---
title: The Ideal Gas, Phase-Space Volume, and the Sackur–Tetrode Entropy
module: The Microcanonical Ensemble
moduleNumber: 3
lessonNumber: 3
order: 303
summary: >
The monatomic ideal gas is the first system whose microcanonical count can be
done in closed form. The momentum integral is the volume of a $3N$-dimensional
ball of radius $\sqrt{2mE}$, the configuration integral is $V^N$, and together
they give the Sackur–Tetrode entropy
$S=Nk[\ln(V/N\lambda^3)+5/2]$ with the thermal wavelength
$\lambda=h/\sqrt{2\pi mkT}$. The formula matches the measured entropy of helium,
fixes the classical regime $n\ll n_Q$, and shows why the $N!$ is needed for
extensivity.
topics: [The Microcanonical Ensemble]
sources:
- book: Schroeder
ref: "Ch. 2 — The Second Law; §2.5 The Ideal Gas, and Ch. 3 — Interactions and Implications; §3.4"
- book: Reif
ref: "Ch. 7 — Simple Applications of Statistical Mechanics; §7.1–7.3"
- book: Pathria & Beale
ref: "Ch. 1 — The Statistical Basis of Thermodynamics; §1.4, and Ch. 2 — Elements of Ensemble Theory; §2.4"
- book: Kardar (Statistical Physics of Particles)
ref: "Ch. 4 — Classical Statistical Mechanics; §4.5"
draft: false
---
The monatomic ideal gas is the system on which the microcanonical machinery first
delivers a closed-form entropy. Its Hamiltonian is a sum of kinetic terms with no
interactions, so the phase-space count factorizes into a configuration part and a
momentum part, each of which can be evaluated exactly. The result is the
Sackur–Tetrode equation, the first entropy formula in which Planck's constant
appears as the size of the phase-space cell, and its numerical value agrees with
calorimetry to the significant figures the constants allow.
## The phase-space volume of the ideal gas
Take $N$ identical point particles of mass $m$ in a container of volume $V$, with
no interactions. The Hamiltonian is purely kinetic,
$$
H(q,p) = \sum_{i=1}^{3N}\frac{p_i^2}{2m},
$$
so the energy constraint $H\le E$ restricts only the momenta, to the interior of a
sphere $\sum_i p_i^2 \le 2mE$ in the $3N$-dimensional momentum space. The
coordinate integral is unconstrained over the box and contributes a factor $V$ per
particle. The enclosed phase-space volume is therefore
$$
\Gamma(E) = \frac{1}{h^{3N}N!}\int_{\text{box}}\!\d^{3N}q
\int_{\sum p_i^2\le 2mE}\!\d^{3N}p
= \frac{V^{N}}{h^{3N}N!}\,\Phi_{3N}\!\left(\sqrt{2mE}\right),
$$
where $\Phi_{D}(R)$ is the volume of a $D$-dimensional ball of radius $R$. The two
factorized pieces are the configuration volume $V^N$ and the momentum-ball volume,
and every ingredient of the ideal-gas entropy comes from evaluating the latter.
$$
% caption: The ideal-gas count factorizes: each particle ranges freely over the configuration volume $V$, contributing $V^N$, while the momenta are confined to a $3N$-ball of radius $\sqrt{2mE}$ set by the fixed energy.
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$$
## The volume of a high-dimensional ball
The volume of a ball of radius $R$ in $D$ dimensions is
$$
\Phi_D(R) = \frac{\pi^{D/2}}{\Gamma\!\left(\tfrac{D}{2}+1\right)}\,R^{D},
$$
with $\Gamma$ the gamma function, the continuous extension of the factorial through
$\Gamma(n+1)=n!$. The familiar low-dimensional cases $\Phi_2=\pi R^2$ and
$\Phi_3=\tfrac{4}{3}\pi R^3$ are the first two entries; the formula holds for any
$D$. Setting $D=3N$ and $R=\sqrt{2mE}$,
$$
\Phi_{3N}\!\left(\sqrt{2mE}\right)
= \frac{\pi^{3N/2}}{\Gamma\!\left(\tfrac{3N}{2}+1\right)}\,(2mE)^{3N/2},
$$
so the enclosed phase-space volume becomes
$$
\Gamma(E) = \frac{V^{N}}{N!}\,
\frac{1}{\Gamma\!\left(\tfrac{3N}{2}+1\right)}
\left(\frac{2\pi mE}{h^{2}}\right)^{3N/2}.
