Sackur-Tetrode Equation for Entropy

Take one mole of neon gas at room temperature and one atmosphere. Ask a simple question: what is its absolute entropy? Classical thermodynamics can only give entropy differences from calorimetry. Statistical mechanics answers the absolute question directly, and the result is the Sackur-Tetrode equation. This equation matters because it connects a purely mechanical description, mass, volume, temperature, to the thermodynamic quantity of entropy, with no adjustable constants. It was one of the first triumphs of quantum statistics applied to a classical-looking gas. It also forces a subtle but essential quantum correction, particle indistinguishability, into an otherwise classical calculation. This article derives the Sackur-Tetrode equation from the translational partition function. It shows why naive classical counting fails without Gibbs correction. It works two full numerical examples. It closes with limiting cases, historical context, and a compact exam sheet.

Key takeaways

The Sackur-Tetrode equation gives the absolute molar entropy of a monatomic ideal gas from $M$ , $T$ , and $p$ alone.
It follows from $S = U/T + k\ln Q$ applied to the translational partition function, with indistinguishability included via $N!$ .
The equation predicts entropy correctly down to moderate temperatures but breaks down as $T \to 0$ , where quantum degeneracy takes over.
It requires only fundamental constants, no fitted parameters, unlike calorimetric third-law entropies which need heat capacity data down to 0 K.
It confirms the Gibbs paradox resolution: dividing by $N!$ is not a convenient trick but a necessary consequence of quantum indistinguishability.
Key idea: absolute entropy of an ideal monatomic gas is computable from first principles once you count indistinguishable translational microstates correctly.

The core insight

The Sackur-Tetrode equation shows that the entropy of a monatomic ideal gas is entirely determined by the number of thermally accessible translational quantum states per particle, correctly counted as indistinguishable, so that $S$ becomes a calculable function of $M$ , $T$ , and $p$ with zero empirical input.

The physical picture

Picture one mole of argon atoms confined to a container of volume $V$ at temperature $T$ . Quantum mechanics says each atom occupies discrete translational energy levels, spaced so closely at ordinary temperatures that they form a near-continuum. The number of these levels that are thermally accessible, weighted by the Boltzmann factor, is the translational partition function $q$ . A larger $q$ means more microstates are available to each atom, and more available microstates means higher entropy, since entropy measures the logarithm of the number of ways the system's energy can be arranged. The subtlety is that atoms of the same element are fundamentally indistinguishable. If you calculate $q$ for one atom, then imagine $N$ atoms independently occupying those states, you overcount configurations that differ only by swapping which atom sits in which state. Gibbs first noticed the resulting paradox: without correcting for this overcounting, entropy fails to be extensive, doubling both $N$ and $V$ at fixed density does not correctly double $S$ . The fix, dividing the naive partition function by $N!$ , was originally an ad hoc patch. It became rigorous once quantum mechanics confirmed identical particles truly are indistinguishable, not just difficult to tell apart. The Sackur-Tetrode equation is what emerges when you push this counting through completely: translational quantum states, Boltzmann statistics, and the $N!$ correction, expressed finally in the macroscopic variables $T$ , $p$ , and molar mass $M$ .

Figure

Translational energy levels become denser as the box grows, raising q and hence S

The core relations

q_{\text{trans}} = \left(\frac{2\pi m k T}{h^2}\right)^{3/2} V

where $m$ is the mass of one particle, $k$ Boltzmann's constant, $h$ Planck's constant, $V$ the volume.

S = \frac{U}{T} + k \ln\!\left(\frac{q^N}{N!}\right)

where $U$ is internal energy, $N$ the number of particles, and $q^N/N!$ the corrected partition function for indistinguishable, non-interacting particles.

S_m = R\left[\ln\!\left(\frac{V_m}{N_A}\left(\frac{2\pi m k T}{h^2}\right)^{3/2}\right) + \frac{5}{2}\right]

where $S_m$ is molar entropy, $V_m$ molar volume, $N_A$ Avogadro's number, $R = N_A k$ .

