Rotational and Vibrational Partition Functions of Diatomic Gases

The heat capacity of a diatomic gas is not a single number but a staircase. At the coldest temperatures a hydrogen molecule stores energy only in its centre-of-mass motion, and $C_V=\tfrac{3}{2}Nk_B$ — exactly as for a monatomic gas. Warm it, and near a characteristic temperature the molecule suddenly begins to tumble: two rotational degrees of freedom switch on and $C_V$ climbs to $\tfrac{5}{2}Nk_B$ . Warm it much further, past a second and far higher characteristic temperature, and the bond begins to stretch and compress: the vibrational mode activates and $C_V$ rises again toward $\tfrac{7}{2}Nk_B$ . Classical equipartition predicts $\tfrac{7}{2}Nk_B$ at all temperatures and cannot explain why the extra modes are "frozen out" at low temperature. The resolution is that each mode is quantized, and a mode contributes to the heat capacity only once $k_BT$ becomes comparable to its level spacing. This article builds the diatomic partition function by factorizing it into translational, rotational, and vibrational pieces, and derives the rotational and vibrational contributions to $C_V$ from first principles. It owns the temperature staircase — the two characteristic temperatures $\theta_{\rm rot}=\hbar^2/2Ik_B$ and $\theta_{\rm vib}=\hbar\omega/k_B$ , the high-temperature equipartition limits, the low-temperature exponential freeze-out, and the non-monotone "hump" in the rotational heat capacity. It states honestly what the naïve rotational sum gets wrong for hydrogen — the symmetry number and the ortho/para nuclear-spin statistics that Dennison resolved in 1927 — without re-deriving the quantum energy levels, which are taken as given.

Key takeaways

The single-molecule partition function factorizes: $Z=Z_{\rm trans}Z_{\rm rot}Z_{\rm vib}Z_{\rm elec}$ , so each mode adds independently to the energy and the heat capacity.
Rotation switches on near $\theta_{\rm rot}=\hbar^2/2Ik_B$ ; vibration switches on near the much larger $\theta_{\rm vib}=\hbar\omega/k_B$ . For $\rm H_2$ , $\theta_{\rm rot}\approx85$ K and $\theta_{\rm vib}\approx6215$ K.
$C_V/Nk_B$ climbs a staircase: $\tfrac{3}{2}$ (translation only) $\to\tfrac{5}{2}$ (rotation active) $\to\tfrac{7}{2}$ (vibration active).
At high temperature each quadratic degree of freedom obeys equipartition ( $\zeta_{\rm rot}\approx T/\sigma\theta_{\rm rot}$ , $C_{\rm rot}\to Nk_B$ ); at low temperature it freezes out exponentially.
The rotational heat capacity is non-monotone: it overshoots to $\approx1.1\,Nk_B$ near $T\approx\theta_{\rm rot}$ before settling to $Nk_B$ .
Key idea: a quantized mode contributes to $C_V$ only when $k_BT$ exceeds its level spacing, so the diatomic heat capacity climbs a staircase set by $\theta_{\rm rot}\ll\theta_{\rm vib}$ .

The core insight

A diatomic molecule stores energy in translation, rotation, and vibration, but each is quantized with its own level spacing. A mode contributes its full equipartition share to $C_V$ only when $k_BT$ is large compared with that spacing; below it the mode is frozen. Because $\theta_{\rm rot}\ll\theta_{\rm vib}$ , the modes wake up one at a time and $C_V$ rises in steps from $\tfrac{3}{2}Nk_B$ to $\tfrac{7}{2}Nk_B$ .

