Sackur-Tetrode Equation

Thermodynamics can tell you how the entropy of a gas changes when you heat it or expand it, but it can never fix the constant — the absolute value of $S$ is beyond its reach. The Sackur–Tetrode equation is the first time in physics that a closed formula gave the absolute entropy of a real substance, the monatomic ideal gas, in terms of nothing but its energy, volume, and particle number. Written compactly, $S/(Nk_B)=\ln\!\big(V/(N\lambda^3)\big)+\tfrac52$ , it is one of the most quietly astonishing results in statistical mechanics: a "classical" gas whose entropy secretly contains Planck's constant $h$ , hidden inside the thermal de Broglie wavelength $\lambda=h/\sqrt{2\pi m k_BT}$ . The equation ties three ideas into one line. The logarithm counts how many quantum phase-space cells of size $\lambda^3$ each particle can roam through inside the volume $V/N$ available to it. The $+\tfrac52$ is an exact pure number — $\tfrac32$ from the kinetic energy $U=\tfrac32 Nk_BT$ and $+1$ from the correct counting of indistinguishable particles. And the division by $N$ inside the logarithm — the fingerprint of the $1/N!$ factor of correct Boltzmann counting — is what makes the entropy extensive, the property without which the Gibbs paradox appears and thermodynamics breaks. What makes the formula worth its own article is that it is not just an entropy value: it is a full fundamental relation. Differentiate $S(U,V,N)$ and the entire thermodynamics of the ideal gas falls out — the caloric law $U=\tfrac32 Nk_BT$ , the equation of state $PV=Nk_BT$ , the heat capacity $C_V=\tfrac32 Nk_B$ . And the numbers are not hand-waving: plug in argon at room temperature and the formula returns $154.8\ \mathrm{J\,mol^{-1}K^{-1}}$ , agreeing with the calorimetric third-law entropy to the last digit — a direct, measurable proof that the classical gas knows about quantum mechanics.

Key takeaways

The Sackur–Tetrode equation gives the absolute entropy of a monatomic ideal gas: $S/(Nk_B)=\ln\!\big(V/(N\lambda^3)\big)+\tfrac52$ , with $\lambda=h/\sqrt{2\pi m k_BT}$ the thermal de Broglie wavelength.
The logarithm counts the quantum phase-space cells (size $\lambda^3$ ) each particle can occupy in its share of volume $V/N$ ; Planck's constant $h$ survives in the absolute entropy even though the gas is classical.
The additive $\tfrac52$ is exact: $\tfrac32$ from $U=\tfrac32 Nk_BT$ plus $1$ from correct Boltzmann counting; it is not an adjustable constant.
The $V/N$ inside the logarithm — the imprint of the $1/N!$ factor — makes $S$ extensive; drop it and the entropy is non-extensive and the Gibbs paradox appears.
It is a full fundamental relation: differentiating $S(U,V,N)$ regenerates $U=\tfrac32 Nk_BT$ , $PV=Nk_BT$ , and $C_V=\tfrac32 Nk_B$ .
For argon at $298\,\mathrm K$ , $1\,\mathrm{atm}$ it predicts $154.8\ \mathrm{J\,mol^{-1}K^{-1}}$ , matching calorimetry exactly.
Key idea: the Sackur–Tetrode equation is the absolute entropy of a classical ideal gas that nonetheless carries $h$ and $1/N!$ — a single relation from which all of ideal-gas thermodynamics follows, and the first quantitative bridge between the microscopic quantum world and macroscopic entropy.

The core insight

The Sackur–Tetrode equation, $S/(Nk_B)=\ln\!\big(V/(N\lambda^3)\big)+\tfrac52$ with $\lambda=h/\sqrt{2\pi m k_BT}$ , is the absolute entropy of a monatomic ideal gas: a full fundamental relation carrying both Planck's constant (via $\lambda$ ) and quantum indistinguishability (via the $V/N$ from $1/N!$ ), from which the entire ideal-gas thermodynamics follows by differentiation.

