Classical Ideal Gas

The classical ideal gas is the hydrogen atom of statistical mechanics — the one model simple enough to solve completely and rich enough to teach every essential lesson. It is a collection of $N$ point particles that do not interact, confined to a volume $V$ at temperature $T$ . From this bare skeleton the full partition function follows in closed form, and from the partition function everything else falls out: the equation of state $PV=Nk_BT$ , the caloric law $U=\tfrac32 Nk_BT$ , the heat capacities $C_V=\tfrac32 Nk_B$ and $C_P=\tfrac52 Nk_B$ , and the absolute entropy given by the Sackur–Tetrode equation. Two subtleties make the model more than a triviality. First, the single-particle partition function $Z_1=V/\lambda_T^3$ carries a quantum length scale — the thermal de Broglie wavelength $\lambda_T=h/\sqrt{2\pi m k_BT}$ — even though the gas is "classical". Planck's constant sneaks in through the correct counting of phase-space cells; it cancels from the equation of state but survives in the absolute entropy. Second, the factor of $1/N!$ in $Z_N=Z_1^N/N!$ — Gibbs's correct Boltzmann counting for indistinguishable particles — is not optional bookkeeping. Without it the entropy is not extensive and the Gibbs paradox (a spurious entropy of mixing when two identical gases merge) appears. The $1/N!$ is the thumbprint of quantum indistinguishability on an otherwise classical theory. Because it is exactly solvable, the ideal gas is the reference point against which every real gas, every interacting system, and every quantum gas is measured. Virial expansions correct it for interactions; Bose and Fermi statistics correct it for degeneracy; but the ideal gas is always the leading term, the baseline that says what "no interactions, no quantum degeneracy" looks like.

Key takeaways

The classical ideal gas is $N$ non-interacting point particles in a box; its single-particle partition function is $Z_1=V/\lambda_T^3$ with $\lambda_T=h/\sqrt{2\pi m k_BT}$ .
Indistinguishability requires $Z_N=Z_1^N/N!$ (correct Boltzmann counting); differentiating $F=-k_BT\ln Z_N$ gives all thermodynamics.
The equation of state is $PV=Nk_BT$ and the internal energy is $U=\tfrac32 Nk_BT$ , so $C_V=\tfrac32 Nk_B$ , $C_P=\tfrac52 Nk_B$ , $\gamma=5/3$ .
The absolute entropy is the Sackur–Tetrode equation, $S/(Nk_B)=\ln(V/N\lambda_T^3)+\tfrac52$ — the $5/2$ is exact, and $h$ survives here.
Without the $1/N!$ the entropy is non-extensive and the Gibbs paradox appears; the factor is quantum indistinguishability leaking into a classical theory.
Key idea: one function, $Z_N=(V/\lambda_T^3)^N/N!$ , regenerates the entire thermodynamics of the ideal gas — the equation of state, the caloric law, and the absolute entropy — with $h$ cancelling from $PV=Nk_BT$ but surviving in the entropy.

The core insight

Every thermodynamic property of the ideal gas is a derivative of one object, the partition function $Z_N=(V/\lambda_T^3)^N/N!$ . The volume derivative gives $PV=Nk_BT$ ; the $\beta$ derivative gives $U=\tfrac32Nk_BT$ ; the temperature derivative of the free energy gives the Sackur–Tetrode entropy. The $1/N!$ makes the entropy extensive; the thermal wavelength $\lambda_T$ carries Planck's constant into the absolute entropy.

Where this lives today

The classical ideal gas is the baseline of essentially all of physical chemistry and much of engineering thermodynamics; every real-gas equation of state (van der Waals, virial, Redlich–Kwong) is written as a correction to it. In modern research it plays a subtler role as the reference system: quantum-gas experiments quote the degeneracy parameter $n\lambda_T^3$ to say how far they are from the classical limit, and the Sackur–Tetrode entropy is used as an absolute-entropy benchmark in precision thermodynamics and even in redefinitions of the temperature scale. The thermal wavelength $\lambda_T$ has become an everyday quantity in cold-atom laboratories, where crossing $n\lambda_T^3\sim1$ marks the onset of Bose–Einstein condensation or Fermi degeneracy — the exact boundary at which the classical ideal gas surrenders to its quantum successors.

