Heat Capacity of Classical Gases

The heat capacity measures how much energy a substance absorbs per unit temperature rise: $C_V=(\partial U/\partial T)_V$ at fixed volume, $C_P=(\partial H/\partial T)_P$ at fixed pressure. For a classical ideal gas the internal energy depends only on temperature, so $C_V$ follows directly from how energy is partitioned among a molecule's microscopic degrees of freedom. The tool is the equipartition theorem: every coordinate or momentum entering the Hamiltonian quadratically carries a mean energy $\tfrac12 k_BT$ , independent of its coupling constant. Counting these quadratic terms gives a strikingly simple result. Three translational momenta yield $\tfrac32 k_BT$ per molecule for any gas. A linear molecule adds two rotational degrees of freedom (rotation about the symmetry axis is inert); a nonlinear molecule adds three. Each vibrational mode contributes two quadratic terms — one kinetic, one potential — and so carries a full $k_BT$ . Summing and multiplying by Avogadro's number gives molar heat capacities that are simple rational multiples of the gas constant $R=N_A k_B$ : $\tfrac32 R$ (monatomic), $\tfrac52 R$ (diatomic with rotation), $\tfrac72 R$ (with vibration active). Two universal relations round out the classical picture. First, Mayer's relation $C_P-C_V=R$ (per mole) for any ideal gas, a direct consequence of $H=U+PV=U+Nk_BT$ . Second, the adiabatic index $\gamma=C_P/C_V$ , which equals $\tfrac53$ , $\tfrac75$ , and $\tfrac97$ for the three cases and sets the speed of sound. The triumph and the failure of this picture are equally instructive: it correctly predicts $\tfrac32 R$ for noble gases and $\tfrac52 R$ for $\mathrm N_2$ and $\mathrm O_2$ near room temperature, but it also predicts these values should never change with temperature — which experiment flatly contradicts. That failure, the freezing-out of degrees of freedom, is the doorway to quantum statistical mechanics and is treated in detail in the companion article on the failure of equipartition.

Key takeaways

Classical equipartition assigns $\tfrac12 k_BT$ per quadratic degree of freedom, making molar heat capacity a sum of temperature-independent integer/half-integer multiples of $R$ .
Monatomic gas: $f=3$ translational $\Rightarrow C_V=\tfrac32 R$ ; diatomic with rotation: $f=5\Rightarrow C_V=\tfrac52 R$ ; with vibration active: $f=7\Rightarrow C_V=\tfrac72 R$ .
A vibrational mode counts as two quadratic terms (kinetic + potential), contributing $k_BT$ , not $\tfrac12 k_BT$ .
Mayer's relation: $C_P-C_V=R$ per mole for any ideal gas, because $H=U+Nk_BT$ .
Adiabatic index $\gamma=C_P/C_V=\tfrac53,\tfrac75,\tfrac97$ for the three cases; it sets the speed of sound $c_s=\sqrt{\gamma RT/M}$ .
Key idea: count the active quadratic degrees of freedom $f$ ; then $C_V=\tfrac f2 R$ , $C_P=C_V+R$ , and $\gamma=(f+2)/f$ — but "active" is temperature-dependent, and that is where classical physics breaks.

The core insight

Heat capacity is bookkeeping on degrees of freedom. Count the quadratic terms $f$ in the molecular Hamiltonian that are thermally active; each contributes $\tfrac12 R$ per mole. Then $C_V=\tfrac f2 R$ , $C_P=C_V+R$ (Mayer), and $\gamma=C_P/C_V=(f+2)/f$ . Monatomic $f=3$ , diatomic $f=5$ (with rotation) or $7$ (with vibration).