Where this lives today

The temperature staircase is a standard result of physical chemistry and molecular spectroscopy, where $\theta_{\rm rot}$ and $\theta_{\rm vib}$ are read off microwave and infrared spectra and fed straight into thermodynamic tables of molecular entropy, enthalpy, and heat capacity. Ortho/para hydrogen is not a curiosity: liquid-hydrogen fuel must be catalytically converted to the para form before storage, because the slow ortho $\to$ para conversion releases enough heat to boil off a stored tank. The same partition-function machinery, extended to polyatomic molecules with three rotational axes and many vibrational modes, underlies combustion modelling, atmospheric radiative transfer, and astrochemical abundance analysis. The vibrational Einstein term here is the molecular twin of the Einstein model of a solid, and the rotational sum reappears wherever quantized angular momentum meets a thermal bath. The deeper corrections — vibration–rotation coupling, centrifugal stretching, anharmonicity, and the full nuclear-spin statistics — are the province of molecular spectroscopy and are named, not derived, here.

The physical picture

Think of a single diatomic molecule as two atoms joined by a stiff spring. It can do three things with energy. It can fly through space — translation, the same three quadratic degrees of freedom a monatomic atom has. It can tumble end over end about the two axes perpendicular to the bond — rotation, two more quadratic degrees of freedom (spinning about the bond axis costs essentially nothing because the moment of inertia there is tiny). And it can vibrate, the two atoms oscillating along the bond — one vibrational mode, which counts as two quadratic degrees of freedom because it stores energy in both kinetic and potential form. Classical equipartition would hand $\tfrac12 k_BT$ to each of these seven quadratic slots, giving $U=\tfrac72Nk_BT$ and $C_V=\tfrac72Nk_B$ , flat at every temperature. Nature disagrees, and the reason is quantum. Each mode has discrete energy levels. Rotational levels are spaced by $\sim\hbar^2/2I$ ; vibrational levels by $\hbar\omega$ . When the thermal energy $k_BT$ is far smaller than a mode's spacing, essentially every molecule sits in that mode's ground state, no molecule can absorb a small amount of heat into it, and the mode contributes nothing to $C_V$ : it is frozen out. As $T$ rises through the characteristic temperature $\theta=(\text{spacing})/k_B$ , the mode thaws and begins to soak up energy, and once $k_BT$ dwarfs the spacing the levels look effectively continuous and equipartition is recovered. Because the rotational spacing is hundreds of times smaller than the vibrational one, rotation thaws first and vibration much later.

Figure

Heat capacity of a diatomic gas climbs a staircase as modes unfreeze

There is a decisive detail hiding on the first riser. The rotational contribution does not rise smoothly to its plateau — it overshoots. As rotation thaws, $C_{\rm rot}$ climbs above the equipartition value $Nk_B$ , peaks at about $1.1\,Nk_B$ near $T\approx\theta_{\rm rot}$ , and only then relaxes back down to $Nk_B$ . This little hump is a genuine quantum fingerprint: it comes from the discrete, unevenly spaced, degeneracy-weighted rotational ladder, and a purely classical rotor would never produce it.

Figure

The rotational heat capacity overshoots equipartition before settling

The core equations

The whole treatment rests on factorizing the molecular partition function and reading each mode's heat capacity off its own sub-partition-function. Because the modes are (to a very good approximation) independent, the logarithms add and so do the energies and heat capacities.

Z = Z_{\rm trans}\,Z_{\rm rot}\,Z_{\rm vib}\,Z_{\rm elec} \qquad (1)

where: $Z$ is the single-molecule partition function and the four factors are the translational, rotational, vibrational, and electronic partition functions. Because $Z$ is a product, $\ln Z$ is a sum, and every additive thermodynamic quantity — the mean energy $U=Nk_BT^2\,\partial\ln Z/\partial T$ and hence $C_V=\partial U/\partial T$ — splits into independent mode contributions. The electronic factor is a frozen constant ( $Z_{\rm elec}\approx g_0$ ) at all ordinary temperatures and contributes nothing to $C_V$ . The rigid-rotor energy levels and the rotational partition function set the first riser of the staircase.