$$
Two features carry over into the entropy. The energy dependence is
$E^{3N/2}$, confirming the exponent $\alpha N=3N/2$ used in the preceding lessons,
so each of the $3N$ momentum degrees of freedom contributes a half-power of the
energy. And the gamma function in the denominator, once expanded by Stirling's
approximation, supplies the density-dependent term that makes the entropy
extensive.
A geometric fact drives the physics: in high dimensions the ball's volume is
dominated by a thin skin just inside its surface, because $\Phi_D\propto R^D$ makes
$\d\Phi/\Phi = D\,\d R/R$ enormous. Almost all momentum configurations of fixed
total energy have very nearly the maximal radius, which is why the enclosed volume
$\Gamma$ and the shell count $\Omega=\d\Gamma/\d E\cdot\delta E$ share the same
logarithm to leading order in $N$.
$$
% caption: The momentum ball has volume $\propto R^{3N}$ with $R=\sqrt{2mE}$; in $3N$ dimensions nearly all of that volume lies in a thin shell just below the surface, so the fixed-energy states cluster at the maximal momentum radius.
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$$
## The Sackur–Tetrode entropy
The entropy is $S=k\ln\Gamma$, valid to leading order in $N$. Taking the logarithm
and applying Stirling's approximation in the form $\ln N!\approx N\ln N-N$ and
$\ln\Gamma(\tfrac{3N}{2}+1)\approx \tfrac{3N}{2}\ln\tfrac{3N}{2}-\tfrac{3N}{2}$,
$$
\ln\Gamma = N\ln V - (N\ln N - N)
- \left(\tfrac{3N}{2}\ln\tfrac{3N}{2} - \tfrac{3N}{2}\right)
+ \frac{3N}{2}\ln\!\frac{2\pi mE}{h^{2}}.
$$
Collecting the terms, the $N\ln N$ combines with $N\ln V$ into $N\ln(V/N)$, the two
$\tfrac{3N}{2}$ pieces combine into $\tfrac{3N}{2}\ln(4\pi mE/3Nh^{2})$, and the
loose constants add to $\tfrac{5N}{2}$:
$$
S(E,V,N) = Nk\left[\ln\!\frac{V}{N}
+ \frac{3}{2}\ln\!\left(\frac{4\pi mE}{3Nh^{2}}\right)
+ \frac{5}{2}\right].
$$
This is the Sackur–Tetrode entropy in its energy form. Every argument of a
logarithm is intensive — $V/N$ is the volume per particle and $E/N$ the energy per
particle — so $S$ is proportional to $N$ at fixed intensities, as an extensive
quantity must be.
The temperature follows from the microcanonical definition
$1/T=(\partial S/\partial E)_{V,N}$. Only the middle term carries the energy, and
its derivative is
$$
\frac{1}{T} = \frac{\partial S}{\partial E}
= Nk\cdot\frac{3}{2}\cdot\frac{1}{E}
= \frac{3Nk}{2E}
\quad\Longrightarrow\quad
E = \frac{3}{2}NkT.
$$
The equipartition result $E=\tfrac{3}{2}NkT$ emerges as a derivative of the count,
not as a separate assumption; each of the $3N$ quadratic momentum terms carries
$\tfrac12 kT$. Substituting $E=\tfrac{3}{2}NkT$ into the entropy replaces the energy
argument by temperature: $4\pi mE/3Nh^{2}=2\pi mkT/h^{2}$, and
$$
S = Nk\left[\ln\!\frac{V}{N} + \frac{3}{2}\ln\!\frac{2\pi mkT}{h^{2}}
+ \frac{5}{2}\right].
$$
## The thermal wavelength and quantum concentration
The combination inside the temperature logarithm has the dimensions of an inverse
volume, and it defines a length scale.
