Classical Gas Partition Functions Revisited

Introduction

The ideal-gas equation can be derived from molecular translation, but a postgraduate treatment must also get entropy and chemical potential right. The distinction between one-molecule q and N-particle Q, including division by N!, is essential. This page follows the partition function through Helmholtz energy, pressure and chemical potential and shows where the classical expression stops being reliable.

Core explanation

For a monatomic particle in a macroscopic three-dimensional box, the translational partition function is q trans = V/Λ³, with Λ = h/sqrt(2πmkT). For N noninteracting identical particles in the classical dilute regime, Q N = q transᴺ/N! = (V/Λ³)ᴺ/N!. The factorial prevents counting a permutation of identical particles as a new physical state. It is an approximate classical correction in the regime where quantum exchange effects on energies and occupations are otherwise negligible.

Take A = −kT ln Q N = −NkT ln(V/Λ³) + kT ln N!. For large N, Stirling's formula gives A ≈ NkT[ln(NΛ³/V) − 1]. This expression is extensive when N and V scale together at fixed density. Without N!, A would contain an unphysical dependence on scaling that leads to a spurious entropy of mixing identical gases, a form of the Gibbs paradox.

Differentiating at fixed T,N gives p = −(∂A/∂V) T,N = NkT/V. The chemical potential per particle is μ = (∂A/∂N) T,V ≈ kT ln(NΛ³/V) = kT ln(n numberΛ³), where n number = N/V is number density. For molar chemical potential, multiply by Avogadro's constant and use R rather than k. The energy reference for internal states can add a temperature-dependent standard term; the displayed expression is for a monatomic translational model with a chosen zero.

Differentiating ln Q with temperature gives U trans = 3NkT/2, provided the chosen energy zero is temperature independent. The entropy is S = (U − A)/T ≈ Nk[5/2 − ln(NΛ³/V)]. This is the translational Sackur–Tetrode form for a monatomic classical ideal gas under the stated conventions. The same Λ appears in pressure-independent entropy differences and in the degeneracy parameter NΛ³/V.

The condition NΛ³/V ≪ 1 means the thermal wave volume is small compared with volume per particle. When it is not small, Bose–Einstein or Fermi–Dirac statistics can matter. The simple Q N then fails even though the particles remain indistinguishable. Thus dividing by N! is not a universal replacement for full quantum statistics.

Step-by-step reasoning

Start from q trans and specify N identical, noninteracting particles. Form Q N = qᴺ/N!, take its logarithm, and apply Stirling only when N is large. Derive A, then obtain p by a V derivative and μ by an N derivative. Use U or a T derivative for entropy. Check that NΛ³/V is small before trusting the classical form.

Visual explanation

Draw N identical dots in a box. Exchange two labels on the dots and show that the physical arrangement is unchanged; that motivates N!. Beneath, put a chain q → Q N → A, branching to p, μ and S. A small inset compares Λ³ with V/N to illustrate the dilute criterion.

Real-world analogy

If identical unnumbered chairs are rearranged by swapping two invisible labels, nobody can tell that a new seating arrangement occurred. Counting both labelings overstates possibilities. The N! correction has similar bookkeeping purpose, although molecular quantum indistinguishability is more fundamental than unmarked furniture.

Real-world example

At ordinary room conditions, a dilute noble gas is well approximated by classical translational statistics. Its pressure follows pV = NkT and its entropy depends on mass through Λ. Cooling a light gas substantially or compressing it raises NΛ³/V, signalling that the simple classical expression may need quantum corrections.

Why?

Available translational states scale with V per particle, giving Q ∝ Vᴺ and thus ideal-gas pressure. The factorial ensures that scaling particle number and volume together produces an extensive A. Thermal wavelength packages the quantum energy-level spacing into a state-density scale, linking microscopic h to bulk entropy.

Common misconception

The factor N! is not optional merely because it disappears from the pressure derivative. It affects entropy and chemical potential. Another error is to use q trans for an entire gas of N molecules without raising it to N. Finally, classical dilute does not mean particles are physically distinguishable; it means their exchange effects beyond the counting correction are negligible.

Worked example

Suppose a model gas has NΛ³/V = 10⁻⁴ at one T and density. Its translational chemical potential per particle is approximately μ = kT ln(10⁻⁴) = −9.21kT relative to the chosen zero. If density doubles at the same T, μ changes by kT ln 2 ≈ 0.693kT. The pressure also doubles because p = (N/V)kT. The negative absolute μ value is reference-dependent; the positive change on compression is the robust conclusion.

Quick check

1. If N and V both double at fixed T, what happens to NΛ³/V and to the approximate A expression per particle? Answer: Density N/V and Λ remain the same, so NΛ³/V and A/N stay the same. Total A doubles, showing extensivity of the corrected expression.

Exam focus

Keep q and Q N distinct, include N!, and use Stirling only for large N. Differentiate at the proper fixed variables. State whether μ is per particle or per mole. Check NΛ³/V before applying the classical formula, and do not use a pressure-only test to judge the entropy correction.

Advanced insight

The classical ideal-gas partition function can be derived by integrating phase space and dividing by h³ᴺN!. The h factors make the count dimensionless; the N! corrects indistinguishable permutations. Quantum statistical mechanics supplies the deeper basis and reveals corrections when occupancy of single-particle states is no longer sparse.

Summary

For N dilute identical monatomic particles, Q N = (V/Λ³)ᴺ/N! and A ≈ NkT[ln(NΛ³/V) − 1]. Derivatives give pV = NkT and μ ≈ kT ln(NΛ³/V), while entropy depends crucially on the N! correction. The formula requires NΛ³/V ≪ 1.

Practice questions

1. Using A ≈ NkT[ln(NΛ³/V) − 1], derive p. Answer: At fixed T,N, Λ is constant and ∂A/∂V = −NkT/V. Thus p = −∂A/∂V = NkT/V. 2. Why does swapping two identical gas particles not create a new physical microstate in the classical corrected count? Answer: Their labels have no observable physical meaning. Counting the swapped arrangement separately would overcount states and spoil extensive entropy. 3. If NΛ³/V rises from 10⁻⁵ to 0.5, which assumption deserves scrutiny? Answer: The classical dilute condition is becoming invalid. Quantum exchange and Bose or Fermi statistics may influence occupations, so Q N = qᴺ/N! may no longer be accurate.