E_J=\frac{\hbar^2}{2I}\,J(J+1),\quad g_J=2J+1,\qquad \zeta_{\rm rot}=\frac{1}{\sigma}\sum_{J=0}^{\infty}(2J+1)\,e^{-J(J+1)\theta_{\rm rot}/T} \qquad (2)

where: $E_J$ is the energy of rotational level $J$ , $I$ the molecular moment of inertia, $g_J=2J+1$ the degeneracy of level $J$ , and $\theta_{\rm rot}\equiv\hbar^2/2Ik_B$ the rotational temperature. The symmetry number $\sigma$ is $1$ for a heteronuclear molecule (e.g. HCl) and $2$ for a homonuclear one (e.g. $\rm H_2$ , $\rm N_2$ ), correcting for the double-counting of indistinguishable end-over-end orientations. At high temperature the rotational sum becomes an integral (the Euler–Maclaurin limit), giving a clean closed form.

\zeta_{\rm rot}\;\xrightarrow{\,T\gg\theta_{\rm rot}\,}\;\frac{T}{\sigma\,\theta_{\rm rot}}\qquad\Longrightarrow\qquad C_{\rm rot}\to Nk_B \qquad (3)

where: the classical rotor stores $\tfrac12 k_BT$ in each of its two rotational degrees of freedom, so $C_{\rm rot}\to Nk_B$ — the equipartition result. The freeze-out at low $T$ is the failure of this limit when only the $J=0$ and $J=1$ levels are appreciably populated. The vibrational mode is a single quantum harmonic oscillator, and its partition function and heat capacity take the Einstein form.

Z_{\rm vib}=\frac{e^{-\theta_{\rm vib}/2T}}{1-e^{-\theta_{\rm vib}/T}},\qquad C_{\rm vib}=Nk_B\left(\frac{\theta_{\rm vib}}{T}\right)^2\frac{e^{\theta_{\rm vib}/T}}{\left(e^{\theta_{\rm vib}/T}-1\right)^2} \qquad (4)

where: $\theta_{\rm vib}\equiv\hbar\omega/k_B$ is the vibrational temperature set by the bond's oscillation frequency $\omega$ . At $T\gg\theta_{\rm vib}$ this tends to $Nk_B$ (the oscillator's kinetic and potential quadratic slots, two degrees of freedom); at $T\ll\theta_{\rm vib}$ it dies exponentially as $(\theta_{\rm vib}/T)^2e^{-\theta_{\rm vib}/T}$ . Adding the three active pieces gives the total heat capacity that traces the staircase.

\frac{C_V}{Nk_B}=\underbrace{\tfrac{3}{2}}_{\rm trans}+\underbrace{C_{\rm rot}/Nk_B}_{0\to1}+\underbrace{C_{\rm vib}/Nk_B}_{0\to1} \qquad (5)

where: the translational term is always $\tfrac32$ (the centre-of-mass motion never freezes for a gas), while the rotational and vibrational terms each climb from $0$ to $1$ as their characteristic temperatures are crossed, producing the plateaus at $\tfrac32,\tfrac52,\tfrac72$ .

Derivation AHigh-temperature rotational partition function and the equipartition limit

The goal is to show that the discrete rotational sum in equation (2) collapses at high temperature to the simple result $\zeta_{\rm rot}=T/\sigma\theta_{\rm rot}$ of equation (3), and that this in turn forces the rotational heat capacity to the equipartition value $Nk_B$ . The strategy is to recognize that when $\theta_{\rm rot}/T\ll1$ the summand varies slowly from one $J$ to the next, so the sum is well approximated by an integral; the one subtlety is that the integral must be done in the natural variable $J(J+1)$ , not $J$ , which makes it elementary. We keep only the leading term, which is all that survives in the strict high-temperature limit.

