Ideal Gas Law and Sackur-Tetrode Entropy

Derivation

Ideal Gas Law and Sackur-Tetrode Entropy

D-239 Home PU-302 Threads matter · energy · chance Depends on The Partition Function and F = -kT ln Z, The Gibbs Paradox and Indistinguishability

Statement

For \(N\) identical, non-interacting point particles of mass \(m\) confined to a volume \(V\) in thermal contact with a reservoir at temperature \(T\), the canonical partition function with Gibbs \(1/N!\) counting is \(Z = \frac{1}{N!}\left(\frac{V}{\lambda^3}\right)^{N}\), where \(\lambda = h/\sqrt{2\pi m k_B T}\) is the thermal de Broglie wavelength. From \(F = -k_B T \ln Z\) this yields simultaneously the mechanical equation of state \(pV = N k_B T\) and the absolute (Sackur–Tetrode) entropy \(S = N k_B\!\left[\ln\!\left(\frac{V}{N}\left(\frac{2\pi m k_B T}{h^2}\right)^{3/2}\right) + \frac{5}{2}\right]\).

Why it matters

This is the point where statistical mechanics stops merely reproducing thermodynamics and starts predicting it. The equation of state \(pV = Nk_BT\) is recovered with no adjustable constants, but the deeper payoff is that the entropy comes out absolute: the argument of the logarithm carries a factor of \(h^{-3}\), so Planck's constant sets the size of the phase-space cell and fixes the additive constant that classical thermodynamics leaves undetermined.

The Sackur–Tetrode formula was the first quantitative confirmation that entropy has an absolute zero-point tied to a quantum of action, and it agrees with calorimetric vapour entropies of the noble gases to better than \(0.1\%\). The \(1/N!\) that makes the entropy extensive is the resolution of the Gibbs paradox, so this single calculation ties together indistinguishability, the third law, and the classical limit.

Assumptions

Non-interacting particles.If interparticle potential energy is retained the partition function no longer factorises into single-particle pieces; \(Z \neq z_1^N/N!\) and one obtains a virial expansion or a fully interacting fluid instead of the ideal gas. Structureless particles (translational degrees of freedom only).If the particles have internal structure, \(z_1\) acquires rotational, vibrational and electronic factors; the equation of state is unchanged but the entropy and heat capacity gain internal contributions absent from Sackur–Tetrode. Non-degenerate (dilute) limit \(n\lambda^3 \ll 1\).If the phase-space occupancy is not small the Gibbs \(1/N!\) is only the leading term of the exact symmetrised counting; quantum statistics (Bose or Fermi) take over and the entropy and pressure receive \(O(n\lambda^3)\) corrections, diverging from the classical result near degeneracy. Continuous single-particle spectrum (thermodynamic limit).If the box is small or \(T\) very low the sum over momentum states cannot be replaced by \(\int d^3p\,\frac{V}{h^3}\); the integral form of \(z_1\) fails and discrete level structure matters. Semiclassical phase-space measure \(d^3q\,d^3p/h^3\).Without the \(h^3\) per cell the partition function is dimensionally undefined and the entropy loses its absolute normalisation, reverting to the classical entropy-up-to-a-constant.

Derivation

\[ z_1 = \sum_{\text{states}} e^{-\beta \varepsilon} \;=\; \frac{1}{h^3}\int_V d^3q \int d^3p \; \exp\!\left(-\beta \frac{\mathbf{p}^2}{2m}\right) \]

Single-particle canonical sum, replacing the discrete sum over quantum states by the semiclassical phase-space integral with measure \(d^3q\,d^3p/h^3\); valid because level spacing \(\ll k_B T\). B

\[ \int_V d^3q = V, \qquad \int_{-\infty}^{\infty} e^{-\beta p_x^2/2m}\,dp_x = \sqrt{\frac{2\pi m}{\beta}} \]

The kinetic energy has no spatial dependence, so the configuration integral gives \(V\); each Cartesian momentum integral is a Gaussian. A

\[ z_1 = \frac{V}{h^3}\left(\frac{2\pi m}{\beta}\right)^{3/2} = \frac{V}{h^3}\left(2\pi m k_B T\right)^{3/2} \equiv \frac{V}{\lambda^3}, \qquad \lambda \equiv \frac{h}{\sqrt{2\pi m k_B T}} \]

Collecting the three momentum integrals and defining the thermal de Broglie wavelength \(\lambda\); the combination \(V/\lambda^3\) counts the number of thermally accessible translational cells. A

\[ Z = \frac{1}{N!}\, z_1^{\,N} = \frac{1}{N!}\left(\frac{V}{\lambda^3}\right)^{N} \]

