Indistinguishability, Gibbs Paradox, and Sackur-Tetrode
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You should be comfortable with the canonical partition function for an ideal gas and the definition of thermal wavelength \(\lambda_{\mathrm{th}}\). Familiarity with the concept of entropy as a state function is assumed.
Why this matters
The resolution of the Gibbs paradox reveals that entropy is fundamentally a measure of information about indistinguishable particles.
In classical statistical mechanics, treating particles as distinct leads to an entropy that is not extensive, and to a spurious entropy increase when two samples of the same gas are mixed. This inconsistency, known as the Gibbs paradox, forces a re-evaluation of what constitutes a "microstate." By correcting for the over-counting of indistinguishable permutations, we derive the Sackur-Tetrode equation, which provides an absolute scale for the entropy of an ideal monatomic gas. This result is not merely a mathematical fix; it is the first concrete instance where quantum mechanics (via the thermal wavelength) imposes a fundamental limit on classical thermodynamics, bridging the gap between molecular chaos and macroscopic observables.
The Problem of Counting
Imagine a box containing \(N\) gas particles. In a purely classical picture, if we label the particles 1, 2, ..., \(N\), then swapping particle 1 and particle 2 creates a new, distinct microstate. However, in reality, identical atoms are indistinguishable. Swapping two identical helium atoms does not change the physical state of the system; it is the same state. If we count every permutation as unique, we over-count the number of accessible microstates by a factor of \(N!\). This over-counting makes the entropy depend on how we choose to label the particles, which is physically nonsensical. The "paradox" arises because classical mechanics cannot naturally explain why this division by \(N!\) is necessary, yet thermodynamics demands it for entropy to be extensive (proportional to the amount of substance).
Correcting the Partition Function
To resolve the over-counting, we modify the canonical partition function \(Z_N\) for \(N\) indistinguishable particles. The partition function for \(N\) independent, distinguishable particles is simply the \(N\)-th power of the single-particle partition function \(z_1\). For indistinguishable particles, we divide by \(N!\) to account for the permutations that do not represent distinct physical states.
This correction is crucial. When we calculate the entropy using the relation \(S = k_B \ln \Omega\) or via the Helmholtz free energy \(F = -k_B T \ln Z_N\), this \(N!\) term introduces a \(-k_B \ln N!\) contribution. Using Stirling's approximation \(\ln N! \approx N \ln N - N\), this term becomes extensive, scaling linearly with \(N\), thereby restoring the additivity of entropy required by the second law.
The Sackur-Tetrode Equation
Applying the corrected partition function to a monatomic ideal gas yields the Sackur-Tetrode equation. The single-particle partition function for a particle in a volume \(V\) with thermal wavelength \(\lambda_{\mathrm{th}} = h / \sqrt{2 \pi m k_B T}\) is \(z_1 = V / \lambda_{\mathrm{th}}^3\). Substituting this into the expression for entropy, we obtain the absolute entropy of the gas.
This result is profound because it provides an absolute value for entropy, not just differences. The term \(\ln(V/N)\) represents the configurational entropy, while the \(\lambda_{\mathrm{th}}^3\) term reflects the momentum space volume accessible to the particles. The constant \(5/2\) arises from the combination of the translational degrees of freedom and the Stirling approximation correction. This equation is valid only when the gas is dilute, such that \(V/N \gg \lambda_{\mathrm{th}}^3\), ensuring the classical limit holds.
Extensivity and the Second Law
The primary physical consequence of the \(1/N!\) correction is the extensivity of entropy. If we double the volume \(V\) and the number of particles \(N\) simultaneously, the entropy \(S\) should also double. Without the \(N!\) correction, the entropy would contain a term proportional to \(N \ln N\), which is super-extensive. The Gibbs paradox highlights that classical mechanics, which treats particles as distinguishable points in phase space, is incomplete without a quantum mechanical input (the thermal wavelength) and a combinatorial correction (indistinguishability). This resolution confirms that the second law of thermodynamics is consistent with statistical mechanics only when we properly account for the indistinguishability of identical particles.