Where this lives today

The mode-counting rules are the backbone of chemical thermodynamics: statistical-mechanics software computes molar heat capacities, entropies, and equilibrium constants by summing translational, rotational, and vibrational partition functions for each molecule, with the classical $\tfrac f2 R$ values as the high-temperature limit and quantum corrections layered on top. In atmospheric science the temperature dependence of air's $\gamma$ (as vibrations begin to thaw at high altitude and in hot exhaust plumes) affects sound propagation and shock structure. The speed-of-sound-to-degree-of-freedom link is used in acoustic gas analysis and in calibrating the thermodynamics of exotic gases. And the Dulong-Petit value $3R$ remains the standard high-temperature benchmark against which the Debye model's low-temperature $T^3$ behavior — the subject of the companion failure-of-equipartition article — is measured.

The physical picture

Think of a gas molecule as a small machine with several ways to store energy: it can translate, rotate, and (as a spring connecting its atoms) vibrate. Equipartition says the thermal bath fills each quadratic storage slot to exactly $\tfrac12 k_BT$ on average, like water finding the same level in connected vessels. Heat capacity is then just the count of open slots times $\tfrac12 R$ per mole — the more ways a molecule can hold energy, the more heat it takes to warm it by a degree. But the slots are not all open at once. Quantum mechanics quantizes each mode, and a mode stays frozen — stuck in its ground state, storing nothing — until $k_BT$ climbs above its level spacing. Translation has an effectively continuous spectrum and is always active; rotation activates at low temperatures (a few kelvin for most molecules, but $\sim85\,\mathrm K$ for light $\mathrm H_2$ ); vibration needs thousands of kelvin. So the heat capacity of a real gas rises in steps as temperature climbs and new slots open, tracing a staircase from $\tfrac32 R$ to $\tfrac52 R$ to $\tfrac72 R$ .

Figure

A diatomic molecule stores energy in translation (3 quadratic terms), rotation (2, about the two perpendicular axes), and vibration (2: kinetic + potential of the bond spring). Rotation about the bond axis is inert. Each active quadratic slot holds ½kT.

The payoff is the staircase itself, plotted for hydrogen. At the lowest temperatures only translation is active and $C_V=\tfrac32 R$ . As $T$ rises past $\Theta_{\rm rot}$ , rotation switches on and $C_V$ climbs to $\tfrac52 R$ . Far higher, past $\Theta_{\rm vib}$ , vibration activates and $C_V$ approaches $\tfrac72 R$ . Each flat tread is a classical equipartition plateau; each riser is a quantum mode thawing. The classical theory correctly predicts the heights of the treads but is silent about the risers — which is precisely its boundary.

Figure

Punchline: the molar heat capacity C_V/R of a diatomic gas climbs in steps — 3/2 (translation) → 5/2 (rotation) → 7/2 (vibration). Classical equipartition predicts the plateau heights exactly; the quantum freeze-out sets the risers.

The core equations

For a classical ideal gas the internal energy is the equipartition sum over active quadratic degrees of freedom $f$ , and the heat capacity is its temperature derivative. Because $U$ is linear in $T$ , $C_V$ is a constant.

U=\frac f2 N k_BT,\qquad C_V=\left(\frac{\partial U}{\partial T}\right)_V=\frac f2 N k_B=\frac f2 R\ \text{(per mole)} \qquad (1)

where: $f$ — number of active quadratic degrees of freedom (3 translational always; +2 or +3 rotational; +2 per active vibrational mode); $R=N_A k_B=8.314\,\mathrm{J\,mol^{-1}K^{-1}}$ . Reading: heat capacity counts storage slots — each active quadratic slot is worth $\tfrac12 R$ per mole. The constant-pressure heat capacity exceeds $C_V$ by exactly $R$ , because at constant pressure some of the added heat does expansion work. This is Mayer's relation.

C_P-C_V=R,\qquad \gamma=\frac{C_P}{C_V}=\frac{f+2}{f} \qquad (2)

where: $C_P$ — molar heat capacity at constant pressure; $\gamma$ — adiabatic index. Reading: $\gamma$ decreases as more degrees of freedom open ( $\tfrac53\to\tfrac75\to\tfrac97$ ), because the same $R$ of expansion work is spread over a larger $C_V$ . The adiabatic index controls the speed of sound, tying an abstract mode count to a directly measurable macroscopic quantity.

c_s=\sqrt{\frac{\gamma R T}{M}} \qquad (3)

where: $c_s$ — adiabatic sound speed; $M$ — molar mass. Reading: measuring the speed of sound in a gas is, in effect, counting its molecular degrees of freedom.