Where this lives today

The Sackur–Tetrode equation is a workhorse of modern precision thermodynamics and physical chemistry. It supplies the absolute standard-state entropy of monatomic gases — argon, helium, the vapours of mercury and the alkali metals — used as reference values in thermochemical tables and in computing equilibrium constants from spectroscopic and molecular data (the statistical-mechanical route to $\Delta G$ ). In metrology it plays a role in the modern realization of the kelvin and in acoustic-gas thermometry, where the known absolute entropy of a noble gas is a fixed anchor. In cold-atom laboratories the same thermal wavelength $\lambda$ that appears in the formula is the everyday diagnostic: experimentalists quote the phase-space density $n\lambda^3$ to say how close a gas is to Bose–Einstein condensation or Fermi degeneracy, the exact point where Sackur–Tetrode surrenders to its quantum successors. And pedagogically it remains the cleanest demonstration that "classical" statistical mechanics is quietly quantum — a formula whose measured agreement with calorimetry cannot be explained without $h$ .

The physical picture

Imagine a single atom of the gas as an explorer of phase space. In real space it can wander anywhere inside the volume it "owns", which for indistinguishable particles is $V/N$ — not the full box $V$ , because the atoms cannot be told apart and dividing by $N$ removes the overcounting. In momentum space it can occupy a Gaussian blur of width set by temperature. The quantum unit of phase-space area is $h$ per degree of freedom, so the natural momentum-space cell has linear size set by the thermal de Broglie wavelength $\lambda=h/\sqrt{2\pi m k_BT}$ . The entropy per particle is essentially the logarithm of how many such cells fit into the accessible region: $\ln\!\big(V/(N\lambda^3)\big)$ , plus a fixed number from the energy and the counting. When the gas is dilute and warm, $V/N\gg\lambda^3$ — there are vastly more cells than particles, each cell nearly empty, and the gas is safely classical. The entropy is large and positive. As you cool or compress, $\lambda$ grows and $V/N$ shrinks; when $V/N\sim\lambda^3$ the cells are as crowded as the particles, quantum statistics take over, and the Sackur–Tetrode formula (which knows nothing of Bose or Fermi bunching) fails. The famous symptom is that at $T\to0$ the classical formula predicts $S\to-\infty$ — an absurdity that is precisely the warning flag that quantum degeneracy has set in.

Figure

A single atom explores V/N in real space (many quantum cells of size λ³). Dilute warm gas: cells vastly outnumber atoms (classical). Cool or compress: λ grows until V/N ≈ λ³ and cells crowd (quantum takes over).

The second figure is the punchline: plot the Sackur–Tetrode entropy against $\ln T$ and it is a straight line of slope $\tfrac32 Nk_B$ (the kinetic contribution), climbing forever as the gas gets hotter. But run it the other way and the line marches straight down through zero into $S<0$ and toward $-\infty$ as $T\to0$ . No real substance has negative entropy — this failure on the plot is the exact signature that the classical description breaks down at low temperature, where quantum statistics and the third law take over.

Figure

Punchline: Sackur-Tetrode S vs ln T is a straight line, slope (3/2)Nk from the kinetic term. Extrapolated to low T it crosses zero and dives to −∞ — the classical formula failing, marking where quantum degeneracy takes over.

The core equations

The central object is the absolute entropy of $N$ monatomic ideal-gas atoms of mass $m$ in a volume $V$ at temperature $T$ . It is usually written in terms of the thermal de Broglie wavelength, which packages the temperature, the mass, and Planck's constant into one length.

\lambda=\frac{h}{\sqrt{2\pi m k_BT}} \qquad (1)

where: $h$ — Planck's constant; $m$ — atomic mass; $k_BT$ — thermal energy. This is the thermal de Broglie wavelength, owned and derived in the companion article; here it is the quantum length scale that sets the phase-space cell. Reading it: $\lambda$ grows as the gas cools, and classicality requires $V/N\gg\lambda^3$ . With $\lambda$ in hand, the Sackur–Tetrode equation states the entropy per particle. The logarithm is a pure ratio of volumes (real-space volume per particle over the momentum-cell volume $\lambda^3$ ), and the additive constant is exact.