The derivation
Sackur-Tetrode Entropy and the Gibbs Paradox
Dividing the phase-space integral by N! and h^{3N} resolves the entropy-of-mixing paradox and yields the absolute entropy of a monatomic ideal gas.
The Sackur-Tetrode equation assumes a classical ideal gas. If the thermal wavelength \(\lambda_{\mathrm{th}}\) becomes comparable to the inter-particle spacing \((V/N)^{1/3}\), the gas becomes degenerate, and quantum statistics (Fermi-Dirac or Bose-Einstein) must replace the Maxwell-Boltzmann distribution. In this regime, the simple \(1/N!\) correction is insufficient, and the entropy calculation requires a full quantum statistical treatment. C
Worked example
Problem. Use the Sackur–Tetrode equation to find the molar entropy of argon (\(M = 0.040\ \mathrm{kg\,mol^{-1}}\)) at 300 K and \(1.0 \times 10^5\ \mathrm{Pa}\). Use \(k_B = 1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}}\), \(h = 6.63 \times 10^{-34}\ \mathrm{J\,s}\), \(N_A = 6.022 \times 10^{23}\ \mathrm{mol^{-1}}\).
- Molar volume: \(V_m = RT/P = 8.314 \times 300/1.0 \times 10^5 = 0.02494\ \mathrm{m^3}\).
- Atomic mass: \(m = 0.040/6.022 \times 10^{23} = 6.64 \times 10^{-26}\ \mathrm{kg}\).
- Thermal wavelength: \(\lambda_{\mathrm{th}} = \dfrac{h}{\sqrt{2\pi m k_BT}} = \dfrac{6.63 \times 10^{-34}}{\sqrt{2\pi \times 6.64 \times 10^{-26} \times 1.38 \times 10^{-23} \times 300}} = 1.595 \times 10^{-11}\ \mathrm{m}\), so \(\lambda_{\mathrm{th}}^3 = 4.06 \times 10^{-33}\ \mathrm{m^3}\).
- Argument of the logarithm: \(\dfrac{V_m}{N_A\lambda_{\mathrm{th}}^3} = \dfrac{0.02494}{6.022 \times 10^{23} \times 4.06 \times 10^{-33}} = 1.02 \times 10^{7}\), whose natural logarithm is 16.14.
- Entropy: \(S_m = R(16.14 + 2.5) = 8.314 \times 18.64 = 155\ \mathrm{J\,K^{-1}\,mol^{-1}}\), in excellent agreement with the calorimetric value for argon (about 155 J K-1 mol-1). The huge ratio \(V/N\lambda^3 \sim 10^7\) also confirms that the classical treatment is valid.
Units check. \(V_m/(N_A\lambda^3)\) is \(\mathrm{m^3\,mol^{-1}}/(\mathrm{mol^{-1}}\,\mathrm{m^3})\), dimensionless, so the logarithm is defined; \(R\) supplies the units of molar entropy.
Three quick tests
- Why does the entropy of an ideal gas become non-extensive if we do not divide the partition function by \(N!\)?
Answer
Without the \(1/N!\) factor, the entropy contains a term \(N \ln N\), which grows faster than linearly with \(N\), violating the additivity required for extensive quantities. - What is the physical significance of the thermal wavelength \(\lambda_{\mathrm{th}}\) in the Sackur-Tetrode equation?
Answer
It represents the quantum mechanical uncertainty in the position of a particle due to its thermal momentum, setting the scale for when classical distinguishability breaks down. - If the temperature of an ideal gas is doubled while keeping the volume and number of particles constant, how does the entropy change?
Answer
The entropy increases because \(\lambda_{\mathrm{th}}\) decreases as \(T^{-1/2}\), so \(\lambda_{\mathrm{th}}^3\) decreases as \(T^{-3/2}\), making the logarithmic argument larger.
Problem set for PU-203 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.