---
title: "Molecular Gases: Rotational and Vibrational Degrees of Freedom"
module: The Classical Ideal Gas
moduleNumber: 5
lessonNumber: 3
order: 503
summary: >
  The internal partition function of a diatomic gas factorizes into translational,
  rotational, vibrational, and electronic parts. The rigid rotor gives a rotational
  temperature $\theta_{\rm rot}$; the harmonic bond gives a vibrational temperature
  $\theta_{\rm vib}$. Each mode contributes to the heat capacity only above its
  characteristic temperature, producing the diatomic $C_V$ staircase from
  $\tfrac32 R$ to $\tfrac52 R$ to $\tfrac72 R$. Homonuclear molecules carry a
  symmetry number, and hydrogen splits into ortho and para species.
topics: [The Classical Ideal Gas]
sources:
  - book: Schroeder
    ref: "Ch. 1 — Energy in Thermal Physics; §1.6, Ch. 6 §6.2 (diatomic)"
  - book: Reif
    ref: "Ch. 7 — Simple Applications of Statistical Mechanics; §7.7"
  - book: Pathria & Beale
    ref: "Ch. 6 — The Theory of Simple Gases; §6.5"
  - book: Tipler & Llewellyn
    ref: "Ch. 8 — Statistical Physics; §8-1 (diatomic heat capacity)"
draft: false
---

A monatomic gas stores energy only in translation. A molecule can also rotate and
vibrate, and each internal motion is a set of quantized levels with its own energy
scale. The heat capacity of a molecular gas therefore depends on temperature: a
mode contributes its equipartition share only when the temperature is high enough to
excite it, and stays frozen below. The observed heat capacity of hydrogen climbs a
staircase from $\tfrac32 R$ at low temperature to $\tfrac52 R$ once rotation is
active and toward $\tfrac72 R$ once vibration turns on. This lesson computes each
step from the partition function, giving the quantitative account behind the
qualitative staircase.

## Factorization of the molecular partition function

The energy of a diatomic molecule separates, to a good approximation, into
independent contributions,

$$
\varepsilon = \varepsilon_{\rm tr} + \varepsilon_{\rm rot} + \varepsilon_{\rm vib} + \varepsilon_{\rm el},
$$

the separation resting on the Born–Oppenheimer picture — electrons move fast enough
to define a potential for the slow nuclear motion, and the bond vibration is fast
compared with the molecular rotation. Because the energy is a sum, the Boltzmann
factor is a product and the single-molecule partition function factorizes,

$$
z = z_{\rm tr}\, z_{\rm rot}\, z_{\rm vib}\, z_{\rm el}.
$$

The $N$-molecule partition function keeps the translational indistinguishability
factor, $Z = z^N/N!$, so $\ln Z = N\ln z - \ln N!$ and each factor of $z$
contributes additively to the free energy, the energy, and the heat capacity. The
heat capacity is thus a sum of independent pieces,

$$
C_V = C_V^{\rm tr} + C_V^{\rm rot} + C_V^{\rm vib} + C_V^{\rm el},
$$

each computed from its own partition function. Translation contributes the
monatomic $C_V^{\rm tr} = \tfrac32 Nk_B$; the electronic factor is a constant $g_0$
for temperatures far below the first electronic excitation, contributing nothing to
$C_V$. Rotation and vibration are the temperature-dependent pieces.

$$
% caption: The single-molecule partition function factorizes over independent energy stores. Translation gives $\tfrac32 Nk_B$; the electronic factor is frozen to its ground degeneracy; rotation and vibration each switch on above a characteristic temperature.
\begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[acc,very thick,fill=acc!12] (2.6,3.0) rectangle (4.2,3.8);
\node[acc] at (3.4,3.4) {$z$};
\foreach \i/\lab in {0/translation,1/rotation,2/vibration,3/electronic}{
  \draw[very thick] (\i*1.75,1.0) rectangle (\i*1.75+1.55,1.8);
  \node[black] at (\i*1.75+0.775,1.4) {\lab};
  \draw[->,black] (3.4,2.95)--(\i*1.75+0.775,1.85);
}
\end{tikzpicture}
$$