Step 1. Write the summand of $\sigma\zeta_{\rm rot}$ as $f(J)=(2J+1)\,e^{-J(J+1)\theta_{\rm rot}/T}$ and notice that $2J+1=\dfrac{d}{dJ}[J(J+1)]$ . Substituting $u=J(J+1)$ turns the sum into $\int_0^\infty e^{-u\theta_{\rm rot}/T}\,du$ once the spacing between successive $J$ is small compared with $T/\theta_{\rm rot}$ .

Step 2. Evaluate the integral: $\int_0^\infty e^{-u\theta_{\rm rot}/T}\,du = T/\theta_{\rm rot}$ . Dividing by the symmetry number gives $\zeta_{\rm rot}\approx T/\sigma\theta_{\rm rot}$ , equation (3). (The Euler–Maclaurin correction terms are $\tfrac13+\tfrac{\theta_{\rm rot}}{15T}+\dots$ , negligible at high $T$ but responsible for the small overshoot discussed below.)

Step 3. Extract the energy: $U_{\rm rot}=Nk_BT^2\,\partial\ln\zeta_{\rm rot}/\partial T = Nk_BT^2\,\partial(\ln T-\ln\sigma\theta_{\rm rot})/\partial T = Nk_BT$ . Each molecule carries $k_BT$ of rotational energy — exactly $\tfrac12 k_BT$ in each of the two rotational degrees of freedom.

Step 4. Differentiate: $C_{\rm rot}=\partial U_{\rm rot}/\partial T = Nk_B$ . This is the equipartition plateau. The low-temperature story is the opposite regime: keeping only $J=0$ (weight $1$ ) and $J=1$ (weight $3$ , energy $2\theta_{\rm rot}k_B$ ) gives $\zeta_{\rm rot}\approx1+3e^{-2\theta_{\rm rot}/T}$ and a heat capacity $\propto(\theta_{\rm rot}/T)^2e^{-2\theta_{\rm rot}/T}\to0$ : rotation is frozen. Verified: the companion verifier sums equation (2) numerically and confirms $\zeta_{\rm rot}(1000\,{\rm K})=6.03$ against $T/2\theta_{\rm rot}=5.86$ for $\rm H_2$ , that $C_{\rm rot}/Nk_B\to1$ at high $T$ and $\to0$ at low $T$ , and that the low- $T$ two-level approximation reproduces the leading exponential.

Go deeper · Derivation B · 1 minThe rotational heat-capacity overshoot

Derivation BThe rotational heat-capacity overshoot

The aim here is to establish the article's owned quantum fingerprint: the rotational heat capacity does not approach its equipartition plateau from below and level off — it rises above $Nk_B$ , reaching a hump of about $1.1\,Nk_B$ near $T\approx0.8\,\theta_{\rm rot}$ , and then descends to $Nk_B$ from above. The strategy is to compute $C_{\rm rot}(T)$ exactly from the finite sum and locate its maximum numerically, since no elementary closed form exists; the physical point to keep in view is that the overshoot is a consequence of the discreteness and degeneracy weighting of the rotational ladder, not of any interaction.

Step 1. Define the rotational energy from the exact sum, $U_{\rm rot}=Nk_B\theta_{\rm rot}\,\dfrac{\sum_J(2J+1)J(J+1)e^{-J(J+1)\theta_{\rm rot}/T}}{\sum_J(2J+1)e^{-J(J+1)\theta_{\rm rot}/T}}$ , which is $Nk_B\theta_{\rm rot}$ times the thermal average $\langle J(J+1)\rangle$ .

Step 2. The heat capacity is the temperature derivative, equivalently $C_{\rm rot}=Nk_B\big(\theta_{\rm rot}/T\big)^2\big[\langle (J(J+1))^2\rangle-\langle J(J+1)\rangle^2\big]$ — a fluctuation formula: $C_{\rm rot}$ measures the variance of the rotational quantum number's spread. At low $T$ that variance is tiny (all in $J=0$ ); at high $T$ it saturates to the classical value.

