The single-particle partition function
Consider one atom of mass m in a box of volume V, with only kinetic energy E=p^2/2m. Its partition function is a sum over microstates — but classically the states form a continuum, so the sum becomes an integral over phase space, divided by h^3 to count in Planck-sized cells (three position, three momentum dimensions):
The one-particle partition function. The position integral gives V; each Gaussian momentum integral gives \sqrt{2\pi m k_BT}/h. The result defines the thermal de Broglie wavelength \lambda_{\text{th}}=h/\sqrt{2\pi m k_BT}.
The combination \lambda_{\text{th}}=h/\sqrt{2\pi m k_BT} is the quantum 'size' of a thermal particle. So z_1=V/\lambda_{\text{th}}^3 counts, roughly, how many quantum wave-packets fit in the box. When V/\lambda_{\text{th}}^3\gg N the gas is dilute and classical; when it approaches N, quantum statistics take over — the entry point to Statistical Mechanics II.
Assembling N particles — and a factor of N!
For N non-interacting particles the energy is a sum, so the Boltzmann factor factorizes and Z_N = z_1^{\,N} — naively. But this overcounts: swapping two identical atoms gives the same physical microstate, not a new one. To avoid counting each arrangement $N!$ times we divide by $N!$:
The N-particle partition function for an ideal gas. The $1/N!$ — Gibbs's correction for indistinguishability — is essential; without it the entropy comes out wrong.
Now turn the crank. From F=-k_BT\ln Z_N and P=-(\partial F/\partial V)_T, only the V^N piece depends on volume, giving F\supset -Nk_BT\ln V and therefore P = Nk_BT/V. We have derived the ideal gas law PV=Nk_BT from nothing but counting Boltzmann factors — with the Boltzmann constant k_B=R/N_A linking it to the gas constant.
Equipartition of energy
The average energy of that gas follows from \langle E\rangle=-\partial\ln Z/\partial\beta. Since z_1\propto\beta^{-3/2}, each particle carries \langle E\rangle = \tfrac{3}{2}k_BT — exactly \tfrac12 k_BT per translational direction. This is a special case of the equipartition theorem:
Equipartition: each quadratic degree of freedom (each term \propto p^2 or \propto q^2 in the energy) contributes \tfrac12 k_BT to the average energy. A monatomic gas has f=3; a diatomic gas adds rotations and, at high T, vibrations.
The Gibbs paradox
Return to that $1/N!. Drop it, and the entropy of an ideal gas comes out *non-extensive*: mix two identical gases by removing a partition and the formula predicts a jump in entropy, as though something irreversible happened — even though nothing did. This is the famous **[[gibbs-paradox|Gibbs paradox]]**. Inserting 1/N! — recognizing the atoms are fundamentally [[identical-particles|indistinguishable]] — cures it exactly, and the entropy becomes properly proportional to N$.
The Sackur–Tetrode equation for the entropy of a monatomic ideal gas. Note the V/N inside the log — the hallmark of the $1/N!$ correction — which makes S correctly extensive (double N and V together, and S doubles).
The resolution is a stunning early hint of quantum mechanics arriving from pure thermodynamics: identical particles are not merely hard to tell apart, they have no individual identity at all. The appearance of h inside \lambda_{\text{th}} in a supposedly classical entropy is the second such hint — nature was quietly quantum all along.
Opening the box: the grand canonical ensemble
Often particles too can flow in and out — a metal exchanging electrons, a gas in contact with a reservoir, adsorption on a surface. Fix T and the chemical potential \mu instead of N, and you have the grand canonical ensemble. Each microstate is now weighted by both its energy and its particle number, and the sum-over-states becomes the grand partition function \mathcal{Z}:
The grand partition function and the grand potential \Phi. The fugacity z=e^{\beta\mu} controls the average particle number; \langle N\rangle and pressure follow by differentiating \Phi.