The physical picture

Think of each molecule as an independent explorer of phase space. Its accessible positions fill the volume $V$ ; its accessible momenta are spread over a Gaussian of width set by temperature. The single-particle partition function $Z_1$ counts the number of quantum cells (each of phase-space volume $h^3$ ) that the molecule can visit — spatial volume $V$ divided by the momentum-space cell $\lambda_T^3$ . When $V/\lambda_T^3\gg N$ , there are vastly more cells than molecules, each cell is essentially empty, and the gas is safely classical. The whole gas is then $N$ independent explorers, so its partition function is $Z_1^N$ — except that swapping two identical molecules is not a new state. Dividing by $N!$ removes the $N!$ ways of relabelling, and this single correction is what makes the entropy add up correctly when you join two boxes. The payoff of all this machinery is spectacular economy: $PV=Nk_BT$ and $U=\tfrac32Nk_BT$ drop out of one logarithm.

Figure

Single-particle phase space: the molecule explores the spatial volume V (left) and a Gaussian momentum cloud of width set by T (right). Z₁ counts how many cells of size h³ fit — V divided by the momentum-space cell λ_T³.

The punchline is the way the partition function converts, in one differentiation each, into the two laws every student memorizes — with a quantum length that leaves no trace in the mechanical law but a permanent one in the entropy.

Figure

Punchline: from the single function ln Z_N, a V-derivative yields PV=NkT (the mechanical law, h cancels) and a β-derivative yields U=3/2 NkT (the caloric law). Planck's constant survives only in the absolute entropy via λ_T.

The core equations

The single-particle partition function is a phase-space integral over one molecule's position and momentum, divided by $h^3$ to count in units of quantum cells. The Gaussian momentum integral produces the thermal wavelength.

Z_1=\frac{1}{h^3}\int d^3r\,d^3p\;e^{-\beta p^2/2m}=\frac{V}{\lambda_T^3},\qquad \lambda_T=\frac{h}{\sqrt{2\pi m k_BT}} \qquad (1)

where: $V$ — container volume; $\lambda_T$ — thermal de Broglie wavelength, the quantum length set by $T$ ; $h$ — Planck's constant. The spatial integral gives $V$ ; the three Gaussian momentum integrals give $\lambda_T^{-3}$ . Classicality requires $V/N\gg\lambda_T^3$ . The $N$ -particle partition function is the product of independent single-particle factors, corrected by $1/N!$ for indistinguishability, and the free energy is its logarithm.

Z_N=\frac{Z_1^N}{N!}=\frac{1}{N!}\left(\frac{V}{\lambda_T^3}\right)^N,\qquad F=-k_BT\ln Z_N \qquad (2)

where: $1/N!$ — Gibbs's correct Boltzmann counting; it removes the $N!$ relabellings of identical particles. $F$ is the Helmholtz free energy, the generating function for the mechanical and thermal properties below. Two derivatives of $F$ (or of $\ln Z_N$ ) give the mechanical and caloric equations of state. Planck's constant, carried by $\lambda_T$ , cancels from both.

P=-\left(\frac{\partial F}{\partial V}\right)_{T,N}=\frac{Nk_BT}{V}\;\Rightarrow\;PV=Nk_BT,\qquad U=-\frac{\partial\ln Z_N}{\partial\beta}=\frac{3}{2}Nk_BT \qquad (3)

where: the pressure follows from the volume dependence $\ln Z_N\ni N\ln V$ ; the internal energy from the $\beta$ dependence $\ln Z_N\ni -\tfrac32 N\ln\beta$ . Both are independent of $h$ — the quantum scale drops out of mechanics. The absolute entropy is the Sackur–Tetrode equation, obtained from $S=-(\partial F/\partial T)_{V,N}$ . Here the $1/N!$ (as $V/N$ ) and $\lambda_T$ (hence $h$ ) both survive.

\frac{S}{Nk_B}=\ln\!\left(\frac{V}{N\lambda_T^3}\right)+\frac{5}{2} \qquad (4)

where: the argument $V/N\lambda_T^3$ is intensive thanks to the $1/N!$ ; the additive $5/2$ is exact (it is $3/2$ from $U/Nk_BT$ plus $1$ from $\ln Z_N/N$ ). This entropy is measurable and matches calorimetry.