\frac{S}{Nk_B}=\ln\!\left(\frac{V}{N\lambda^3}\right)+\frac52 \qquad (2)

where: $V/N$ — the volume per particle (the $1/N$ is the imprint of correct Boltzmann counting, $1/N!$ ); $\lambda^3$ — the volume of one quantum phase-space cell; the $+\tfrac52$ is exact, decomposing as $\tfrac32$ (from the kinetic energy) plus $1$ (from the counting). Reading it: the entropy is large when the gas is dilute ( $V/N\gg\lambda^3$ ), and the formula becomes unphysical ( $S<0$ ) when the argument of the log drops below $e^{-5/2}$ — the classical–quantum boundary. Because $\lambda\propto T^{-1/2}$ , the entropy can be written as an explicit function of $U$ , $V$ , and $N$ using $U=\tfrac32 Nk_BT$ , turning it into a fundamental relation from which everything else follows.

S(U,V,N)=Nk_B\left[\ln\!\left(\frac{V}{N}\left(\frac{4\pi mU}{3Nh^2}\right)^{3/2}\right)+\frac52\right] \qquad (3)

where: the substitution $\lambda^{-3}=(2\pi m k_BT/h^2)^{3/2}=(4\pi mU/3Nh^2)^{3/2}$ (using $k_BT=2U/3N$ ) replaces $T$ by $U$ . Reading it: this is the entropy representation of the ideal gas; its partial derivatives are the intensive variables, so all of thermodynamics is one differentiation away. The two first derivatives of equation (3) are the payoff — they return the caloric law and the equation of state.

\left(\frac{\partial S}{\partial U}\right)_{V,N}=\frac{3Nk_B}{2U}=\frac1T,\qquad \left(\frac{\partial S}{\partial V}\right)_{U,N}=\frac{Nk_B}{V}=\frac PT \qquad (4)

where: the first gives $U=\tfrac32 Nk_BT$ (caloric law), the second gives $PV=Nk_BT$ (ideal-gas law). That both emerge from one $S(U,V,N)$ is the sense in which the Sackur–Tetrode equation contains the complete thermodynamics of the gas.

Derivation AAssembling the Sackur–Tetrode entropy from the partition function

The goal is to obtain the entropy $S/(Nk_B)=\ln\!\big(V/(N\lambda^3)\big)+\tfrac52$ from the ideal-gas partition function, showing exactly where the $\tfrac52$ and the $V/N$ come from. The strategy is to start from the free energy $F=-k_BT\ln Z_N$ with $Z_N=(V/\lambda^3)^N/N!$ , apply Stirling's approximation to the $1/N!$ , and use $S=(U-F)/T$ . We deliberately recall the $1/N!$ from correct Boltzmann counting and $\lambda$ from the thermal de Broglie wavelength rather than re-derive them — the assembly is what this article owns.

Step 1. Take the logarithm of the partition function $Z_N=(V/\lambda^3)^N/N!$ : $\ln Z_N=N\ln(V/\lambda^3)-\ln N!$ . Apply Stirling $\ln N!\approx N\ln N-N$ (recalled from the Stirling-approximation article), giving $\ln Z_N=N\ln(V/\lambda^3)-N\ln N+N$ .

Step 2. Combine the two $N\ln$ terms: $N\ln(V/\lambda^3)-N\ln N=N\ln\!\big(V/(N\lambda^3)\big)$ . So $\ln Z_N=N\ln\!\big(V/(N\lambda^3)\big)+N$ . The $-N\ln N$ from the $1/N!$ is precisely what turns the extensive-breaking $V/\lambda^3$ into the intensive $V/(N\lambda^3)$ .

Step 3. The entropy is $S=k_B\ln Z_N+U/T$ . The kinetic energy is $U=\tfrac32 Nk_BT$ (recalled from equipartition), so $U/T=\tfrac32 Nk_B$ . Therefore $S/(Nk_B)=\ln\!\big(V/(N\lambda^3)\big)+1+\tfrac32$ . Why the constant is exactly $\tfrac52$ . The $+1$ comes from the $+N$ left over from Stirling on $1/N!$ (the counting), and the $+\tfrac32$ comes from $U/(Nk_BT)$ (the kinetic energy). Their sum $\tfrac52$ is a pure number with no free parameter — it is fixed once you accept $h$ as the phase-space cell size and $1/N!$ as the counting.

