Molar heat capacities and equipartition

Heat is not a state function, so “the heat capacity of nitrogen” is undefined until you say what was held fixed while the heat went in. Hold the volume fixed and no work is done, so the first law gives Q = ΔU and C_V = (1/n)(∂U/∂T)_V. Hold the pressure fixed and the gas expands against its surroundings, so Q = ΔU + pΔV; the combination U + pV is the enthalpy H, and Cₚ = (1/n)(∂H/∂T)ₚ. The partial derivatives matter: U and H are functions of two variables, and the subscript says which one is being held still. For an ideal gas one fact is decisive. Joule's free expansion showed that U depends on T alone, not on V, so dU = nC_V dT holds on every path — isobaric, adiabatic, any curve on the p-V plane — even though C_V was defined at constant volume. Argon has C_V = 12.47 J K⁻¹ mol⁻¹, so 2.00 mol warming by 25 K gains 624 J of internal energy however it was heated.

Heat 1 mol of ideal gas by dT at constant pressure. With work counted as done by the gas, the first law reads dQ = dU + p dV. The first term is C_V dT on any path. The second is the expansion work, and the equation of state fixes it: with p constant, p dV = d(pV) = R dT. So dQ = (C_V + R) dT, and Cₚ − C_V = R with no assumption about the molecules at all. That R is the work per kelvin per mole spent pushing the atmosphere back: at 101 kPa, one mole heated by 10 K at constant pressure swells by ΔV = RΔT/p = 8.2 × 10⁻⁴ m³, about 0.82 L, and does 83 J of work. For a general substance the gap is Cₚ − C_V = TVα²/κ_T, with α the volume expansivity in K⁻¹ and κ_T the isothermal compressibility in Pa⁻¹; copper at 298 K gives 0.7 J K⁻¹ mol⁻¹, about 3% of its C_V, which is why solids are usually quoted with one heat capacity and gases must always be quoted with two. Dimensions check: K × m³ × K⁻² × Pa = J K⁻¹, per mole once V is molar.

How large C_V is for a particular molecule is a counting question, and the kinetic-theory topic Equipartition and molar heat capacities of gases (UPIII-06.08) has answered it: ½kT per active quadratic term, so C_V = (f/2)R, 3R/2 for a monatomic gas and 5R/2 for a diatomic whose vibration is frozen. The first law uses that number through the ratio γ = Cₚ/C_V = 1 + R/C_V, and uses it hardest where no heat flows. On a reversible adiabat, with work counted as done by the gas, 0 = nC_V dT + p dV: every joule of compression work stays in U, so a gas with a larger C_V warms less for the same squeeze. Differentiate pV = nRT to get p dV + V dp = nR dT, eliminate dT, and C_V V dp = −(C_V + R)p dV. Mayer's relation turns the bracket into Cₚ, so dp/p = −γ dV/V: the adiabat is steeper than the isotherm through the same point, where dp/p = −dV/V, by exactly the factor γ. Integrating that into pV^γ = constant, and finding its work, belongs to the next topic; what matters here is that γ can be measured. Sound is a compression too quick for heat to cross half a wavelength, so the stiffness it feels is the adiabatic γp rather than the isothermal p. The isothermal estimate √(RT/M) gives 290 m s⁻¹ for air at 293 K; the factor √1.40 = 1.18 lifts it to the measured 343 m s⁻¹. Invert the relation and a sound speed is a heat-capacity measurement, C_V = R/(γ − 1). Nitrogen at 300 K has γ = 1.40, so C_V = 20.8 J K⁻¹ mol⁻¹, the f = 5 value, where all seven classical terms would give 29.1 J K⁻¹ mol⁻¹ and γ = 9/7 = 1.29. Two terms are missing.

The classical Boltzmann average behind equipartition lets a vibrating bond take any energy at all. A quantum oscillator cannot: its energies are ħω(v + ½), spaced by ħω, and thermal agitation lifts it out of the ground state only if kT is comparable to that spacing. Boltzmann-weighting the ladder gives a mean thermal energy ⟨ε⟩ = ħω/(eˣ − 1) with x = ħω/kT = θ_vib/T, and differentiating with respect to T gives the mode's heat capacity, C_vib = R x² eˣ/(eˣ − 1)² per mole. Both limits check. For x → 0, eˣ − 1 ≈ x and the bracket tends to 1: C_vib → R, the classical pair of terms, so the quantum result contains the classical one as it must. For x → ∞ it tends to R x² e^(−x), which vanishes exponentially: the mode is frozen. Nitrogen's stretch has ħω = 0.29 eV and θ_vib = 3390 K. At 300 K, x = 11.3 and C_vib = 0.0016 R — a part in a thousand, invisible to a calorimeter. At 1000 K, x = 3.39 and the mode already carries 0.41 R, so C_V = 2.91 R = 24.2 J K⁻¹ mol⁻¹ and γ has fallen to 1.34. Nothing switched; the mode thawed.

Rotation has its own quantum. A rigid rotor's levels are E_J = J(J + 1)kθᵣₒₜ with θᵣₒₜ = ħ²/(2Ik), and the first excited level sits 2θᵣₒₜ above the ground state. For most gases θᵣₒₜ is tiny — 2.9 K for N₂, 2.1 K for O₂ — so rotation is fully classical at any temperature at which the substance is a gas. Hydrogen is the exception: its moment of inertia is so small that θᵣₒₜ = 85 K, and near 25 K hydrogen gas measures C_V ≈ 3R/2 and γ ≈ 1.66, behaving like a monatomic gas because its molecules cannot afford to spin. The running total is what γ = 1 + R/C_V reports, and the speed of sound reads it directly: v = √(γRT/M) gives 353 m s⁻¹ in nitrogen at 300 K with γ = 1.40, and would give 338 m s⁻¹ if γ were 9/7. A resonance tube is therefore a thermometer for degrees of freedom. Read C_V(T) as a staircase on a logarithmic temperature axis — 3R/2, then 5R/2 above a few θᵣₒₜ, then 7R/2 above a few θ_vib — with each riser smeared over roughly a decade of temperature, never a step at one point.

A crystal is the limiting case of a molecule whose terms are all vibrational. Each atom sits in a three-dimensional harmonic well, three kinetic and three potential terms, f = 6, so equipartition predicts C_V = 3R = 24.9 J K⁻¹ mol⁻¹ for every monatomic solid — the Dulong–Petit law, and copper, silver and lead all sit within a few per cent of it at room temperature. Diamond does not: its stiff, light lattice has vibrational temperatures near 2000 K, so at 298 K it holds only 6.1 J K⁻¹ mol⁻¹, a quarter of the classical value, and every solid's heat capacity falls towards zero as T → 0, as the third law requires. The rule is the same everywhere: a mode contributes its classical share once kT exceeds its level spacing by a factor of a few, and almost nothing when kT is a few times smaller. Equipartition is not wrong; it is the high-temperature limit of the quantum count, and the temperature at which a given mode joins in is set by ħω/k, which spectroscopy hands you directly.