Derivation AEquation of state and internal energy from the partition function

The goal is to show that both equations of state of the ideal gas — the mechanical $PV=Nk_BT$ and the caloric $U=\tfrac32Nk_BT$ — are simple derivatives of the single logarithm $\ln Z_N$ , and that Planck's constant cancels from both. The strategy is to assemble $\ln Z_N$ using Stirling's approximation for $\ln N!$ , then differentiate with respect to volume for the pressure and with respect to $\beta$ for the energy.

Step 1. Take the logarithm of (2): $\ln Z_N=N\ln(V/\lambda_T^3)-\ln N!\approx N\ln(V/\lambda_T^3)-(N\ln N-N)$ , using Stirling $\ln N!\approx N\ln N-N$ .

Step 2. The free energy is $F=-k_BT\ln Z_N$ . Only the $N\ln V$ term depends on $V$ , so $P=-(\partial F/\partial V)_T=k_BT\,\partial(\ln Z_N)/\partial V=k_BT\cdot N/V$ . Hence $PV=Nk_BT$ — the ideal-gas law, with $\lambda_T$ (and $h$ ) absent.

Step 3. For the energy, isolate the $\beta$ dependence. Since $\lambda_T\propto\beta^{1/2}$ , we have $\lambda_T^3\propto\beta^{3/2}$ , so $\ln Z_N=-\tfrac32N\ln\beta+(\beta\text{-independent})$ . Why the $\beta^{3/2}$ . Each of the three momentum integrals contributes $\beta^{-1/2}$ ; the three together give $\lambda_T^{-3}\propto\beta^{-3/2}$ , the origin of the $\tfrac32$ in the energy.

Step 4. Differentiate: $U=-\partial\ln Z_N/\partial\beta=-\partial(-\tfrac32N\ln\beta)/\partial\beta=\tfrac32N/\beta=\tfrac32Nk_BT$ . Again $h$ cancels.

Step 5. The heat capacity is $C_V=(\partial U/\partial T)_V=\tfrac32Nk_B$ , and with $C_P-C_V=Nk_B$ (Mayer) $C_P=\tfrac52Nk_B$ , giving $\gamma=C_P/C_V=5/3$ — matching the noble gases. Verified: symbolic differentiation of $\ln Z_N=N\ln(V/\lambda_T^3)-(N\ln N-N)$ gives $P=N/(\beta V)$ (so $PV=Nk_BT$ ) and $U=\tfrac32N/\beta=\tfrac32Nk_BT$ , with $C_P-C_V=Nk_B$ and $\gamma=5/3$ — all in verify_aq-sm-classical-ideal-gas.py.

Go deeper · Derivation B · 1 minThe Sackur–Tetrode entropy and why the 1/N! is essential

Derivation BThe Sackur–Tetrode entropy and why the 1/N! is essential

The goal is to derive the absolute entropy of the ideal gas and to show that the indistinguishability factor $1/N!$ is what makes it extensive, resolving the Gibbs paradox. The strategy is to compute $S=-(\partial F/\partial T)_{V,N}=k_B\ln Z_N+U/T$ with the $1/N!$ in place, then repeat without it to expose the non-extensive failure.

Step 1. Write $S=k_B\ln Z_N+U/T$ . With $\ln Z_N=N\ln(V/\lambda_T^3)-(N\ln N-N)$ and $U/T=\tfrac32Nk_B$ , the entropy per particle is $S/(Nk_B)=\ln(V/\lambda_T^3)-\ln N+1+\tfrac32$ .

Step 2. Combine the logarithms: $\ln(V/\lambda_T^3)-\ln N=\ln(V/N\lambda_T^3)$ . The additive constant is $1+\tfrac32=\tfrac52$ . Hence $S/(Nk_B)=\ln(V/N\lambda_T^3)+\tfrac52$ , the Sackur–Tetrode equation (4).

Step 3. The argument $V/N\lambda_T^3$ is intensive — it depends only on the density $N/V$ and temperature — so $S$ scales linearly with $N$ : it is extensive, as an entropy must be.