Derivation AMonatomic and diatomic heat capacities by counting quadratic terms

The goal is to derive the molar heat capacities $\tfrac32 R$ , $\tfrac52 R$ , $\tfrac72 R$ from the equipartition count of quadratic terms in the molecular Hamiltonian, and to fix the subtle point that a vibrational mode counts as two quadratic terms while the axial rotation of a linear molecule counts as none. This is the calculation that turns molecular structure into a measurable heat capacity and adiabatic index.

Step 1. Monatomic gas. The Hamiltonian per atom is purely translational, $H=(p_x^2+p_y^2+p_z^2)/2m$ : three quadratic momentum terms, $f=3$ . By equipartition $\langle H\rangle=\tfrac32 k_BT$ , so $U=\tfrac32 Nk_BT$ and $C_V=\tfrac32 R$ per mole.

Step 2. Diatomic gas, rotation active. Add rotation about the two axes perpendicular to the bond: $H_{\rm rot}=L_1^2/2I_1+L_2^2/2I_2$ , two quadratic terms. Why only two, not three: rotation about the bond axis has a negligible moment of inertia (the mass is concentrated on the axis), so its rotational quantum is enormous and the mode is always frozen — it contributes nothing. Total $f=3+2=5$ , so $C_V=\tfrac52 R$ .

Step 3. Diatomic gas, vibration active. Add the bond vibration, modeled as a harmonic oscillator $H_{\rm vib}=p^2/2\mu+\tfrac12\mu\omega^2 x^2$ : two quadratic terms (kinetic and potential). Why a vibration is worth $k_BT$ , not $\tfrac12 k_BT$ : both the kinetic and potential energies are quadratic, and each earns $\tfrac12 k_BT$ , so a fully active vibrational mode contributes $k_BT$ . Total $f=3+2+2=7$ , so $C_V=\tfrac72 R$ .

Step 4. Adiabatic indices: $\gamma=(f+2)/f$ gives $\tfrac53$ ( $f=3$ ), $\tfrac75=1.4$ ( $f=5$ ), $\tfrac97\approx1.29$ ( $f=7$ ). Why $\gamma$ falls: $\gamma=1+R/C_V=1+2/f$ , so more degrees of freedom mean a smaller $\gamma$ . Verified: the mode counts $f=3,5,7$ , the heat capacities $\tfrac32 R,\tfrac52 R,\tfrac72 R$ , and the indices $\gamma=\tfrac53,\tfrac75,\tfrac97$ are confirmed symbolically, along with the numeric $\tfrac52 R=20.79\,\mathrm{J\,mol^{-1}K^{-1}}$ , in verify_aq-sm-heat-capacity-classical-gases.py.

Go deeper · Derivation B · 1 minMayer's relation C_P − C_V = R from the enthalpy

Derivation BMayer's relation C_P − C_V = R from the enthalpy

The goal is to prove that for any ideal gas the constant-pressure heat capacity exceeds the constant-volume one by exactly the gas constant $R$ per mole, regardless of the number of internal degrees of freedom, and to interpret the difference as the expansion work done when heating at constant pressure. This universal relation lets one measure $\gamma$ and infer $f$ from either heat capacity.

Step 1. Definitions: $C_V=(\partial U/\partial T)_V$ and $C_P=(\partial H/\partial T)_P$ , where $H=U+PV$ is the enthalpy. For an ideal gas $U=U(T)$ depends only on temperature (no interactions), and $PV=Nk_BT$ .