The particles are non-interacting so the \(N\)-body Hamiltonian is a sum of single-particle terms and \(Z\) factorises; the Gibbs \(1/N!\) corrects for the indistinguishability overcounting (prior result: Gibbs paradox and indistinguishability). B

\[ \ln Z = N\ln\!\left(\frac{V}{\lambda^3}\right) - \ln N! \;\approx\; N\ln\!\left(\frac{V}{\lambda^3}\right) - N\ln N + N \]

Take the logarithm and apply Stirling's approximation \(\ln N! \approx N\ln N - N\) (valid for \(N \gg 1\); the retained \(+N\) is what produces the \(5/2\)). B

\[ \ln Z = N\left[\ln\!\left(\frac{V}{N\lambda^3}\right) + 1\right] \]

Combine \(N\ln(V/\lambda^3) - N\ln N = N\ln\!\big(V/(N\lambda^3)\big)\); the result is manifestly of the form \(N\times(\text{intensive})\), i.e. extensive — the analytic signature that \(1/N!\) has cured the Gibbs paradox. C

\[ F = -k_B T \ln Z = -N k_B T\left[\ln\!\left(\frac{V}{N\lambda^3}\right) + 1\right] \]

Definition of the Helmholtz free energy \(F = -k_B T \ln Z\) (prior result: canonical partition function and free energy). A

\[ p = -\left(\frac{\partial F}{\partial V}\right)_{T,N} = N k_B T \,\frac{\partial}{\partial V}\ln V = \frac{N k_B T}{V} \]

Mechanical equation of state; only the \(\ln V\) term depends on \(V\) (with \(\lambda\) fixed at constant \(T\)), giving \(\partial_V \ln V = 1/V\). A

\[ F = -N k_B T\left[\ln\!\left(\frac{V}{N}\Big(\tfrac{2\pi m k_B}{h^2}\Big)^{3/2}\right) + \tfrac{3}{2}\ln T + 1\right] \]

Rewrite \(\lambda^{-3} = (2\pi m k_B T/h^2)^{3/2}\) and isolate the temperature dependence, so that \(\ln(V/N\lambda^3) = \text{const} + \tfrac{3}{2}\ln T\); this exposes every place \(T\) enters before differentiating. B

\[ S = -\left(\frac{\partial F}{\partial T}\right)_{V,N} = N k_B\left[\ln\!\left(\frac{V}{N\lambda^3}\right) + 1\right] + N k_B T\cdot\frac{3}{2T} \]

Differentiate \(F = -Nk_BT\,g(T)\) by the product rule; the explicit \(-Nk_BT\) factor gives back \(Nk_B\,g(T)\), and the \(\tfrac{3}{2}\ln T\) inside \(g\) contributes \(-Nk_BT\cdot\tfrac{3}{2T}\) with the outer minus sign flipping it to \(+\). C

\[ S = N k_B\left[\ln\!\left(\frac{V}{N\lambda^3}\right) + \frac{5}{2}\right] = N k_B\left[\ln\!\left(\frac{V}{N}\left(\frac{2\pi m k_B T}{h^2}\right)^{3/2}\right) + \frac{5}{2}\right] \]

Collect the constant: \(1 + \tfrac{3}{2} = \tfrac{5}{2}\), and substitute \(\lambda^{-3}\) explicitly. This is the Sackur–Tetrode equation. A

\[ U = -\frac{\partial \ln Z}{\partial \beta} = \frac{3N}{2\beta} = \frac{3}{2}N k_B T \;\Rightarrow\; k_B T = \frac{2U}{3N},\quad S = N k_B\!\left[\ln\!\left(\frac{V}{N}\Big(\tfrac{4\pi m U}{3N h^2}\Big)^{3/2}\right)+\frac{5}{2}\right] \]

Cross-check via the internal energy: \(U = -\partial_\beta \ln Z\) gives the equipartition value \(\tfrac{3}{2}Nk_BT\); eliminating \(T\) in favour of \(U\) recovers the microcanonical form of Sackur–Tetrode, confirming ensemble equivalence. C

Result

\[ pV = N k_B T, \qquad S = N k_B\left[\ln\!\left(\frac{V}{N}\left(\frac{2\pi m k_B T}{h^2}\right)^{3/2}\right) + \frac{5}{2}\right] \]

Reading. The classical ideal-gas law and the absolute entropy of a monatomic gas fall out of one partition function. The entropy per particle is \(k_B\) times the logarithm of the number of thermal cells available to each particle, \(V/(N\lambda^3) = 1/(n\lambda^3)\), plus a fixed \(\tfrac{5}{2}\). The pressure knows nothing of \(h\) or \(m\); the entropy knows both, because counting states requires the quantum measure \(h^{-3}\) and the thermal wavelength shrinks with \(\sqrt{m}\).