Step 4. Now delete the $1/N!$ : then $\ln Z_N=N\ln(V/\lambda_T^3)$ with no $-\ln N$ , and $S/(Nk_B)=\ln(V/\lambda_T^3)+\tfrac32$ , whose argument $V/\lambda_T^3$ grows with $V$ at fixed density. This $S$ is not extensive. Why it matters. A non-extensive entropy predicts a spurious increase when two identical gas samples merge (the Gibbs paradox): $\Delta S_{\text{mix}}=2Nk_B\ln2\neq0$ for identical gases. The $1/N!$ turns $V^N$ into $(V/N)^N$ and the paradox vanishes.

Step 5. Check numerically: for argon at $298\,\mathrm{K}$ , $1\,\mathrm{atm}$ , plugging $m=39.95\,\mathrm{u}$ into $\lambda_T$ and the density into (4) gives a molar entropy of $154.8\,\mathrm{J\,mol^{-1}K^{-1}}$ — in exact agreement with the calorimetric value. Verified: $S/(Nk_B)=\ln(V/N\lambda_T^3)+\tfrac52$ symbolically (the $5/2$ confirmed), the argon molar entropy computes to $154.8\,\mathrm{J\,mol^{-1}K^{-1}}$ , and dropping $1/N!$ yields a non-extensive $\ln(V/\lambda_T^3)$ — verify_aq-sm-classical-ideal-gas.py.

Limiting cases

The ideal gas is a limit itself — the leading term of every more complete theory.

Case	Result	Meaning
Dilute, $V/N\gg\lambda_T^3$	Ideal-gas laws hold	Cells vastly outnumber particles; classical, non-degenerate.
Density up, $n\lambda_T^3\to1$	Quantum corrections appear	Approaching degeneracy; Bose/Fermi statistics take over.
Interactions turned on	Virial expansion $PV/Nk_BT=1+B_2(T)n+\ldots$	Ideal gas is the $n\to0$ leading term.
$T\to0$ (classically)	$S\to-\infty$	Sackur–Tetrode goes unphysical — signals quantum degeneracy sets in first.
Drop the $1/N!$	Non-extensive $S$ , Gibbs paradox	Indistinguishability is required, not optional.

Worked examples

Example 1 — Thermal wavelength and classicality of air (enumerate, apply, boundary number). Check that nitrogen at room temperature is safely classical, and find the temperature where it stops being so.

Step 1. Enumerate the inputs: $m(\mathrm{N_2})=28.0\,\mathrm{u}=4.65\times10^{-26}\,\mathrm{kg}$ , $T=300\,\mathrm{K}$ , number density at $1\,\mathrm{atm}$ is $n=P/k_BT=101325/(1.381\times10^{-23}\cdot300)=2.45\times10^{25}\,\mathrm{m^{-3}}$ .

Step 2. Apply (1): $\lambda_T=h/\sqrt{2\pi m k_BT}=6.626\times10^{-34}/\sqrt{2\pi(4.65\times10^{-26})(1.381\times10^{-23})(300)}\approx\mathbf{1.9\times10^{-11}\,m}$ .

Step 3. Degeneracy parameter $n\lambda_T^3=(2.45\times10^{25})(1.9\times10^{-11})^3\approx1.7\times10^{-7}\ll1$ . Air is classical by seven orders of magnitude.

Step 4. Boundary number: degeneracy ( $n\lambda_T^3\sim1$ ) at fixed density would need $\lambda_T$ about $180\times$ larger, i.e. $T$ smaller by $\sim180^2\approx3\times10^4$ , so of order $T\sim0.01\,\mathrm{K}$ — far below where nitrogen has long since liquefied. This is why quantum statistics of gases needs either ultracold atoms or the huge densities inside stars.

Example 2 — Molar entropy of argon (Sackur–Tetrode versus calorimetry). Compute the standard molar entropy of argon and compare with the tabulated value.

Step 1. Inputs: $m=39.95\,\mathrm{u}=6.63\times10^{-26}\,\mathrm{kg}$ , $T=298.15\,\mathrm{K}$ , $P=1\,\mathrm{atm}$ , so $n=P/k_BT=2.46\times10^{25}\,\mathrm{m^{-3}}$ and $V/N=1/n=4.06\times10^{-26}\,\mathrm{m^3}$ .