## The rigid rotor

A diatomic molecule rotating about an axis perpendicular to its bond is a rigid
rotor with moment of inertia $I = \mu r_0^2$, where $\mu$ is the reduced mass and
$r_0$ the bond length. Its quantized energy levels are

$$
\varepsilon_J = \frac{\hbar^2}{2I}\,J(J+1) = k_B\,\theta_{\rm rot}\,J(J+1),
\qquad
\theta_{\rm rot} \equiv \frac{\hbar^2}{2 I k_B},
$$

with degeneracy $2J+1$ for the $2J+1$ orientations of the angular momentum. The
**rotational temperature** $\theta_{\rm rot}$ sets the scale; for hydrogen it is
about $85\ \text{K}$, and for heavier molecules it is smaller. The rotational
partition function is

$$
z_{\rm rot} = \sum_{J=0}^{\infty}(2J+1)\,e^{-\theta_{\rm rot} J(J+1)/T}.
$$

$$
% caption: Rigid-rotor levels $\varepsilon_J\propto J(J+1)$ with degeneracy $2J+1$. The spacing grows with $J$, and the Boltzmann weight $e^{-\varepsilon_J/k_BT}$ cuts off the sum near $J(J+1)\sim T/\theta_{\rm rot}$.
\begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0,0)--(0,4.2) node[above,black]{energy};
% levels at J(J+1) scaled by 0.28
\foreach \J/\y/\deg in {0/0/1,1/0.56/3,2/1.68/5,3/3.36/7}{
  \draw[acc,very thick] (0.2,\y)--(2.4,\y);
  \node[black,left] at (0.15,\y) {$J{=}\J$};
  \node[black,right] at (2.5,\y) {$2J{+}1{=}\deg$};
}
\end{tikzpicture}
$$

Two limits are needed. At **high temperature**, $T\gg\theta_{\rm rot}$, the levels
are closely spaced on the scale $k_BT$ and the sum becomes an integral. Substituting
$u = J(J+1)$, $\d u = (2J+1)\,\d J$,

$$
z_{\rm rot} \approx \int_0^{\infty}(2J+1)\,e^{-\theta_{\rm rot}J(J+1)/T}\,\d J
= \int_0^{\infty} e^{-\theta_{\rm rot}u/T}\,\d u = \frac{T}{\theta_{\rm rot}}.
$$

Then $\ln z_{\rm rot} = \ln T + \text{const}$, so $U_{\rm rot} = Nk_BT$ and

$$
C_V^{\rm rot} \xrightarrow{T\gg\theta_{\rm rot}} Nk_B,
$$

the two rotational degrees of freedom each contributing $\tfrac12 k_BT$, exactly
the equipartition value. At **low temperature**, $T\ll\theta_{\rm rot}$, only the
first two terms survive,

$$
z_{\rm rot} \approx 1 + 3\,e^{-2\theta_{\rm rot}/T},
$$

and the heat capacity falls to zero exponentially, $C_V^{\rm rot}\sim Nk_B\,(2\theta_{\rm rot}/T)^2\,e^{-2\theta_{\rm rot}/T}$.
Rotation is frozen out below $\theta_{\rm rot}$, and its full $Nk_B$ is available
only above it.

## The vibrating bond

The bond stretches as a one-dimensional harmonic oscillator of angular frequency
$\omega$, with levels $\varepsilon_n = \hbar\omega\,(n+\tfrac12)$ and a **vibrational
temperature** $\theta_{\rm vib} = \hbar\omega/k_B$, of order $6000\ \text{K}$ for
hydrogen. The partition function is a geometric series,

$$
z_{\rm vib} = \sum_{n=0}^{\infty} e^{-\theta_{\rm vib}(n+1/2)/T}
= \frac{e^{-\theta_{\rm vib}/2T}}{1 - e^{-\theta_{\rm vib}/T}}.
$$

