The Ideal Gas Partition Function and the Gibbs Paradox

The monatomic ideal gas is the first system where the canonical machinery is carried all the way to a closed thermodynamic potential. Its Hamiltonian is purely kinetic, so the partition function factorizes into single-particle pieces and the integrals are Gaussian. The result, the Sackur–Tetrode entropy, exposes two features that the classical theory alone cannot supply: a length scale set by $\hslash$ that fixes the additive constant of the entropy, and a factor $N!$ that makes the entropy extensive. The missing $N!$ is the Gibbs paradox, and its origin is the indistinguishability of identical particles.

The single-particle partition function

A single structureless particle of mass $m$ confined to a box of volume $V$ has the Hamiltonian $H=\vec{p}{}^2/2m$, with no potential energy inside the box. In the classical phase-space form of the canonical ensemble, each quantum state occupies a cell of volume $h^3$ in the six-dimensional single-particle phase space, so the partition function is the phase-space integral divided by $h^3$,¹

The configuration integral is trivial: the integrand is independent of position, so ${\int }_V{\mathbf{d}}^{3}q=V$. The momentum integral factorizes into three identical Gaussians,

Collecting the factors,

The quantity $\lambda$ is the thermal de Broglie wavelength. It is the de Broglie wavelength of a particle whose kinetic energy is of order $k_{B}T$, and it sets the scale on which quantum coherence between particles matters. With $\lambda$ in hand, $z_1=V/{\lambda }^3$ has a direct reading: it counts how many thermal cells of volume ${\lambda }^3$ fit inside the container.

Two limits fix the meaning. At high temperature or low density, $\lambda$ is small compared with the mean interparticle spacing $n^{-1/3}$, where $n=N/V$; the particles are far apart on the scale of their thermal wavelength and behave classically. When $\lambda$ grows to the interparticle spacing, the wave packets overlap and quantum statistics take over. The classical treatment of this lesson requires the nondegeneracy condition

The N-particle partition function and the factor N!

For $N$ noninteracting particles the Hamiltonian is a sum, $H={\sum }_{i=1}^N{\vec{p}}_i{}^2/2m$, so the Boltzmann factor factorizes and the phase-space integral separates into $N$ identical single-particle integrals. If the particles were labelled and distinguishable, this would give $z_1^N$. They are not. Two configurations that differ only by a permutation of identical particles are the same physical state, and the labelled integral counts each physical state $N!$ times over. Correcting the overcount divides by $N!$:

The division by $N!$ is correct Boltzmann counting. It is exact only in the nondegenerate regime, where the probability that any single-particle state holds more than one particle is negligible; then every occupied state is occupied once and the permutations of the $N$ occupied states are all distinct, giving exactly $N!$ relabellings per physical configuration. When occupancies are not small the count of permutations that produce genuinely distinct arrangements is more subtle, and the correct bookkeeping is Bose or Fermi statistics. For the classical gas, $n{\lambda }^3\ll 1$ guarantees the simple $N!$.

Thermodynamics from the free energy

The Helmholtz free energy follows from $F=-k_{B}T\ln Z_N$. Using Stirling's approximation $\ln N!=N\ln N-N+O(\ln N)$,

so

Every equilibrium property comes from derivatives of $F$. The combination $V/N{\lambda }^3=1/(n{\lambda }^3)$ appearing inside the logarithm is the reciprocal of the degeneracy parameter, large in the classical regime.

Pressure. The mechanical equation of state is $P=-(\partial F/\partial V{)}_{T,N}$. Only the $\ln V$ term carries volume dependence,

The ideal-gas law is recovered, and it is independent of $\lambda$: the pressure knows nothing about $\hslash$ or the mass, because those enter only the additive constant of $F$.

Internal energy. With $\lambda \propto T^{-1/2}$, the temperature dependence of $\ln Z_N$ sits in $\ln {\lambda }^{-3}=\frac{3}{2}\ln T+\text{const}$, so $\ln Z_N=\frac{3}{2}N\ln T+(T\text{-independent})$. Writing $\beta =1/k_{B}T$,

which is $\frac{1}{2}{k}_{B}T$ per translational degree of freedom, three per particle. The heat capacity is $C_V=(\partial U/\partial T{)}_V=\frac{3}{2}N{k}_B$, constant.

Entropy. The entropy is $S=-(\partial F/\partial T{)}_{V,N}=(U-F)/T$. Since $\ln {\lambda }^{-3}=\frac{3}{2}\ln T+\text{const}$ contributes $\frac{3}{2}N{k}_B$ from $-\partial F/\partial T$ acting on the explicit $T$ prefactor and another $\frac{3}{2}N{k}_B$ from the $\ln T$ inside the bracket, the two pieces combine with the constant to give

This is the Sackur–Tetrode equation for the monatomic ideal gas.² Its structure is worth reading term by term: the logarithm measures the number of thermal cells per particle, and the additive $5/2$ collects the Gaussian normalization and the Stirling remainder. The appearance of $\lambda$, and hence of $h$, inside the logarithm means the classical entropy carries an absolute zero-point set by quantum mechanics — a purely classical calculation would leave the additive constant undetermined.