Step 4. Collect: $S/(Nk_B)=\ln\!\big(V/(N\lambda^3)\big)+\tfrac52$ , equation (2). This is the Sackur–Tetrode equation. Verified: the $\tfrac52=\tfrac32+1$ decomposition, and the identity $\int e^{-\beta p^2/2m}\,d^3p/h^3=1/\lambda^3$ with $\lambda=h/\sqrt{2\pi m k_BT}$ , are confirmed in verify_aq-sm-sackur-tetrode-equation.py.

Go deeper · Derivation B · 1 minWhy Sackur–Tetrode is a fundamental relation, and the role of the 1/N!

Derivation BWhy Sackur–Tetrode is a fundamental relation, and the role of the 1/N!

The goal is to show that equation (3), $S(U,V,N)$ , is a genuine fundamental relation: differentiating it recovers the entire thermodynamics of the ideal gas, and the $V/N$ (from $1/N!$ ) is exactly what makes $S$ extensive. The strategy is to differentiate $S(U,V,N)$ with respect to $U$ and $V$ , identify $1/T$ and $P/T$ , and then test extensivity by scaling $N,V,U$ together.

Step 1. From equation (3), $S=Nk_B[\ln(V/N)+\tfrac32\ln U+\text{const}(m,N)+\tfrac52]$ where the constant collects the $U$ - and $V$ -independent pieces. Differentiate with respect to $U$ at fixed $V,N$ : $(\partial S/\partial U)_{V,N}=Nk_B\cdot\tfrac32\cdot\tfrac1U=\tfrac{3Nk_B}{2U}$ .

Step 2. Set this equal to $1/T$ : $\tfrac{3Nk_B}{2U}=\tfrac1T$ gives $U=\tfrac32 Nk_BT$ — the caloric law, recovered purely by differentiation.

Step 3. Differentiate with respect to $V$ at fixed $U,N$ : only the $\ln V$ term contributes, $(\partial S/\partial V)_{U,N}=Nk_B/V$ . Setting this equal to $P/T$ gives $P/T=Nk_B/V$ , i.e. $PV=Nk_BT$ — the ideal-gas equation of state, again by one derivative. Why the $1/N!$ is not optional. The volume enters equation (3) only through $\ln(V/N)$ . That $/N$ came from the $-N\ln N$ of Stirling on $1/N!$ . Scale the system by $\alpha$ ( $N\to\alpha N,\ V\to\alpha V,\ U\to\alpha U$ ): the argument $V/N$ and the density-dependent pieces are unchanged (intensive), and the overall $Nk_B$ scales by $\alpha$ , so $S\to\alpha S$ — extensive. Delete the $/N$ and the log picks up an extra $\ln\alpha$ , breaking extensivity: that is the Gibbs paradox.

Step 4. The heat capacity follows too: $C_V=(\partial U/\partial T)_V=\partial(\tfrac32 Nk_BT)/\partial T=\tfrac32 Nk_B$ . So $U$ , $P$ , and $C_V$ all descend from the single $S(U,V,N)$ — the defining property of a fundamental relation. Verified: $(\partial S/\partial U)_V=3Nk/2U=1/T\Rightarrow U=\tfrac32 NkT$ , $(\partial S/\partial V)_U=Nk/V=P/T\Rightarrow PV=NkT$ , $C_V=\tfrac32 Nk$ , and the extensivity scaling $S(\alpha N,\alpha V,\alpha U)=\alpha S$ (which fails without $1/N!$ ), are all confirmed symbolically in the verify script.

Limiting cases

The Sackur–Tetrode equation is exact for a classical monatomic ideal gas; each row shows how it specializes or where it fails.

Case	Result	Meaning
Dilute, warm ( $V/N\gg\lambda^3$ )	$S>0$ , extensive	Classical regime; formula exact and matches calorimetry.
Isothermal expansion $V_1\to V_2$	$\Delta S=Nk_B\ln(V_2/V_1)$	Only the log- $V$ term changes; $\lambda,\ \tfrac52$ drop out.
Isochoric heating $T_1\to T_2$	$\Delta S=\tfrac32 Nk_B\ln(T_2/T_1)$	Only the log- $T$ (via $\lambda^{-3}$ ) term changes.
$T\to0$ (classical)	$S\to-\infty$	Unphysical; signals quantum degeneracy and the third law.
$V/N\to\lambda^3$	formula fails	Quantum concentration reached; Bose/Fermi statistics required.

