molar specific heat at constant volume follows by a single differentiation, $C_v = \mathrm{d}U/\mathrm{d}T = \tfrac{f}{2}R$. Everything in this topic is that one line applied to different values of $f$. Counting degrees of freedom correctly is the whole game.

Monatomic gases

A monatomic molecule — helium, neon, argon — is a single atom. It has only the three translational degrees of freedom, $f=3$, with no axis to rotate meaningfully about and no bond to vibrate along. The average energy per atom is $\tfrac32 k_B T$, so for one mole NCERT writes

$$ U = \frac{3}{2}\,k_B T \cdot N_A = \frac{3}{2}\,RT,\qquad C_v = \frac{\mathrm{d}U}{\mathrm{d}T} = \frac{3}{2}R \approx 12.5~\text{J mol}^{-1}\text{K}^{-1}. $$

This is the floor of the whole topic. No gas has a smaller $C_v$, because three translational modes is the minimum any free molecule can have.

Mayer's relation and gamma

Specific heat at constant pressure exceeds that at constant volume because, on heating at constant pressure, the gas also does expansion work $P\,\Delta V = R\,\Delta T$ per mole. The extra heat needed to supply that work is exactly $R$, giving Mayer's relation, valid for every ideal gas:

$$ C_p - C_v = R. $$

The adiabatic ratio $\gamma$ then follows directly. Dividing $C_p = C_v + R$ by $C_v$ and substituting $C_v = \tfrac{f}{2}R$:

$$ \gamma = \frac{C_p}{C_v} = 1 + \frac{R}{C_v} = 1 + \frac{2}{f}. $$

For a monatomic gas $C_p = \tfrac32 R + R = \tfrac52 R$ and $\gamma = 1 + \tfrac23 = \tfrac53 \approx 1.67$. The compact formula $\gamma = 1 + 2/f$ is the single most useful identity in this topic — it lets you read off $\gamma$ from the degree-of-freedom count without computing $C_v$ and $C_p$ separately.

Diatomic gases

A diatomic molecule such as $\text{O}_2$ or $\text{N}_2$, treated as a rigid dumbbell, adds two rotational degrees of freedom to the three translational ones — rotation about the two axes perpendicular to the bond. Rotation along the bond axis has a negligible moment of inertia and does not count. So $f = 3 + 2 = 5$ at ordinary temperatures, and NCERT gives

$$ U = \frac{5}{2}RT,\quad C_v = \frac{5}{2}R,\quad C_p = \frac{7}{2}R,\quad \gamma = \frac{7}{5} = 1.40. $$

At higher temperatures the molecule can also vibrate along its bond. A vibrational mode is special: it stores both kinetic and potential energy, so it contributes two quadratic terms, i.e. a full $k_B T$ rather than $\tfrac12 k_B T$. Switching vibration on raises the count to $f = 7$:

$$ C_v = \frac{7}{2}R,\quad C_p = \frac{9}{2}R,\quad \gamma = \frac{9}{7} \approx 1.29. $$

NEET overwhelmingly uses the rigid (room-temperature) values $C_v = \tfrac52 R$, $\gamma = 1.40$ unless the question explicitly invokes vibration or a higher temperature.

Polyatomic gases

A general polyatomic molecule has three translational and three rotational degrees of freedom, plus some number of vibrational modes. Ignoring vibration (the basis of NCERT Table 12.1), $f = 3 + 3 = 6$, giving the triatomic/polyatomic entry

$$ U = 3RT,\quad C_v = 3R,\quad C_p = 4R,\quad \gamma = \frac{4}{3} \approx 1.33. $$

NCERT also records the general result with $f$ vibrational modes included: $C_v = (3+f)R$, $C_p = (4+f)R$ and $\gamma = (4+f)/(3+f)$. As $f$ grows, $\gamma$ falls steadily toward 1 — heavier, floppier molecules approach $\gamma = 1$. Across the three idealised gas types the trend in $\gamma$ is monotonic, which the next figure makes visual before the master table consolidates the numbers.

Master comparison table

The four idealised cases differ only in $f$; every other column is generated from it. Memorise this grid — almost every NEET specific-heat question is a lookup into one of its rows. Values match NCERT Table 12.1 (vibrational modes ignored).

The $C_p-C_v$ column would be identical in every row — exactly $R = 8.31~\text{J mol}^{-1}\text{K}^{-1}$ — which is why it is left out. That invariance is itself the most common test point.

Specific heat of solids

Equipartition extends naturally to a crystalline solid. Each atom oscillates about its lattice site, and a one-dimensional oscillation stores both kinetic and potential energy, hence $2 \times \tfrac12 k_B T = k_B T$. In three dimensions each atom carries $3 k_B T$, so a mole of solid has $U = 3RT \times N_A/N_A = 3RT$. Because a solid expands negligibly on heating, $\Delta Q = \Delta U$ and

$$ C = \frac{\Delta U}{\Delta T} = 3R \approx 25~\text{J mol}^{-1}\text{K}^{-1}. $$

This is the Dulong–Petit law. NCERT notes that the prediction agrees with experiment at ordinary temperatures for most solids, with carbon (diamond) a striking exception — its anomalously low value foreshadows the quantum freezing-out of modes that classical equipartition cannot capture.

Liquid water is treated by the same atom-counting logic. An $\text{H}_2\text{O}$ molecule has three atoms; assigning each the solid-like $3 k_B T$ gives $3 \times 3 = 9$ modes per molecule, so

$$ U \approx 9RT,\qquad C \approx 9R \approx 75~\text{J mol}^{-1}\text{K}^{-1}, $$

which matches the measured molar heat capacity of water. NCERT presents this as an extension of the equipartition counting from a single oscillating atom to a polyatomic molecule in the condensed phase.

Worked examples