The mean vibrational energy and heat capacity follow from $U = -\partial\ln Z/\partial\beta$,

$$
U_{\rm vib} = N k_B\theta_{\rm vib}\left[\frac12 + \frac{1}{e^{\theta_{\rm vib}/T}-1}\right],
\qquad
C_V^{\rm vib} = N k_B\left(\frac{\theta_{\rm vib}}{T}\right)^2
\frac{e^{\theta_{\rm vib}/T}}{\left(e^{\theta_{\rm vib}/T}-1\right)^2}.
$$

The high-temperature limit is $C_V^{\rm vib}\to Nk_B$, the full equipartition value
for a mode with both a kinetic and a potential quadratic term ($\tfrac12 k_BT$ each).
The low-temperature limit is $C_V^{\rm vib}\to Nk_B(\theta_{\rm vib}/T)^2\,e^{-\theta_{\rm vib}/T}\to 0$,
frozen below $\theta_{\rm vib}$. Because $\theta_{\rm vib}\gg\theta_{\rm rot}$ for
typical molecules, the two modes switch on at well-separated temperatures.

## The heat-capacity staircase

Adding the pieces gives the temperature dependence of the molar heat capacity,
expressed per mole with $Nk_B = R$:

- **Below $\theta_{\rm rot}$:** only translation is active, $C_V = \tfrac32 R$.
- **Between $\theta_{\rm rot}$ and $\theta_{\rm vib}$:** translation and rotation,
  $C_V = \tfrac32 R + R = \tfrac52 R$.
- **Above $\theta_{\rm vib}$:** all three, $C_V = \tfrac52 R + R = \tfrac72 R$.

The heat capacity rises in two steps, each smoothed over about a decade of
temperature around the relevant $\theta$. For hydrogen the lower step sits near
$85\ \text{K}$ and the upper near $6000\ \text{K}$; in practice the molecule
dissociates before the vibrational plateau is fully reached, so the $\tfrac72 R$
value is approached rather than cleanly attained. The staircase is the direct,
measurable signature of quantized internal levels: a classical molecule with
continuous rotation and vibration would show the flat $\tfrac72 R$ at all
temperatures, and its absence was one of the early failures of classical statistical
mechanics that the equipartition lesson flagged.

$$
% caption: The diatomic heat capacity climbs from $\tfrac32 R$ (translation only) to $\tfrac52 R$ once rotation unfreezes near $\theta_{\rm rot}$, and toward $\tfrac72 R$ once vibration unfreezes near $\theta_{\rm vib}$. Each riser is centered on the corresponding characteristic temperature; the horizontal axis is logarithmic.
\begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0.8,0)--(6.6,0) node[right,black]{$\log T$};
\draw[->,black] (0.8,0)--(0.8,4.4) node[above,black]{$\frac{C_V}{R}$};
\draw[black,dashed] (0.8,1.5)--(6.4,1.5);
\draw[black,dashed] (0.8,2.5)--(6.4,2.5);
\draw[black,dashed] (0.8,3.5)--(6.4,3.5);
\node[black,left] at (0.75,1.5) {$\tfrac32$};
\node[black,left] at (0.75,2.5) {$\tfrac52$};
\node[black,left] at (0.75,3.5) {$\tfrac72$};
\draw[acc,very thick]
  (1.0,1.5)--(2.0,1.5)
  .. controls (2.35,1.5) and (2.05,2.5) .. (2.8,2.5)
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  .. controls (4.75,2.5) and (4.45,3.5) .. (5.2,3.5)
  --(6.2,3.5);
\node[black] at (2.4,0.35) {rotation};
\node[black] at (4.8,0.35) {vibration};
\draw[black,dashed] (2.4,0.55)--(2.4,2.05);
\draw[black,dashed] (4.8,0.55)--(4.8,3.05);
\end{tikzpicture}
$$