S_m = R\left[\ln\!\left(\frac{kT}{p}\left(\frac{2\pi m k T}{h^2}\right)^{3/2}\right) + \frac{5}{2}\right]

where $p$ is pressure, using $V_m/N_A = kT/p$ from the ideal gas law.

Derivation AFrom the translational partition function to Sackur-Tetrode

Goal. Obtain a closed-form expression for the molar entropy of a monatomic ideal gas using only $M$ , $T$ , $p$ , and fundamental constants.

Step 1: Start from the statistical entropy formula. For an ideal gas of $N$ indistinguishable, non-interacting, structureless (monatomic) particles, the canonical partition function is $Q = q^N/N!$ , where $q$ is the single-particle translational partition function. The Helmholtz energy is $A = -kT\ln Q$ , and entropy follows from

S = -\left(\frac{\partial A}{\partial T}\right)_{V,N} = k\ln Q + kT\left(\frac{\partial \ln Q}{\partial T}\right)_{V,N}.

Since $U = kT^2(\partial \ln Q/\partial T)_{V,N}$ , this is equivalently $S = U/T + k\ln Q$ .

Step 2: Insert the translational partition function. For a particle of mass $m$ in a box of volume $V$ ,

q = \left(\frac{2\pi m k T}{h^2}\right)^{3/2} V \equiv \frac{V}{\Lambda^3},

where $\Lambda = h/\sqrt{2\pi m k T}$ is the thermal de Broglie wavelength. This form comes directly from summing the Boltzmann factors over the particle-in-a-box translational levels and replacing the sum by an integral, valid whenever $\Lambda \ll$ the box dimensions.

Step 3: Apply Stirling's approximation to remove $N!$ .

\ln Q = N\ln q - \ln N! \approx N\ln q - N\ln N + N.

This step is where indistinguishability enters explicitly: without dividing by $N!$ , the $-\ln N! \approx -N\ln N + N$ term would be absent, and $S$ would not scale correctly with system size (the Gibbs paradox).

Step 4: Get the internal energy. Translational motion in three dimensions contributes, by equipartition (confirmed by $q \propto T^{3/2}$ ),

U = \frac{3}{2}NkT, \qquad \frac{U}{T} = \frac{3}{2}Nk.

Step 5: Combine and simplify.

S = \frac{3}{2}Nk + Nk\ln q - Nk\ln N + Nk = Nk\left[\ln\!\left(\frac{q}{N}\right) + \frac{5}{2}\right].

Substituting $q = V\Lambda^{-3}$ and setting $N = N_A$ , $Nk = R$ , $V = V_m$ gives the molar form,

S_m = R\left[\ln\!\left(\frac{V_m}{N_A \Lambda^3}\right) + \frac{5}{2}\right].

Step 6: Eliminate $V_m$ in favor of pressure. Using $pV_m = RT$ , so $V_m/N_A = kT/p$ ,

\boxed{S_m = R\left[\ln\!\left(\frac{kT}{p}\left(\frac{2\pi m k T}{h^2}\right)^{3/2}\right) + \frac{5}{2}\right]}

This is the Sackur-Tetrode equation. Every quantity on the right is measurable or fundamental: no fitted constants appear anywhere.

Worked examples

Example 1: Molar entropy of neon gas at standard conditions. Find $S_m$ for neon at $T = 298.15\ \text{K}$ , $p = 1.00\times10^{5}\ \text{Pa}$ (standard pressure, IUPAC convention). Mass of one neon atom:

m = \frac{M}{N_A} = \frac{20.18\times10^{-3}\ \text{kg mol}^{-1}}{6.022\times10^{23}\ \text{mol}^{-1}} = 3.351\times10^{-26}\ \text{kg}.

Compute the thermal wavelength quantity $2\pi m k T/h^2$ :

2\pi m k T = 2\pi (3.351\times10^{-26})(1.381\times10^{-23})(298.15) = 8.680\times10^{-46}\ \text{J kg}.

h^2 = (6.626\times10^{-34})^2 = 4.390\times10^{-67}\ \text{J}^2\text{s}^2.