Common trapThe ideal gas is classical, so its entropy has no Planck's constant

The most seductive misconception is that a "classical" gas should have a purely classical entropy. It does not. Planck's constant $h$ enters through the thermal wavelength $\lambda=h/\sqrt{2\pi m k_BT}$ and survives in the absolute entropy — it cancels only from the mechanical and caloric equations of state ( $PV=Nk_BT$ , $U=\tfrac32 Nk_BT$ ). Without $h$ there is no way to fix the additive constant, and the entropy would be defined only up to an arbitrary reference. A second trap is treating the $1/N!$ (equivalently the $V/N$ inside the log) as harmless bookkeeping: it drops out of $U$ and $P$ , so it is easy to forget, but it is essential for the entropy. Omit it and $S$ becomes non-extensive, doubling the system fails to double the entropy, and the Gibbs paradox appears. The Sackur–Tetrode equation is the place where the quantum imprint on a classical gas is unavoidable.

Worked examples

Example 1 — The absolute molar entropy of argon (enumerate, apply, boundary number). Compute the standard molar entropy of argon gas at $T=298.15\,\mathrm K$ and $P=1\,\mathrm{atm}=101325\,\mathrm{Pa}$ from the Sackur–Tetrode equation, and compare with the calorimetric third-law value.

Step 1. Enumerate the inputs. Argon is monatomic, $m=39.948\ \mathrm u=39.948\times1.6605\times10^{-27}=6.633\times10^{-26}\,\mathrm{kg}$ . Constants: $h=6.626\times10^{-34}\,\mathrm{J\,s}$ , $k_B=1.381\times10^{-23}\,\mathrm{J/K}$ , $N_A=6.022\times10^{23}$ .

Step 2. Thermal wavelength: $\lambda=h/\sqrt{2\pi m k_BT}=6.626\times10^{-34}/\sqrt{2\pi(6.633\times10^{-26})(1.381\times10^{-23})(298.15)}=\mathbf{1.60\times10^{-11}\,m}$ , so $\lambda^3=4.10\times10^{-33}\,\mathrm{m^3}$ .

Step 3. Volume per particle from $PV=Nk_BT$ : $V/N=k_BT/P=(1.381\times10^{-23})(298.15)/101325=4.06\times10^{-26}\,\mathrm{m^3}$ . The log argument is $V/(N\lambda^3)=4.06\times10^{-26}/4.10\times10^{-33}=9.9\times10^{6}$ , so $\ln\!\big(V/(N\lambda^3)\big)=16.11$ .

Step 4. Apply the formula: $S/(Nk_B)=16.11+2.5=18.61$ . Multiply by $R=N_Ak_B=8.314\,\mathrm{J\,mol^{-1}K^{-1}}$ : $S_{\text{molar}}=18.61\times8.314=\mathbf{154.7\ J\,mol^{-1}K^{-1}}$ . The calorimetric third-law entropy of argon is $154.8\ \mathrm{J\,mol^{-1}K^{-1}}$ — agreement to the last figure, a direct experimental confirmation that a classical gas's entropy contains $h$ .

Example 2 — Entropy of isothermal expansion. One mole of argon at $298.15\,\mathrm K$ is allowed to double its volume isothermally. Find $\Delta S$ .

Step 1. At fixed $T$ , the only $V$ -dependence in Sackur–Tetrode is through $\ln V$ : $S/(Nk_B)=\ln V+(V\text{-independent})$ . So $\Delta S=Nk_B\ln(V_2/V_1)$ .

Step 2. For doubling, $V_2/V_1=2$ , and per mole $Nk_B\to R$ : $\Delta S=R\ln 2=8.314\times0.6931=\mathbf{5.76\ J\,mol^{-1}K^{-1}}$ .