Units check. \(\lambda = h/\sqrt{2\pi m k_B T}\) has units \(\mathrm{J\,s}/\sqrt{\mathrm{kg}\cdot \mathrm{J}} = \sqrt{\mathrm{J\,s^2/kg}} = \mathrm{m}\), so \(V/\lambda^3\) is dimensionless and the logarithm is well defined. \(pV\): \(\mathrm{Pa\cdot m^3 = J = N k_B T}\). \(S\): \(N k_B\) carries \(\mathrm{J\,K^{-1}}\) and the bracket is dimensionless. \(\checkmark\)

Limiting cases

Isothermal expansion \(V \to 2V\): \(\Delta S = N k_B \ln 2\), independent of \(m\) and \(T\) — the classical mixing/expansion entropy, recovered exactly.
High \(T\) or low \(n\): \(n\lambda^3 \to 0\), \(S/Nk_B \to \infty\) logarithmically; the gas becomes ever more classical and the phase-space occupancy vanishes.
Approach to degeneracy \(n\lambda^3 \to 1\): the bracket \(\to \tfrac{5}{2}\) and \(S \to \tfrac{5}{2}Nk_B\); the formula predicts its own breakdown, since below this the argument of the log would drop under 1 and \(S\) would fall through zero — impossible classically, the cue for quantum statistics.
Heat capacity: \(C_V = (\partial U/\partial T)_V = \tfrac{3}{2}N k_B\), constant, from \(U = \tfrac{3}{2}Nk_BT\); \(C_p = C_V + Nk_B = \tfrac{5}{2}Nk_B\), so \(\gamma = 5/3\).

Breaks when

Quantum degeneracy, \(n\lambda^3 \gtrsim 1\). At low \(T\) or high density the \(1/N!\) is no longer the correct symmetry factor; multiple occupancy of momentum cells matters and one must use Bose–Einstein or Fermi–Dirac statistics. Sackur–Tetrode then predicts a negative entropy, a manifest failure — as in liquid helium, conduction electrons, or a trapped BEC near condensation.
Interacting gas. Once the interparticle potential is non-negligible (dense gas, near a phase transition), \(Z\) no longer factorises as \(z_1^N/N!\); real-gas corrections enter through the virial expansion \(pV/Nk_BT = 1 + B_2(T)n + \dots\) and the entropy acquires configurational contributions.
Polyatomic / internal structure. For molecular gases the rotational and vibrational modes contribute to \(U\), \(C_V\) and \(S\); \(\gamma \neq 5/3\) and the \(\tfrac{3}{2}Nk_BT\) energy and the \(\tfrac{5}{2}\) constant no longer hold, though \(pV=Nk_BT\) survives.
Relativistic or massless particles. When \(k_B T \gtrsim mc^2\) the dispersion \(\varepsilon = p^2/2m\) is wrong; the Gaussian momentum integral must be replaced (e.g. \(z_1 \propto V T^3\) for photons/ultrarelativistic gas) and both the energy and entropy scalings change.

Failure modes

Dropping the \(1/N!\). Using \(Z = z_1^N\) gives \(S = Nk_B[\ln(V/\lambda^3)+\tfrac{3}{2}]\), which is non-extensive and produces a spurious entropy of mixing for identical gases — the Gibbs paradox.
Forgetting the Stirling \(+N\). Approximating \(\ln N! \approx N\ln N\) only drops the \(+N\), turning the physical \(\tfrac{5}{2}\) into \(\tfrac{3}{2}\); the entropy is then wrong by \(Nk_B\).
Differentiating \(S\) with \(\lambda\) held fixed in \(T\). Treating \(\lambda\) as \(T\)-independent when computing \((\partial F/\partial T)\) loses the \(+\tfrac{3}{2}Nk_B\) term; \(\lambda \propto T^{-1/2}\) must be differentiated.
Confusing \(V/N\) with \(V\) inside the log. Writing \(\ln(V(\dots)^{3/2})\) instead of \(\ln((V/N)(\dots)^{3/2})\) breaks extensivity again — the \(N\) inside the logarithm is essential.
Using \(h\) vs \(\hbar\) inconsistently. \(\lambda = h/\sqrt{2\pi m k_B T}\) uses \(h\); slipping in \(\hbar\) shifts \(S\) by \(3Nk_B\ln(2\pi)\), a large systematic error in the absolute entropy.
Applying it to a diatomic gas. Using \(\tfrac{5}{2}\) and \(C_V=\tfrac{3}{2}Nk_B\) for \(\mathrm{N_2}\) or \(\mathrm{O_2}\) ignores rotational modes (\(C_V=\tfrac{5}{2}Nk_B\) at room temperature).