\frac{2\pi m k T}{h^2} = \frac{8.680\times10^{-46}}{4.390\times10^{-67}} = 1.977\times10^{21}\ \text{m}^{-2}.

Raise to the 3/2 power:

(1.977\times10^{21})^{3/2} = 8.79\times10^{31}\ \text{m}^{-3}.

Compute $kT/p$ :

\frac{kT}{p} = \frac{(1.381\times10^{-23})(298.15)}{1.00\times10^{5}} = 4.117\times10^{-26}\ \text{m}^3.

Multiply:

\frac{kT}{p}\left(\frac{2\pi mkT}{h^2}\right)^{3/2} = (4.117\times10^{-26})(8.79\times10^{31}) = 3.619\times10^{6}.

Take the logarithm and add 5/2:

\ln(3.619\times10^{6}) = 15.102, \qquad 15.102 + 2.500 = 17.602.

Multiply by $R$ :

S_m = (8.314\ \text{J K}^{-1}\text{mol}^{-1})(17.602) = 146.3\ \text{J K}^{-1}\text{mol}^{-1}.

This compares well with the tabulated third-law value for neon, $S_m^\circ \approx 146.3\ \text{J K}^{-1}\text{mol}^{-1}$ , an excellent check on the theory.

Example 2: How much does entropy change on isothermal expansion, and what temperature makes argon's entropy equal neon's at the same pressure? Part (a). Using the general $\ln(V_m)$ dependence, expanding one mole of neon isothermally at 298.15 K from $p_1 = 1.00\times10^5$ Pa to $p_2 = 0.500\times10^5$ Pa (doubling the volume) changes entropy by

\Delta S_m = R\ln\!\left(\frac{p_1}{p_2}\right) = (8.314)\ln(2.00) = (8.314)(0.6931) = 5.76\ \text{J K}^{-1}\text{mol}^{-1}.

This matches the standard ideal-gas isothermal expansion formula exactly, since only the $\ln V_m$ term inside Sackur-Tetrode depends on $p$ at fixed $T$ ; the check confirms internal consistency between statistical and classical thermodynamics. Part (b). Argon has $M_{\text{Ar}} = 39.95\ \text{g mol}^{-1}$ , roughly double neon's. At what temperature does argon's molar entropy at 1 bar equal neon's value at 298.15 K, 146.3 J/(K·mol)? Write $S_m(T) = R[\ln C + \tfrac32\ln T + \tfrac52]$ where $C$ collects the $T$ -independent factors:

C = \frac{k}{p}\left(\frac{2\pi m k}{h^2}\right)^{3/2}.

For argon, $m_{\text{Ar}} = 39.95\times10^{-3}/6.022\times10^{23} = 6.634\times10^{-26}\ \text{kg}$ , exactly $1.980\times$ the neon mass. Since $C \propto m^{3/2}$ , argon's $C$ is $(1.980)^{3/2} = 2.787$ times neon's, at the same pressure. Set neon's value at 298.15 K equal to argon's value at unknown $T$ :

R\ln\!\left(C_{\text{Ne}}(298.15)^{3/2}\right) = R\ln\!\left(2.787\, C_{\text{Ne}}\, T^{3/2}\right)

(the $5/2$ and prefactors cancel identically), giving

(298.15)^{3/2} = 2.787\, T^{3/2} \quad\Rightarrow\quad T^{3/2} = \frac{(298.15)^{3/2}}{2.787}.

Compute $(298.15)^{3/2} = 5149.7$ , so $T^{3/2} = 5149.7/2.787 = 1847.7$ , giving

T = (1847.7)^{2/3} = 148.9\ \text{K}.

Sanity check: heavier atoms have smaller $\Lambda$ at fixed $T$ , hence larger $q$ and larger $S_m$ ; to match neon's entropy, argon must be cooled well below 298 K. A drop to about 149 K, roughly half of room temperature, is physically reasonable given the mass ratio of two.