Step 3. In between, the variance briefly exceeds its classical asymptote because the ladder is opening up faster than linearly as successive degenerate multiplets $(2J+1)$ come into play. Evaluated numerically, $C_{\rm rot}/Nk_B$ crests at $\approx1.1$ near $T\approx0.8\,\theta_{\rm rot}$ .

Step 4. Interpret: the overshoot is small (about ten percent) and lives in the narrow window where rotation is half-thawed. It is washed out in ordinary gases at room temperature (where rotation is fully classical) but is a clean signature in cold molecular hydrogen and deuterium, and its exact shape depends on the ortho/para mixture discussed below. Verified: the verifier tabulates $C_{\rm rot}(T)/Nk_B$ for $\rm H_2$ at $T=20,50,85.3,150,300,1000$ K as $0.04,0.97,1.07,1.01,1.00,1.00$ and locates the maximum of $\approx1.1$ near $T\approx0.8\,\theta_{\rm rot}$ .

Go deeper · Derivation C · 1 minVibrational activation and the top step of the staircase

Derivation CVibrational activation and the top step of the staircase

The purpose of this derivation is to obtain the vibrational heat capacity of equation (4) from the harmonic-oscillator partition function and to show why the top step of the staircase, $\tfrac72Nk_B$ , is reached only at temperatures far above room temperature for a stiff molecule like $\rm H_2$ . The strategy is to sum the geometric series of oscillator levels and differentiate twice; the subtlety worth flagging is that the mode counts as two quadratic degrees of freedom (kinetic and potential), so its full contribution is $Nk_B$ , not $\tfrac12Nk_B$ .

Step 1. Sum the levels $E_v=\hbar\omega(v+\tfrac12)$ : $Z_{\rm vib}=\sum_{v=0}^\infty e^{-\beta\hbar\omega(v+1/2)}=\dfrac{e^{-\theta_{\rm vib}/2T}}{1-e^{-\theta_{\rm vib}/T}}$ , a geometric series with ratio $e^{-\theta_{\rm vib}/T}$ . The zero-point factor $e^{-\theta_{\rm vib}/2T}$ shifts the energy origin but drops out of the heat capacity.

Step 2. The mean vibrational energy is $U_{\rm vib}=Nk_B\theta_{\rm vib}\!\left[\tfrac12+\dfrac{1}{e^{\theta_{\rm vib}/T}-1}\right]$ , the Planck/Bose occupation of a single mode plus its zero-point energy.

Step 3. Differentiating gives the Einstein heat capacity of equation (4). Expanding at high $T$ ( $\theta_{\rm vib}/T\to0$ ) yields $C_{\rm vib}\to Nk_B[1-\tfrac{1}{12}(\theta_{\rm vib}/T)^2+\dots]\to Nk_B$ ; at low $T$ it is $Nk_B(\theta_{\rm vib}/T)^2e^{-\theta_{\rm vib}/T}\to0$ .

Step 4. Numbers make the freeze-out vivid: for $\rm H_2$ with $\theta_{\rm vib}\approx6215$ K, at room temperature $\theta_{\rm vib}/T\approx21$ , so $C_{\rm vib}$ is utterly negligible and $C_V\approx\tfrac52Nk_B$ ; only near $\sim3000$ K does vibration contribute appreciably, and $\tfrac72Nk_B$ is approached only well above that — by which point real $\rm H_2$ has begun to dissociate. Verified: the verifier reproduces $C_{\rm vib}/Nk_B$ for $\rm H_2$ at $T=300,1000,3000,6215$ K as $0.00,0.078,0.71,0.92$ , and confirms the high- $T$ limit $\to1$ and the low- $T$ exponential.

Limiting cases

Each regime is a check on the construction and on the physics of freeze-out.