> **Definition (Thermal de Broglie wavelength).** The thermal wavelength of a
> particle of mass $m$ at temperature $T$ is
> $$
> \lambda = \frac{h}{\sqrt{2\pi mkT}},
> \qquad \frac{1}{\lambda^{3}} = \left(\frac{2\pi mkT}{h^{2}}\right)^{3/2}.
> $$
Physically $\lambda$ is the de Broglie wavelength of a particle whose kinetic
energy is of order $kT$; it is the size of a particle's quantum-mechanical
wavepacket at temperature $T$. With this substitution the Sackur–Tetrode entropy
takes its most compact form,
$$
S = Nk\left[\ln\!\left(\frac{V}{N\lambda^{3}}\right) + \frac{5}{2}\right].
$$
The argument $V/N\lambda^{3}$ compares the volume per particle $V/N=1/n$ to the
thermal volume $\lambda^{3}$. Writing the **quantum concentration**
$n_Q=1/\lambda^{3}=(2\pi mkT/h^{2})^{3/2}$, the entropy reads
$S=Nk[\ln(n_Q/n)+\tfrac52]$. The classical treatment is self-consistent only where
its argument is large and positive, $n\ll n_Q$: the particles are dilute compared
to the quantum concentration, their wavepackets do not overlap, and their
indistinguishability has not yet forced the quantum statistics that a later module
develops. When $n$ approaches $n_Q$ the wavepackets overlap, the entropy formula
would predict a negative or near-zero entropy, and the classical count breaks down.
$$
% caption: The classical regime is set by comparing the thermal wavelength $\lambda$ to the interparticle spacing $n^{-1/3}$; when $\lambda\ll n^{-1/3}$ the wavepackets are far apart and the Sackur–Tetrode formula holds, while $\lambda\sim n^{-1/3}$ marks the onset of quantum degeneracy.
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$$
The circles depict the particle wavepackets, each of size $\lambda$; on the left,
where $\lambda\ll n^{-1/3}$, they are far apart and the classical count holds, while
on the right they overlap once $n$ reaches $n_Q$ and the classical treatment fails.
> **Worked example.** Helium gas at $T=300\ \mathrm{K}$ and $P=1\ \mathrm{atm}$.
> The atomic mass is $m=4.003\,\mathrm{u}=6.646\times10^{-27}\ \mathrm{kg}$, so the
> thermal wavelength is
> $$
> \lambda = \frac{h}{\sqrt{2\pi mkT}}
> = \frac{6.626\times10^{-34}}{\sqrt{2\pi(6.646\times10^{-27})(1.381\times10^{-23})(300)}}
> = 5.0\times10^{-11}\ \mathrm{m}.
> $$
> The number density from $n=P/kT$ is
> $n=(1.013\times10^{5})/[(1.381\times10^{-23})(300)]=2.45\times10^{25}\ \mathrm{m^{-3}}$,
> giving an interparticle spacing $n^{-1/3}=3.4\times10^{-9}\ \mathrm{m}$. The ratio
> $\lambda/n^{-1/3}\approx0.015\ll1$ confirms the classical regime. The volume per
> particle is $V/N=1/n=4.09\times10^{-26}\ \mathrm{m^{3}}$, so
> $V/N\lambda^{3}=3.2\times10^{5}$ and $\ln(V/N\lambda^{3})=12.7$. The molar entropy
> is
> $$
> \frac{S}{n_{\text{mol}}} = R\left[\ln\!\frac{V}{N\lambda^{3}}+\frac52\right]
> = (8.314)(12.7+2.5) = 126\ \mathrm{J\,K^{-1}\,mol^{-1}},
> $$
> in agreement with the calorimetric standard molar entropy of helium,
> $126.2\ \mathrm{J\,K^{-1}\,mol^{-1}}$.[^nist-he]
$$
% caption: The Sackur–Tetrode molar entropy rises logarithmically with temperature at fixed pressure (left) and falls logarithmically with density at fixed temperature (right); the marked point is helium at $300\ \mathrm{K}$, $1\ \mathrm{atm}$.