Limiting cases & common traps

Situation	What happens	Why
$T \to 0$	Equation gives $S_m \to -\infty$	Unphysical; classical/continuum treatment fails, third law violated
Very low $p$ , fixed $T$	$S_m$ grows without bound	Formally correct for an ideal gas; entropy diverges as $V \to \infty$
Heavier atom, same $T,p$	$S_m$ increases	Smaller $\Lambda$ , more states per particle available
Polyatomic molecules	Formula incomplete	Must add rotational and vibrational partition function contributions
High density / low $T$	Equation overestimates or gives negative $S$	Signals breakdown of the classical (non-degenerate) approximation, $q/N \gg 1$ required

The chemist behind it

The equation is named for Otto Sackur and Hugo Tetrode, who independently derived it around 1911–1912, several years before modern quantum mechanics was fully developed. They combined the emerging idea of phase-space quantization, cells of size $h^3$ , with classical statistical mechanics to compute absolute gas entropies for the first time. Their work was an early, prescient use of Planck's constant far outside blackbody radiation, and it played a role in convincing physicists that quantization was a general feature of nature, not a special rule for cavity radiation. Sackur died young, in a laboratory accident in 1914; Tetrode's career touched both physics and, briefly, statistical thermodynamics of chemical systems before he shifted to work influenced by relativity.

Origins · 1 minOtto Sackur and how this was discovered

Otto Sackur

1880-1914 · German physical chemist who derived the absolute entropy formula for ideal gases using early quantum phase-space arguments

Ask the tutor: How does the Sackur-Tetrode result change once rotational and vibrational partition functions are added for a diatomic gas?

Why does the Sackur-Tetrode equation only apply to monatomic gases

The derivation assumes the only energy a particle carries is translational kinetic energy, captured entirely by $q_{\text{trans}}$ . Monatomic gases like argon or neon have no rotational or vibrational degrees of freedom near room temperature (electronic excitation is far too energetic to matter), so translation is the whole story. Polyatomic molecules need additional partition function factors for rotation and vibration, each contributing its own term to $S_m$ , layered on top of the translational term derived here.

Does the Sackur-Tetrode equation agree with experimentally measured (calorimetric) entropies

Yes, remarkably well for monatomic gases at ordinary temperatures. Third-law calorimetric entropies, obtained by integrating $C_p/T$ from near 0 K up to the temperature of interest, agree with Sackur-Tetrode predictions typically to within experimental error, often better than 1 J/(K·mol) for noble gases. This agreement was historically important evidence that quantized translational states and the $N!$ correction are physically real, not mathematical artifacts.

What breaks the equation at very low temperature

The derivation assumes the sum over translational energy levels can be replaced by an integral, valid when the thermal wavelength $\Lambda$ is much smaller than the mean interparticle spacing. As $T \to 0$ , $\Lambda$ grows and eventually becomes comparable to that spacing, the classical continuum approximation collapses, and quantum statistics, Bose-Einstein or Fermi-Dirac, take over. The equation's divergence to $-\infty$ as $T \to 0$ signals this breakdown rather than a real physical violation of the third law.

How is Sackur-Tetrode related to the translational partition function and entropy formulas covered earlier

It is the direct specialization of the general formula $S = U/T + k\ln Q$ to a monatomic ideal gas, using the specific translational partition function $q = V/\Lambda^3$ derived from particle-in-a-box quantization. Nothing new is assumed beyond those two building blocks, plus Stirling's approximation for $N!$ . Once you have $q_{\text{trans}}$ and the general statistical entropy formula, Sackur-Tetrode falls out algebraically.

Why does entropy increase with molar mass at fixed T and p

Heavier particles have a shorter thermal de Broglie wavelength at the same temperature, since $\Lambda \propto m^{-1/2}$ . A shorter $\Lambda$ means more translational quantum states fit within the thermally accessible energy range, so $q$ is larger. More accessible microstates per particle means higher entropy, which is why, for example, xenon has higher standard molar entropy than helium at the same $T$ and $p$ .