## The symmetry number and ortho/para hydrogen

For a **homonuclear** molecule the two nuclei are identical, and a rotation by
$\pi$ maps the molecule onto itself. Counting all rotational orientations then
double-counts every physical configuration, and the high-temperature partition
function must be divided by the **symmetry number** $\sigma$,

$$
z_{\rm rot} \xrightarrow{T\gg\theta_{\rm rot}} \frac{T}{\sigma\,\theta_{\rm rot}},
\qquad \sigma = \begin{cases} 1 & \text{heteronuclear}\\ 2 & \text{homonuclear.}\end{cases}
$$

The symmetry number changes the additive constant of the entropy but not the heat
capacity, which depends only on $\ln T$. Its microscopic origin is the exchange
symmetry of identical nuclei, and for hydrogen this origin has an observable
consequence. The two protons are spin-$\tfrac12$ fermions, so the total molecular
wavefunction must be antisymmetric under their exchange. Exchange combines a spatial
part, whose parity under nuclear interchange is $(-1)^J$, with a nuclear-spin part.
The symmetric nuclear-spin states (total nuclear spin $1$, three states, **ortho**
hydrogen) must pair with odd $J$; the antisymmetric nuclear-spin state (total spin
$0$, one state, **para** hydrogen) pairs with even $J$.

$$
% caption: Nuclear-spin statistics of $\mathrm{H}_2$. Even-$J$ rotational states carry the single para nuclear-spin state; odd-$J$ states carry the three ortho states. The $3{:}1$ statistical weight favors odd $J$, and at high temperature the average recovers the symmetry number $\sigma=2$.
\begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% even J column (para)
\node[acc] at (1.2,3.9) {even $J$};
\node[black] at (1.2,3.45) {para, weight $1$};
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% odd J column (ortho)
\begin{scope}[xshift=4.6cm]
\node[black!70] at (1.2,3.9) {odd $J$};
\node[black] at (1.2,3.45) {ortho, weight $3$};
\foreach \J/\y in {1/0.6,3/2.0}{
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  \filldraw[black!70] (3.1,\y) circle (2pt);
}
\end{scope}
\end{tikzpicture}
$$

At high temperature the two species populate odd and even $J$ in the $3:1$ ratio of
their nuclear-spin weights, and the effective symmetry number $\sigma = 2$ emerges
as the average over that mixture. Because interconversion between ortho and para is
slow — it requires flipping a nuclear spin, weakly coupled to the rotational
motion — cooled hydrogen behaves as a nonequilibrium mixture of two gases with
distinct rotational partition functions, and its low-temperature heat capacity
depends on the ortho/para ratio. This is a direct thermodynamic fingerprint of
nuclear-spin statistics, and it was historically decisive evidence that the proton
has spin $\tfrac12$.

## Summary

- The molecular partition function factorizes, $z = z_{\rm tr}z_{\rm rot}z_{\rm vib}z_{\rm el}$,
  so $C_V$ is a sum of independent mode contributions.
- The rigid rotor has $\varepsilon_J = k_B\theta_{\rm rot}J(J+1)$ with degeneracy
  $2J+1$; above $\theta_{\rm rot}$, $z_{\rm rot}\to T/\theta_{\rm rot}$ and
  $C_V^{\rm rot}\to Nk_B$, frozen below.
- The harmonic bond has $\theta_{\rm vib} = \hbar\omega/k_B$; $C_V^{\rm vib}\to Nk_B$
  above $\theta_{\rm vib}$ and freezes exponentially below.
- With $\theta_{\rm rot}\ll\theta_{\rm vib}$, the molar heat capacity climbs a
  staircase $\tfrac32 R\to\tfrac52 R\to\tfrac72 R$; for hydrogen the steps sit near
  $85\ \text{K}$ and $6000\ \text{K}$.
- Homonuclear molecules carry a symmetry number $\sigma = 2$; for hydrogen the
  underlying nuclear-spin statistics split the gas into ortho ($3$, odd $J$) and
  para ($1$, even $J$) species, observable in the low-temperature heat capacity.