Sackur-Tetrode Equation: Absolute Quantum Molar Entropy of Monatomic Gases

How Hugo Tetrode (age 17!) and Otto Sackur put Planck’s constant h inside ideal-gas thermodynamics!

Classical thermodynamics could only measure entropy CHANGES ΔS—never absolute entropy—and suffered Gibbs’ paradox until Otto Sackur and 17-year-old Hugo Tetrode (1912) quantized phase space into cells of volume h^3 and divided by N! indistinguishable-particle permutations! Their exact quantum statistical formula S_m = R * (LN((k_B * T / P) * (2 * π * m * k_B * T / h^2)^1.5) + 2.5) predicts NIST JANAF calorimetric entropies of monatomic noble gases to 4+ significant figures with ZERO adjustable parameters!

Collecting CODATA constants (k_B, h, N_A) for molar mass M in g/mol, temperature T in K, and pressure P in bar (10^5 Pa), the constant LN(C_ST) + 2.5 equals -3.65171 + 2.5 = -1.15171: S_m = 8.31446 * (1.5 * LN(M) + 2.5 * LN(T) - LN(P_bar) - 1.15171) J/(mol·K)! Check standard state 298.15 K and 1.00 bar: Argon (M = 39.948 g/mol) evaluates to 154.85 J/(mol·K) (NIST: 154.84) and lighter Helium (M = 4.0026 g/mol, larger thermal de Broglie wavelength) evaluates to 126.15 J/(mol·K) (NIST: 126.15)!

Sackur-Tetrode Absolute Molar Entropy S_m (J/(mol·K))

Evaluates 8.31446 * (1.5 * LN(M_g_mol) + 2.5 * LN(T_K) - LN(P_bar) - 1.15171) from molar mass B2 (g/mol), temperature B3 (K), and pressure B4 (bar).

=IF(OR(B2 <= 0, B3 <= 0, B4 <= 0), "Invalid Thermodynamic State", ROUND(8.31446 * (1.5 * LN(B2) + 2.5 * LN(B3) - LN(B4) - 1.15171), 2))

Dimensionless Reduced Entropy S_m / R

Returns S_m / R directly in Boltzmann/gas-constant units.

=IF(OR(B2 <= 0, B3 <= 0, B4 <= 0), "Invalid Thermodynamic State", ROUND(1.5 * LN(B2) + 2.5 * LN(B3) - LN(B4) - 1.15171, 4))

Further reading

• Microsoft Support — LN function