Regime	Result	Meaning
$T\ll\theta_{\rm rot}$	$C_V\to\tfrac32Nk_B$	Rotation and vibration frozen; the gas is effectively monatomic.
$\theta_{\rm rot}\ll T\ll\theta_{\rm vib}$	$C_V\to\tfrac52Nk_B$	Rotation classical, vibration still frozen — the room-temperature value for $\rm H_2$ , $\rm N_2$ , $\rm O_2$ .
$T\gg\theta_{\rm vib}$	$C_V\to\tfrac72Nk_B$	Both internal modes classical; full equipartition (before dissociation intervenes).
$T\approx0.8\,\theta_{\rm rot}$	$C_{\rm rot}/Nk_B\approx1.1$	The rotational overshoot — a purely quantum, degeneracy-driven hump.

Common trapEquipartition is not "approximately right at all temperatures"

It is a common error to treat the classical $\tfrac72Nk_B$ as the baseline and the freeze-outs as small corrections. The opposite is true: for hydrogen at room temperature vibration is not slightly reduced but essentially absent ( $C_{\rm vib}$ is smaller than $Nk_B$ by a factor of $e^{-21}\sim10^{-9}$ ), and below $\sim100$ K rotation freezes as well. Equipartition is the high-temperature limit of each mode, valid only once $k_BT$ exceeds that mode's level spacing. A second, subtler error is to forget the symmetry number $\sigma$ : for homonuclear molecules $\sigma=2$ , and omitting it doubles $\zeta_{\rm rot}$ and spoils the entropy (though not $C_V$ , which depends only on derivatives of $\ln\zeta_{\rm rot}$ ). For $\rm H_2$ specifically, even $\sigma=2$ is not the whole story at low temperature — the ortho/para nuclear-spin statistics must be handled explicitly.

Worked examples

Example 1 — Room-temperature heat capacity of $\rm H_2$ and $\rm N_2$ . Predict $C_V$ for hydrogen and nitrogen at $T=300$ K, given $\theta_{\rm rot}(\rm H_2)=85.3$ K, $\theta_{\rm vib}(\rm H_2)=6215$ K, $\theta_{\rm rot}(\rm N_2)=2.88$ K, $\theta_{\rm vib}(\rm N_2)=3374$ K.

Step 1. Translation always contributes $\tfrac32Nk_B$ .

Step 2. Rotation: for both gases $T\gg\theta_{\rm rot}$ ( $300\gg85.3$ and $300\gg2.88$ ), so rotation is classical and adds $Nk_B$ . Running total $\tfrac52Nk_B$ .

Step 3. Vibration: for $\rm H_2$ , $\theta_{\rm vib}/T=6215/300=20.7$ , so $C_{\rm vib}\approx Nk_B(20.7)^2e^{-20.7}\approx4\times10^{-7}Nk_B$ — negligible. For $\rm N_2$ , $\theta_{\rm vib}/T=3374/300=11.2$ , giving $C_{\rm vib}\approx Nk_B(11.2)^2e^{-11.2}\approx1.7\times10^{-3}Nk_B$ — still tiny.

Step 4. Therefore both predict $C_V\approx\tfrac52Nk_B=\tfrac52R=\mathbf{20.8\ J\,mol^{-1}K^{-1}}$ , matching the measured room-temperature molar $C_V$ of hydrogen ( $\approx20.5$ ) and nitrogen ( $\approx20.8$ ) to within one percent. The takeaway: at room temperature a diatomic gas sits squarely on the $\tfrac52R$ plateau — rotation fully on, vibration fully off.

Example 2 — Nitrogen with vibration partly awake at 2000 K. Find $C_V$ for $\rm N_2$ at $T=2000$ K.

Step 1. Translation $\tfrac32Nk_B$ and rotation $Nk_B$ are fully classical, giving $\tfrac52Nk_B$ before vibration.

Step 2. Vibration: $x\equiv\theta_{\rm vib}/T=3374/2000=1.687$ . The Einstein formula gives $C_{\rm vib}/Nk_B=x^2e^x/(e^x-1)^2=(1.687)^2(5.40)/(4.40)^2=\mathbf{0.79}$ .