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$$
## Extensivity and the role of the $N!$
The $N!$ inserted in the phase-space measure is what makes the argument of the
first logarithm the intensive density $V/N$ rather than the extensive volume $V$.
Dropping the $1/N!$ and repeating the calculation removes the $-\ln N!$ term, and
the entropy becomes
$$
S_{\text{labelled}} = Nk\left[\ln V + \frac{3}{2}\ln\!\frac{2\pi mkT}{h^{2}}
+ \frac{3}{2}\right].
$$
The offending term is $Nk\ln V$. Doubling the system at fixed density sends
$N\to2N$ and $V\to2V$, and $Nk\ln V$ becomes $2Nk\ln(2V)=2Nk\ln V+2Nk\ln2$, which
exceeds twice the original by $2Nk\ln2$. The labelled entropy is superextensive:
the entropy per particle grows without bound with system size, which no
thermodynamic entropy can do. The corrected Sackur–Tetrode entropy has
$Nk\ln(V/N)$ instead, and doubling gives exactly $2Nk\ln(V/N)$, so $S$ scales
linearly.
The physical failure the $N!$ repairs is the spurious entropy of mixing two
samples of the _same_ gas. Remove a partition between two equal volumes of
identical gas at the same temperature and density; nothing observable happens, and
the entropy must not change. The labelled count predicts an increase
$2Nk\ln2$ — the same it would predict for mixing two _different_ gases — because it
treats the interchange of identical atoms across the removed partition as new
microstates. Dividing by $N!$ removes exactly this overcount, so the entropy of
mixing vanishes for identical species and survives only for distinct ones. This is
the Gibbs paradox, resolved here by indistinguishability and treated in full when
the canonical ideal gas is built.
## Summary
- The ideal-gas phase-space volume factorizes into the configuration part $V^N$
and the momentum-ball volume $\Phi_{3N}(\sqrt{2mE})=\pi^{3N/2}(2mE)^{3N/2}/\Gamma(\tfrac{3N}{2}+1)$,
giving $\Gamma(E)=V^N(2\pi mE/h^2)^{3N/2}/[N!\,\Gamma(\tfrac{3N}{2}+1)]$.
- Taking $S=k\ln\Gamma$ with Stirling's approximation yields the Sackur–Tetrode
entropy $S=Nk[\ln(V/N\lambda^3)+\tfrac52]$, and
$1/T=\partial S/\partial E$ recovers $E=\tfrac32 NkT$.
- The thermal wavelength $\lambda=h/\sqrt{2\pi mkT}$ and quantum concentration
$n_Q=1/\lambda^3$ set the classical regime $n\ll n_Q$; helium at $300\ \mathrm{K}$
and $1\ \mathrm{atm}$ has $\lambda/n^{-1/3}\approx0.015$ and molar entropy
$126\ \mathrm{J\,K^{-1}\,mol^{-1}}$, matching experiment.
- The $N!$ makes the entropy extensive by converting $\ln V$ into $\ln(V/N)$ and
cancels the spurious entropy of mixing for identical gases, previewing the Gibbs
paradox.
[^schroeder-st]: Schroeder, _An Introduction to Thermal Physics_ (Addison-Wesley, 2000), Ch. 2 §2.5 derives the Sackur–Tetrode equation for the monatomic ideal gas; companion page .
[^pathria-st]: Pathria & Beale, _Statistical Mechanics_ (4th ed., Elsevier, 2021), §1.4 and §2.4; the $3N$-ball volume and the entropy of the ideal gas. Reif, _Fundamentals of Statistical and Thermal Physics_ (Waveland reprint, 2009), §7.1–7.3; Kardar, _Statistical Physics of Particles_ (Cambridge, 2007), §4.5, MIT OCW 8.333, .
[^nist-he]: The standard molar entropy of monatomic helium gas at $298.15\ \mathrm{K}$ is $126.2\ \mathrm{J\,K^{-1}\,mol^{-1}}$; the constants $k$, $h$, $N_A$, and $R$ are the CODATA/SI-2019 values, NIST, .