Step 2. Write the enthalpy of one mole: $H=U+PV=U(T)+RT$ . Why $PV=RT$ contributes: at constant pressure, warming the gas by $\mathrm dT$ expands it, and the term $RT$ is the work-plus-flow-energy bookkeeping that enthalpy carries.

Step 3. Differentiate at constant pressure: $C_P=\dfrac{\mathrm dH}{\mathrm dT}=\dfrac{\mathrm dU}{\mathrm dT}+R=C_V+R$ . Since $U$ depends only on $T$ , $\mathrm dU/\mathrm dT=C_V$ regardless of the process. Hence $C_P-C_V=R$ , equation (2), independent of $f$ .

Step 4. Physical reading: at constant volume all added heat raises $U$ ; at constant pressure the gas also does expansion work $P\,\mathrm dV=R\,\mathrm dT$ per mole, so more heat is needed for the same $\mathrm dT$ — exactly $R$ more. Verified: $C_P=C_V+R$ from $H=U+RT$ (symbolic), and the resulting $\gamma$ values, are confirmed in the verify script.

Limiting cases

Case	$C_V$	$\gamma$	Example
Monatomic ( $f=3$ )	$\tfrac32 R=12.47$	$5/3\approx1.667$	He, Ar, Ne
Diatomic, rotation ( $f=5$ )	$\tfrac52 R=20.79$	$7/5=1.400$	$\mathrm N_2,\mathrm O_2$ at 300 K
Diatomic, +vibration ( $f=7$ )	$\tfrac72 R=29.10$	$9/7\approx1.286$	$\mathrm N_2$ above $\sim3000\,\mathrm K$
Nonlinear polyatomic ( $f=6$ , rot only)	$3R=24.94$	$4/3\approx1.333$	$\mathrm{H_2O,CO_2}$ (bent) rot part
Solid (Dulong-Petit, $f=6$ )	$3R=24.94$	—	Cu, Ag, Pb above $\Theta_D$

Worked examples

Example 1 — Predicting the heat capacity and adiabatic index of nitrogen at room temperature (count, apply, check). $\mathrm N_2$ at $T=300\,\mathrm K$ : at this temperature rotation is fully active but vibration ( $\Theta_{\rm vib}\approx3340\,\mathrm K$ ) is frozen.

Step 1. Count active quadratic terms: 3 translational + 2 rotational (linear molecule) + 0 vibrational = $f=5$ .

Step 2. Apply equation (1): $C_V=\tfrac52 R=\tfrac52(8.314)=\mathbf{20.79\,J\,mol^{-1}K^{-1}}$ ; $C_P=C_V+R=\tfrac72 R=29.10\,\mathrm{J\,mol^{-1}K^{-1}}$ ; $\gamma=C_P/C_V=7/5=\mathbf{1.400}$ .

Step 3. Boundary check: the measured $C_V$ of $\mathrm N_2$ at $300\,\mathrm K$ is $20.8\,\mathrm{J\,mol^{-1}K^{-1}}$ and $\gamma=1.40$ — an essentially exact match, confirming that vibration is frozen out and only 5 degrees of freedom are active. If vibration were active, $C_V$ would be $\tfrac72 R=29.1$ and $\gamma$ would drop to $1.29$ .

Example 2 — The speed of sound in air from the adiabatic index. Air is mostly diatomic ( $\mathrm N_2,\mathrm O_2$ ), so $\gamma=1.40$ and mean molar mass $M=29\,\mathrm{g/mol}=0.029\,\mathrm{kg/mol}$ .

Step 1. From equation (3), $c_s=\sqrt{\gamma RT/M}$ . At $T=293\,\mathrm K$ (20 °C), $\gamma RT=1.40(8.314)(293)=3410\,\mathrm{J/mol}$ .

Step 2. Divide by $M$ : $c_s=\sqrt{3410/0.029}=\sqrt{1.176\times10^5}=\mathbf{343\,m/s}$ .