Step 3. Boundary check. This is exactly the entropy of free expansion into a doubled volume — independent of the gas's identity (mass, wavelength, the $\tfrac52$ all cancel), depending only on the volume ratio. It matches the thermodynamic result $\Delta S=nR\ln(V_2/V_1)$ derived without any microscopic input, confirming the Sackur–Tetrode formula reproduces the classical entropy change while additionally fixing the absolute value.

Origins · 1 minsackur-tetrode equation and how this was discovered

sackur-tetrode equation

derived 1911–1912 · Otto Sackur (1880–1914) and Hugo Tetrode (1895–1931)

The Sackur–Tetrode equation was derived nearly simultaneously and independently by two physicists whose lives ended in tragedy. Otto Sackur, a German physical chemist, published the absolute entropy of a monatomic gas in 1911, fixing the additive constant with Planck's new quantum of action $h$ ; he died in 1914 at age 34 in a laboratory explosion while researching chemical warfare agents. Hugo Tetrode, a reclusive Dutch prodigy, derived the same result in 1912 at roughly seventeen years old, working almost entirely in isolation and corresponding rarely; he died of tuberculosis in 1931 at 35. Their equation was the first to give a real substance's absolute entropy from first principles, and the exact agreement of its predictions with calorimetric measurements was an early, decisive vindication of the quantum hypothesis in thermodynamics.

History

1900

— Planck introduces the quantum of action h, providing the natural phase-space cell size.

1902

— Gibbs formulates the correct counting of indistinguishable particles (the 1/N! factor).

1911

— Sackur derives the absolute entropy of a monatomic ideal gas, fixing the constant with h.

1912

— Tetrode independently derives the same result and compares it to mercury-vapour data.

1912

— The exact match with calorimetric entropies confirms the quantum phase-space cell.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., §1.5–1.6, §2.4 (Gibbs paradox, Sackur–Tetrode) · Reif, Fundamentals of Statistical and Thermal Physics, §7.3 · Kardar, Statistical Physics of Particles, §4.4 · Callen, Thermodynamics, §16 · McQuarrie, Statistical Mechanics, Ch. 5.

Provenance. The Sackur–Tetrode equation $S/(Nk_B)=\ln(V/N\lambda^3)+\tfrac52$ with $\lambda=h/\sqrt{2\pi m k_BT}$ is standard (Pathria §2.4, Reif §7.3). A verifier should confirm: (i) the $\tfrac52=\tfrac32+1$ decomposition; (ii) $(\partial S/\partial U)_V=3Nk/2U=1/T\Rightarrow U=\tfrac32 NkT$ ; (iii) $(\partial S/\partial V)_U=Nk/V=P/T\Rightarrow PV=NkT$ ; (iv) $C_V=\tfrac32 Nk$ ; (v) extensivity $S(\alpha N,\alpha V,\alpha U)=\alpha S$ (failing without $1/N!$ ); (vi) $\int e^{-\beta p^2/2m}d^3p/h^3=1/\lambda^3$ ; (vii) argon at $298.15\,\mathrm K$ , $1\,\mathrm{atm}$ gives $S_{\text{molar}}=154.7\ \mathrm{J\,mol^{-1}K^{-1}}$ , matching the calorimetric $154.8$ . All checks in verify_aq-sm-sackur-tetrode-equation.py. Constants: $h=6.626\times10^{-34}\,\mathrm{J\,s}$ , $k_B=1.380649\times10^{-23}\,\mathrm{J/K}$ , $m(\mathrm{Ar})=39.948\,\mathrm u$ , $R=8.314\,\mathrm{J\,mol^{-1}K^{-1}}$ .

Builds on: Classical Ideal Gas · Correct Boltzmann Counting · Thermal de Broglie Wavelength · Equipartition Theorem Leads to: Gibbs Paradox · Classical Limit of Quantum Statistics · Virial Equation of State

Why does a "classical" ideal gas have Planck's constant in its entropy?