Step 2. Thermal wavelength: $\lambda_T=6.626\times10^{-34}/\sqrt{2\pi(6.63\times10^{-26})(1.381\times10^{-23})(298.15)}=1.60\times10^{-11}\,\mathrm{m}$ , so $\lambda_T^3=4.1\times10^{-33}\,\mathrm{m^3}$ .

Step 3. Apply (4): $S/(Nk_B)=\ln\big[(4.06\times10^{-26})/(4.1\times10^{-33})\big]+2.5=\ln(9.9\times10^{6})+2.5=16.1+2.5=18.6$ .

Step 4. Multiply by $R=N_Ak_B=8.314\,\mathrm{J\,mol^{-1}K^{-1}}$ : $S_{\text{molar}}=18.6\times8.314\approx\mathbf{154.8\,J\,mol^{-1}K^{-1}}$ . The calorimetric (Third-Law) value for argon is $154.8\,\mathrm{J\,mol^{-1}K^{-1}}$ — agreement to the last figure, a striking confirmation that the "classical" ideal gas secretly knows about $h$ .

Origins · 1 minideal gas and how this was discovered

ideal gas

thermodynamic model · Boyle 1662, Gay-Lussac, Clapeyron 1834

The ideal gas is the limiting model of a gas whose particles have negligible size and no mutual interactions. Its empirical laws were assembled over two centuries — Boyle's $PV=$ const (1662), Charles's and Gay-Lussac's temperature laws, and Avogadro's hypothesis — and unified by Clapeyron (1834) into $PV=nRT$ . Statistical mechanics later derived this from the partition function and, crucially, supplied what thermodynamics alone could not: the absolute entropy (Sackur–Tetrode, 1912), complete with Planck's constant and the indistinguishability factor. The ideal gas remains the universal reference state — the $n\to0$ , $T\to\infty$ corner of the phase diagram against which every correction (virial, quantum, interacting) is measured.

History

1662

— Boyle's law: at fixed temperature, PV is constant for a gas.

1834

— Clapeyron combines the gas laws into PV = nRT, the ideal-gas equation of state.

1902

— Gibbs's Elementary Principles derives gas thermodynamics from the partition function and introduces the 1/N! counting.

1911

— Sackur, and independently 1912 Tetrode, derive the absolute entropy of a monatomic ideal gas, fixing the constant with Planck's h.

1924

— Bose and Einstein extend ideal-gas theory to quantum statistics, showing where the classical model breaks down.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., Ch. 1–3 · Reif, Fundamentals of Statistical and Thermal Physics, §7.2–7.5 · Kardar, Statistical Physics of Particles, §4 (the ideal gas from the canonical ensemble) · Callen, Thermodynamics, Ch. 3 · Landau & Lifshitz, Statistical Physics Part 1, §41–45.

Provenance. The chain $Z_1=V/\lambda_T^3$ , $Z_N=Z_1^N/N!$ , $PV=Nk_BT$ , $U=\tfrac32Nk_BT$ , and the Sackur–Tetrode $S/(Nk_B)=\ln(V/N\lambda_T^3)+\tfrac52$ is standard (Pathria §1.4–2.4). A verifier should confirm: (i) $\lambda_T=h/\sqrt{2\pi m k_BT}$ gives $1.6\times10^{-11}\,\mathrm{m}$ for argon at $300\,\mathrm{K}$ ; (ii) $P=-\partial F/\partial V=Nk_BT/V$ and $U=-\partial\ln Z_N/\partial\beta=\tfrac32Nk_BT$ symbolically; (iii) the Sackur–Tetrode $5/2$ constant; (iv) the argon molar entropy $154.8\,\mathrm{J\,mol^{-1}K^{-1}}$ ; (v) $C_P-C_V=Nk_B$ , $\gamma=5/3$ ; (vi) dropping $1/N!$ makes $S$ non-extensive. All in verify_aq-sm-classical-ideal-gas.py. Constants: $k_B=1.380649\times10^{-23}\,\mathrm{J/K}$ , $h=6.62607015\times10^{-34}\,\mathrm{J\,s}$ , $R=8.314\,\mathrm{J\,mol^{-1}K^{-1}}$ .