Step 3. Boundary check: the measured speed of sound in dry air at $20\,^\circ\mathrm C$ is $343\,\mathrm{m/s}$ — exact. Had we (wrongly) used the isothermal relation $\gamma=1$ , we would get $290\,\mathrm{m/s}$ , the value Newton originally computed and which puzzled physicists for a century until Laplace supplied the adiabatic $\gamma$ . The audible speed of sound is a direct measurement of the molecular degree-of-freedom count.

Origins · 1 mindulong-petit law and how this was discovered

dulong-petit law

statistical-mechanics concept · Dulong & Petit 1819; Mayer 1842

The Dulong-Petit law (Pierre Louis Dulong and Alexis Thérèse Petit, 1819) states that the molar heat capacity of a crystalline solid is $3R\approx24.9\,\mathrm{J\,mol^{-1}K^{-1}}$ , independent of the material — the equipartition result for $3N$ oscillators, each with two quadratic terms. It was one of the first quantitative successes of the kinetic-molecular picture and let chemists estimate atomic weights. Its low-temperature failure (heat capacities falling toward zero as $T^3$ ) was, like the diatomic-gas freeze-out, a decisive clue that classical equipartition was incomplete, resolved only by the Einstein and Debye quantum models. For gases the analogous universal relation is Mayer's ( $C_P-C_V=R$ , Julius Robert von Mayer 1842), which follows from the ideal-gas enthalpy and holds for any number of internal degrees of freedom.

History

1819

— Dulong and Petit find the molar heat capacity of solids is nearly

3R

, independent of substance.

1842

— Mayer states

C_P-C_V=R

and estimates the mechanical equivalent of heat.

1859

— Maxwell derives equipartition of translational and rotational energy.

1876

— Boltzmann formalizes equipartition for arbitrary quadratic degrees of freedom.

1877

— the anomalously low room-temperature heat capacity of diatomic gases is recognized as an open puzzle.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., §3.7, §6.5 · Kardar, Statistical Physics of Particles, Ch. 4 · Reif, Fundamentals of Statistical and Thermal Physics, §7.5–7.7 · Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, Ch. 6 · Landau & Lifshitz, Statistical Physics Part 1, §44–§47.

Provenance. All numbers are standard textbook values: $R=8.314\,462\,\mathrm{J\,mol^{-1}K^{-1}}$ , $k_B=1.380649\times10^{-23}\,\mathrm{J\,K^{-1}}$ , $N_A=6.022\,140\,76\times10^{23}\,\mathrm{mol^{-1}}$ (CODATA/SI). Heat-capacity values $\tfrac32 R,\tfrac52 R,\tfrac72 R$ and indices $\gamma=5/3,7/5,9/7$ are derivable by mode counting (Reif §7.5, Pathria §3.7). Characteristic temperatures ( $\Theta_{\rm rot}\approx85\,\mathrm K$ for $\mathrm H_2$ , $\approx2.9\,\mathrm K$ for $\mathrm N_2$ ; $\Theta_{\rm vib}\approx3340\,\mathrm K$ for $\mathrm N_2$ ) are standard molecular-spectroscopy values (Pathria §6.5, Kardar Ch. 4). Sound speed in air $\approx343\,\mathrm{m\,s^{-1}}$ at $20\,^\circ\mathrm C$ follows from $c_s=\sqrt{\gamma RT/M}$ with $\gamma=1.40$ , $M=29\,\mathrm{g\,mol^{-1}}$ . A verifier should recompute the Gaussian integral behind equipartition, confirm Mayer's relation from $H=U+Nk_BT$ , and check the numeric heat capacities. All checks are in verify_aq-sm-heat-capacity-classical-gases.py.

Builds on: Equipartition Theorem · Classical Ideal Gas · Generalized Equipartition Theorem · Maxwell-Boltzmann Velocity Distribution Leads to: Sackur-Tetrode Equation · Virial Equation of State · Thermal de Broglie Wavelength

Why is a diatomic gas 5/2 R and not 7/2 R at room temperature?