Extensivity and the Gibbs paradox

Entropy is extensive: doubling a system at fixed intensive parameters must double $S$. Scale $V\to \alpha V$ and $N\to \alpha N$ at fixed $T$. Then $\lambda$ is unchanged, $V/N{\lambda }^3$ is unchanged, and

Extensivity holds precisely because the $N!$ turned $V$ inside the logarithm into $V/N$. Drop the $N!$ and the entropy becomes

which fails the test: under $V\to \alpha V,N\to \alpha N$ it picks up a spurious $\alpha N{k}_B\ln \alpha$. The non-extensive form assigns a larger entropy to a combined system than to the sum of its parts, in violation of thermodynamics.

The physical face of this failure is the Gibbs paradox. Take two gas samples, each of $N$ particles in volume $V$ at the same temperature and pressure, separated by a partition, and remove the partition. Two cases must be distinguished.

Distinct species. If the two samples are different gases, say species $A$ on the left and $B$ on the right, each expands into the full volume $2V$. Each contributes an entropy increase $N{k}_B\ln 2$ from the volume doubling, so the total entropy of mixing is

This is real and measurable; interdiffusion of distinct gases is irreversible, and unmixing them costs work.

Identical species. If the two samples are the same gas, removing the partition changes nothing: the equilibrium state of $2N$ particles in $2V$ at the same $T,P$ is identical before and after, and reinserting the partition restores the original state exactly. The entropy change must be zero. Sackur–Tetrode, with the $N!$, delivers exactly this. Before,

and after, with $2N$ particles in $2V$,

so $\Delta S=0$. The factors of two inside the logarithm cancel because both $V$ and $N$ doubled. Without the $N!$, the same calculation gives $\Delta S=2N{k}_B\ln 2$ even for identical gases — the paradox, an entropy of mixing where nothing has mixed.

The resolution is that the $N!$ is not an ad hoc fix for one paradox but the statement that permuting identical particles does not produce a new microstate. Once identical particles are indistinguishable, the mixing of identical gases is a non-event and the entropy of mixing vanishes automatically, while genuinely distinct species still mix irreversibly. The same indistinguishability, followed to its quantum conclusion, becomes Bose and Fermi statistics; the classical $N!$ is its leading, dilute-limit shadow.

Chemical potential and the fugacity

The chemical potential follows from $\mu =(\partial F/\partial N{)}_{T,V}$, or more cleanly from $F=-N{k}_{B}T[\ln (V/N{\lambda }^3)+1]$ by differentiation,

In the classical regime $n{\lambda }^3\ll 1$, so the logarithm is large and negative: the chemical potential of a dilute classical gas is negative, and it grows toward zero as the gas is compressed or cooled toward degeneracy. The fugacity$z\equiv e^{\beta \mu }=n{\lambda }^3$ is then small, and it is the natural expansion parameter for the corrections beyond the ideal gas. Consistency with the free energy is immediate through the Euler relation $F=-PV+\mu N$: substituting $P=n{k}_{B}T$ and $\mu =k_{B}T\ln (n{\lambda }^3)$ reproduces $F=N{k}_{B}T[\ln (n{\lambda }^3)-1]=-N{k}_{B}T[\ln (V/N{\lambda }^3)+1]$.

Summary

• The single-particle partition function is $z_1=V/{\lambda }^3$ with the thermal wavelength $\lambda =h/\sqrt{2\pi m{k}_{B}T}$; it counts thermal cells of volume ${\lambda }^3$ in the box. Classical validity requires $n{\lambda }^3\ll 1$.
• Indistinguishability forces $Z_N=z_1^N/N!$ (correct Boltzmann counting), exact when no single-particle state is multiply occupied.
• From $F=-N{k}_{B}T[\ln (V/N{\lambda }^3)+1]$ come $PV=N{k}_{B}T$, $U=\frac{3}{2}N{k}_{B}T$, $C_V=\frac{3}{2}N{k}_B$, and the Sackur–Tetrode entropy $S=N{k}_B[\ln (V/N{\lambda }^3)+\frac{5}{2}]$.
• The $N!$ makes $S$ extensive and resolves the Gibbs paradox: identical gases mix with $\Delta S=0$, while distinct species mix with $\Delta S=2N{k}_B\ln 2$.
• The chemical potential $\mu =k_{B}T\ln (n{\lambda }^3)$ is negative for a dilute gas; the fugacity $z=n{\lambda }^3$ is the small parameter of the classical limit.