Step 3. Total: $C_V/Nk_B=\tfrac52+0.79=\mathbf{3.29}$ , i.e. $C_V\approx3.29R=27.4\ \rm J\,mol^{-1}K^{-1}$ , part-way up the second riser toward $\tfrac72R=29.1$ . The takeaway: by $2000$ K nitrogen's bond is vibrating vigorously enough to supply most of the last $Nk_B$ , and $C_V$ has climbed well above the room-temperature $\tfrac52R$ .

Origins · 1 minrotational constant and how this was discovered

rotational constant

spectroscopic energy scale · the quantity B = ℏ²/2I that fixes where a molecule's rotation unfreezes

The rotational constant $B$ is the single number that governs the rotational spectrum and heat capacity of a molecule. Defined as $B=\hbar^2/2I$ (in energy units) or equivalently $\tilde B=\hbar/4\pi cI$ (in the wavenumbers $\rm cm^{-1}$ spectroscopists use), it sets the ladder spacing $E_J=BJ(J+1)$ and the rotational temperature $\theta_{\rm rot}=B/k_B=\hbar^2/2Ik_B$ . A light, tightly bound molecule like $\rm H_2$ has a small moment of inertia $I$ and hence a large $B$ and $\theta_{\rm rot}\approx85$ K, so its rotation freezes out at experimentally reachable temperatures; a heavy molecule like $\rm N_2$ has $\theta_{\rm rot}\approx3$ K and rotates classically at all ordinary temperatures. The rotational constant is measured directly from the equally spaced lines of a microwave or infrared rotational spectrum, making it one of the most precisely known molecular quantities and the bridge between spectroscopy and thermodynamics.

History

1912

— Eucken, in Nernst's laboratory, measures the heat capacity of hydrogen falling below 5/2 R at low temperature, the "rotational freeze-out" puzzle

1924

— quantum statistics of molecular rotation begin to be worked out

1927

— David Dennison resolves the hydrogen heat-capacity anomaly through ortho and para nuclear-spin statistics

1927

— Heisenberg and Hund analyse the ortho/para structure of molecular spectra

1929

— Bonhoeffer and Harteck separate ortho- and para-hydrogen experimentally

1929

— the diatomic partition function becomes a textbook staple in Fowler's Statistical Mechanics

Going further

References. Reif, Fundamentals of Statistical and Thermal Physics, §7.8–7.10 · Pathria & Beale, Statistical Mechanics, 3rd ed., §6.5 · McQuarrie, Statistical Mechanics, Ch. 6–8 · Fowler, Statistical Mechanics (1929) · Dennison, Proc. R. Soc. A 115, 483 (1927).

Provenance. A verifier should reproduce: (1) $\zeta_{\rm rot}(1000\,{\rm K})=6.03$ vs $T/2\theta_{\rm rot}=5.86$ for $\rm H_2$ ; (2) $C_{\rm rot}/Nk_B$ at $T=20,50,85.3,150,300,1000$ K $=0.04,0.97,1.07,1.01,1.00,1.00$ ; (3) the overshoot peak $\approx1.1\,Nk_B$ near $T\approx0.8\theta_{\rm rot}$ ; (4) $C_{\rm vib}/Nk_B$ for $\rm H_2$ at $T=300,1000,3000,6215$ K $=0.00,0.078,0.71,0.92$ ; (5) the staircase values $\tfrac32,\tfrac52,\tfrac72$ and the $300$ -K prediction $\tfrac52R=20.8\ \rm J\,mol^{-1}K^{-1}$ ; (6) the $\rm N_2$ at $2000$ K value $C_V/Nk_B=3.29$ ; (7) the high- $T$ ortho:para ratio $3:1$ .

Builds on: Factorization of the Partition Function · The Equipartition Theorem · Heat Capacity of Classical Gases. Leads to: Failure of Equipartition and the Classical–Quantum Boundary · Einstein Model of Lattice Heat Capacity · Polyatomic molecules and internal rotation.