Builds on: Canonical Ensemble / Partition Function · Correct Boltzmann Counting · Thermal de Broglie Wavelength Leads to: Sackur-Tetrode Equation · Maxwell-Boltzmann Velocity Distribution · Virial Equation of State · Classical Limit of Quantum Statistics

Why does a "classical" gas involve Planck's constant at all?

Because counting states requires a unit of phase-space volume, and that unit is $h^3$ per degree of freedom — a purely quantum input. In the mechanical and caloric equations of state ( $PV=Nk_BT$ , $U=\tfrac32Nk_BT$ ) the $h$ -dependence cancels between numerator and denominator, so those look fully classical. But the absolute entropy cannot be fixed without choosing the phase-space cell size, and there $h$ survives explicitly through $\lambda_T$ . The Sackur–Tetrode entropy is the classical thermodynamic quantity that most directly betrays the quantum graininess of phase space.

What exactly is the thermal de Broglie wavelength?

It is the quantum length $\lambda_T=h/\sqrt{2\pi m k_BT}$ , roughly the de Broglie wavelength of a particle with thermal momentum $\sqrt{2\pi m k_BT}$ . Physically it is the size of a particle's quantum "wave packet" at temperature $T$ . When the average interparticle spacing $(V/N)^{1/3}$ is much larger than $\lambda_T$ — equivalently $n\lambda_T^3\ll1$ — the wave packets do not overlap, the particles behave as distinguishable classical objects, and the ideal-gas laws hold. When $n\lambda_T^3\sim1$ they overlap and quantum statistics takes over.

Where does the 3/2 in U = (3/2)NkT come from?

From three translational degrees of freedom, each contributing $\tfrac12k_BT$ by equipartition — or, in the partition-function language, from the three Gaussian momentum integrals, each of which gives a factor $\beta^{-1/2}$ in $Z_1$ . So $Z_1\propto\beta^{-3/2}$ , $\ln Z_N\ni-\tfrac32N\ln\beta$ , and $U=-\partial\ln Z_N/\partial\beta=\tfrac32N/\beta=\tfrac32Nk_BT$ . For a diatomic gas with rotational and vibrational modes active the coefficient rises to $5/2$ and then $7/2$ ; the monatomic $3/2$ is just the translational part.

Is the ideal-gas entropy really measurable, or just a formula?

It is genuinely measurable. The Third Law fixes $S=0$ at $T=0$ for a perfect crystal, and integrating the calorimetric heat capacity $\int C_P\,dT/T$ up through melting and vaporization to the gas phase gives the absolute molar entropy. For argon and other monatomic gases this calorimetric value agrees with the Sackur–Tetrode prediction to within experimental error (both about $154.8\,\mathrm{J\,mol^{-1}K^{-1}}$ for argon at standard conditions). This agreement is one of the classic experimental confirmations that statistical mechanics — including its quantum phase-space counting — is correct.

Does the ideal gas ever fail, and what replaces it?

Yes, in two ways. When interactions matter (dense or cold-but-not-degenerate gases) the virial expansion corrects $PV=Nk_BT$ with density-dependent terms $B_2(T),B_3(T),\ldots$ set by the interparticle potential. When quantum degeneracy matters ( $n\lambda_T^3\gtrsim1$ , i.e. ultracold or ultradense) the classical counting fails and you must use Bose–Einstein or Fermi–Dirac statistics, giving degeneracy pressure, Bose–Einstein condensation, and the electronic heat capacity of metals. In both regimes the ideal gas remains the leading term of the correct theory.

Why divide by N! and not by something else?

Because there are exactly $N!$ ways to permute the labels of $N$ identical particles, and in classical mechanics these permuted configurations are counted as distinct microstates even though they are physically identical. Dividing $Z_1^N$ by $N!$ removes this overcounting. It is an approximation to the exact quantum symmetrization (which would give the Bose or Fermi count), valid in the non-degenerate regime where multiple occupancy of a single state is negligible — precisely the classical regime $n\lambda_T^3\ll1$ where the ideal-gas model lives.