Because vibration is frozen out. A diatomic molecule has 3 translational, 2 rotational, and 2 vibrational quadratic degrees of freedom, which would give $f=7$ and $C_V=\tfrac72 R$ if all were active. But the vibrational quantum $\hbar\omega$ corresponds to a characteristic temperature of thousands of kelvin (e.g. $\Theta_{\rm vib}\approx3340\,\mathrm K$ for $\mathrm N_2$ ), far above room temperature, so at 300 K almost no molecules are vibrationally excited and the mode contributes nothing. Only translation and rotation are active, giving $f=5$ and $C_V=\tfrac52 R$ . Heat a diatomic gas to a few thousand kelvin and $C_V$ does climb toward $\tfrac72 R$ .

Why does a linear molecule have only two rotational degrees of freedom?

Because rotation about the molecular (bond) axis has a vanishingly small moment of inertia — nearly all the mass sits on that axis, so spinning about it barely stores any energy. Quantum-mechanically, a tiny moment of inertia means a huge rotational level spacing ( $\hbar^2/2I$ is enormous), so the mode's characteristic temperature is astronomically high and it is always frozen. Only the two rotations about axes perpendicular to the bond are active. A nonlinear molecule, by contrast, has three genuine rotational axes with comparable moments of inertia, so it gets three rotational degrees of freedom.

Where does Mayer's relation C_P − C_V = R come from physically?

At constant volume, all the heat you add goes into raising the internal energy, so $C_V=\mathrm dU/\mathrm dT$ . At constant pressure, the gas also expands as it warms, doing work $P\,\mathrm dV$ against its surroundings; you must supply that work in addition to raising $U$ . For an ideal gas $PV=RT$ per mole, so $P\,\mathrm dV=R\,\mathrm dT$ , meaning you need exactly $R$ more joules per kelvin at constant pressure than at constant volume. Hence $C_P=C_V+R$ . The relation is universal for ideal gases because it depends only on $PV=RT$ , not on the internal structure of the molecule.

How does the speed of sound reveal the number of degrees of freedom?

Sound is an adiabatic (fast, no heat exchange) compression wave, so its speed depends on the adiabatic index $\gamma=C_P/C_V$ , via $c_s=\sqrt{\gamma RT/M}$ . Since $\gamma=(f+2)/f$ depends only on the active degree-of-freedom count $f$ , measuring the sound speed at known $T$ and $M$ pins down $f$ . This is exactly how the puzzle of Newton's too-low sound-speed prediction was resolved: Newton assumed isothermal compression ( $\gamma=1$ ) and got $290\,\mathrm{m/s}$ ; Laplace realized the compression is adiabatic and used $\gamma=1.4$ , obtaining the correct $343\,\mathrm{m/s}$ .

Does the heat capacity of a monatomic gas ever change with temperature?

Not for its translational motion — a point atom has only 3 translational degrees of freedom, and translation has an essentially continuous energy spectrum, so it is always active and $C_V=\tfrac32 R$ from near absolute zero up to where the atom ionizes or dissociates. This is why noble gases are the cleanest test of equipartition: $C_V=\tfrac32 R$ and $\gamma=5/3$ hold over an enormous temperature range. Electronic excitation can add a contribution at very high temperatures, but for ordinary conditions a monatomic gas has a genuinely temperature-independent heat capacity — the one case where the classical prediction of constancy is actually correct.

Why does classical equipartition predict constant heat capacity when experiments show it rising with temperature?

Because equipartition is purely classical: it treats every quadratic mode as always available, with no notion of a minimum excitation energy. Real molecular modes are quantized, and a mode contributes to the heat capacity only once $k_BT$ exceeds its level spacing. Below that threshold the mode is "frozen" and contributes nothing; as temperature rises it thaws and its $\tfrac12 R$ or $R$ switches on. So the true heat capacity is a staircase rising with temperature, while classical equipartition sees only the flat treads and predicts each is permanent. The risers — the temperature dependence — are a quantum effect, treated fully in the failure-of-equipartition article.