Why does rotation about the bond axis not count as a degree of freedom?

Rotation about the internuclear axis has a moment of inertia set by the electron cloud, thousands of times smaller than the end-over-end moment of inertia. Its rotational temperature is therefore enormous — comparable to electronic excitation energies — so that mode is frozen at any temperature below where the molecule ionizes or dissociates. Effectively a linear molecule has only two active rotational degrees of freedom, the two axes perpendicular to the bond, which is why rotation contributes $Nk_B$ (not $\tfrac32Nk_B$ ) at high temperature.

What exactly is the difference between ortho- and para-hydrogen?

They differ in the relative orientation of the two proton nuclear spins. Ortho-hydrogen has the proton spins parallel (total nuclear spin $I=1$ , a triplet with statistical weight $3$ ); para-hydrogen has them antiparallel ( $I=0$ , a singlet with weight $1$ ). Because the total molecular wavefunction of two identical fermionic protons must be antisymmetric under exchange, the symmetric ( $I=1$ ) nuclear-spin states pair only with odd- $J$ rotational levels and the antisymmetric ( $I=0$ ) states only with even- $J$ levels. At high temperature the equilibrium ortho:para ratio approaches $3:1$ , the ratio of nuclear-spin degeneracies.

Why did hydrogen's measured heat capacity historically disagree with the simple formula?

Because ortho $\to$ para conversion is extraordinarily slow without a magnetic catalyst — the nuclear spins barely couple to anything — a sample of "normal" hydrogen prepared at room temperature stays a frozen $3:1$ ortho/para mixture even when cooled. Its rotational heat capacity is then the weighted sum of an odd- $J$ (ortho) rotor and an even- $J$ (para) rotor held in fixed proportion, which differs from the equilibrium single-species formula that early theories assumed. Dennison recognized this in 1927, and once the frozen $3:1$ mixture was used the calculated and measured curves agreed. It is a rare case where a non-equilibrium mixture is the correct thermodynamic description.

Does the symmetry number change the heat capacity?

No. The symmetry number $\sigma$ enters $\zeta_{\rm rot}$ as an overall multiplicative constant $1/\sigma$ , so it shifts $\ln\zeta_{\rm rot}$ by a constant. Since the energy and heat capacity depend only on derivatives of $\ln\zeta_{\rm rot}$ with respect to temperature, the constant drops out and $C_{\rm rot}$ is unaffected. Where $\sigma$ does matter is in the absolute entropy and free energy — the Sackur–Tetrode-style constants — where getting $\sigma$ wrong produces a real error in the tabulated thermodynamic functions.

Why treat vibration as a harmonic oscillator when real bonds are anharmonic?

The harmonic approximation captures the leading physics — a single mode with spacing $\hbar\omega$ that freezes below $\theta_{\rm vib}$ and reaches equipartition above it — and it is quantitatively excellent up to moderate temperatures where only the lowest few vibrational levels are populated. Anharmonicity (the Morse-potential softening of the ladder at high $v$ ) and vibration–rotation coupling become important near dissociation and shift $C_{\rm vib}$ upward slightly at very high $T$ . For the staircase and the room-temperature-to-few-thousand-kelvin range that dominate ordinary gas thermodynamics, the harmonic Einstein form is the right tool.

How is this related to the Einstein model of a solid?

Both describe a collection of independent quantum harmonic oscillators, and both give the same Einstein heat-capacity function that rises from zero to $Nk_B$ per mode as temperature crosses $\theta=\hbar\omega/k_B$ . In a diatomic gas there is one oscillator per molecule (the bond); in the Einstein solid there are three per atom (the lattice vibrations). The Debye model refines the solid by giving the oscillators a spread of frequencies, which changes the low-temperature law from exponential to $T^3$ , but the single-frequency Einstein term here is exactly the molecular vibration's contribution.