Because you cannot count phase-space states without a unit of phase-space volume, and that unit is $h$ per degree of freedom. The number of distinct quantum cells a particle can occupy is (real-space volume) $\times$ (momentum-space volume) divided by $h^3$ , and this ratio sets the absolute entropy. Planck's constant cancels from the equation of state and the internal energy — quantities that involve only entropy differences at fixed $T$ — but it cannot cancel from the absolute entropy, which counts states directly. The Sackur–Tetrode formula makes this explicit: $h$ sits inside $\lambda$ , and the measured agreement with calorimetry proves it is really there.

What does the 5/2 actually mean?

It is an exact pure number that splits into two physically distinct pieces. The $\tfrac32$ comes from the kinetic energy: $U=\tfrac32 Nk_BT$ , and $U/T$ contributes $\tfrac32 Nk_B$ to the entropy $S=k_B\ln Z+U/T$ . The $+1$ comes from the correct counting of indistinguishable particles: Stirling's approximation of $\ln N!$ leaves a $+N$ that becomes the $+1$ per particle. So $\tfrac52=\tfrac32+1$ is "kinetic energy plus counting". There is nothing to adjust; once you fix $h$ as the cell size and use $1/N!$ , the constant is determined.

Why is the volume divided by N inside the logarithm?

Because the particles are indistinguishable, and the $1/N!$ factor of correct Boltzmann counting shows up as a $/N$ inside the log after Stirling's approximation. Physically, each particle "owns" a share $V/N$ of the box, not the whole box $V$ . This is not cosmetic: it makes the entropy extensive. If you wrote $\ln V$ instead of $\ln(V/N)$ , then doubling the system (double $N$ , $V$ , $U$ ) would not double the entropy — an extra $\ln 2$ per particle would appear. That extensivity failure is exactly the Gibbs paradox, and the $V/N$ is its resolution.

Why does the entropy go to −∞ as T → 0?

Because the thermal wavelength $\lambda=h/\sqrt{2\pi m k_BT}$ grows without bound as $T\to0$ , so $V/(N\lambda^3)\to0$ and its logarithm diverges to $-\infty$ . But no real substance has negative entropy — the third law requires $S\to$ constant (usually zero) at absolute zero. The divergence is a warning: long before $T=0$ , the gas reaches quantum concentration ( $V/N\sim\lambda^3$ ), where Bose or Fermi statistics take over and the classical Sackur–Tetrode formula ceases to apply. In reality the gas has condensed or become degenerate well before the formula predicts negative entropy.

How well does the formula actually match experiment?

Remarkably well for monatomic gases. For argon at $298.15\,\mathrm K$ and $1\,\mathrm{atm}$ the formula gives $154.7\ \mathrm{J\,mol^{-1}K^{-1}}$ , and the calorimetric third-law entropy (built up by integrating $C_p/T$ from near absolute zero through the solid, liquid, and gas phases) is $154.8\ \mathrm{J\,mol^{-1}K^{-1}}$ — agreement to about $0.1\%$ . Similar accuracy holds for the other noble gases and mercury vapour. This match was historically one of the strongest early confirmations that Planck's quantum of action genuinely governs the counting of thermodynamic states, not just blackbody radiation.

Does the Sackur–Tetrode equation apply to molecular gases like N₂?

Only for the translational part. The formula gives the entropy of the center-of-mass motion of monatomic (or, for the translational contribution, any) gas. For a molecular gas you must add the rotational and vibrational (and electronic) contributions to the entropy, computed from their own partition functions, since those degrees of freedom store additional entropy. So for nitrogen the full standard entropy is the Sackur–Tetrode translational term plus rotational and (frozen-out at room temperature) vibrational terms. The Sackur–Tetrode equation is exact for the translational piece, which dominates and is common to all gases.

Is the 1/N! really quantum, or just a classical bookkeeping trick?

It is the classical shadow of quantum indistinguishability. In a fully classical world, identical particles would still be labelable in principle, and there would be no reason to divide by $N!$ — leading to the non-extensive entropy and the Gibbs paradox. Quantum mechanics says identical particles are fundamentally indistinguishable: swapping two of them gives the same state, not a new one. In the classical (non-degenerate) limit, where at most one particle occupies each state, this indistinguishability reduces exactly to the $1/N!$ overcounting correction. So the factor is "put in by hand" classically but derived quantum-mechanically — it is the fingerprint of quantum statistics on the classical gas.