Discussion

The physical content sits entirely in the dimensionless combination \(n\lambda^3\), the degeneracy parameter. It is the ratio of the quantum volume a particle "occupies" (a cube of side \(\lambda\)) to the volume actually available per particle \(1/n = V/N\). When \(n\lambda^3 \ll 1\) the wavepackets do not overlap, particles are effectively distinguishable by position, and the classical counting with a simple \(1/N!\) is exact. The Sackur–Tetrode entropy is nothing but \(k_B\ln(1/n\lambda^3)\) per particle plus \(\tfrac{5}{2}k_B\), so entropy literally measures how many thermal cells each particle has to roam in.

That the same free energy delivers both \(pV=Nk_BT\) and an absolute \(S\) is a first illustration of the power of the potentials: \(p = -(\partial F/\partial V)_T\) and \(S=-(\partial F/\partial T)_V\) are two derivatives of one function, so mechanical and thermal behaviour are not independent inputs but consequences of a single generating quantity. The appearance of \(h\) only in \(S\) and not in \(pV\) is the statement that thermodynamic forces are classical while the counting of states is irreducibly quantum.

Historically, Sackur (1911) and Tetrode (1912) used the vapour-pressure and entropy data of mercury and argon to measure \(h\) from thermodynamics, obtaining a value consistent with Planck's blackbody \(h\) — striking independent evidence, years before quantum mechanics, that a quantum of action underlies entropy. The formula's agreement with calorimetric entropies of the monatomic gases to \(\sim 0.1\%\) is one of the quieter triumphs of statistical mechanics.

More deeply, the extensivity restored by \(1/N!\) foreshadows quantum field theory's resolution of the same issue: identical particles are not "labelled objects miscounted," they are excitations of a single field, and the symmetric/antisymmetric structure of many-body states makes the \(N!\) exact only in the non-degenerate limit. The Gibbs factor is the classical shadow of second quantisation; the very fact that we must put it in by hand classically, yet it emerges automatically from Bose/Fermi state-counting, is a signpost that indistinguishability is a quantum, not a bookkeeping, phenomenon. Sackur–Tetrode is thus the leading term of an expansion whose next order, \(\pm\, n\lambda^3/2^{5/2}\), already encodes the statistics of the particle species.

Common misconceptions. The \(1/N!\) is not a small correction or an approximation — it is exact in the classical limit and its omission changes the physics qualitatively (extensivity). It is also not "because the particles are moving"; even a static configuration of identical particles requires it. And Sackur–Tetrode is not "the classical entropy": it is a semiclassical result, quantum through the \(h^{-3}\) measure, which is precisely why it can be an absolute entropy while purely classical statistical mechanics cannot.

Worked examples

Example 1 — Thermal wavelength and validity of the classical limit for argon at STP.

\[ \lambda = \frac{h}{\sqrt{2\pi m k_B T}}, \qquad m = 39.95\,\mathrm{u} = 39.95\times 1.6605\times10^{-27}\,\mathrm{kg} = 6.634\times10^{-26}\,\mathrm{kg} \]

Set up the thermal de Broglie wavelength; convert the atomic mass of argon to kg. A

\[ 2\pi m k_B T = 2\pi (6.634\times10^{-26})(1.381\times10^{-23})(298\,\mathrm{K}) = 1.716\times10^{-45}\,\mathrm{kg\,J} \]

Insert \(k_B = 1.381\times10^{-23}\,\mathrm{J\,K^{-1}}\), \(T = 298\,\mathrm{K}\). A

\[ \lambda = \frac{6.626\times10^{-34}}{\sqrt{1.716\times10^{-45}}} = \frac{6.626\times10^{-34}}{4.143\times10^{-23}} = 1.60\times10^{-11}\,\mathrm{m} \approx 16\,\mathrm{pm} \]

Evaluate; \(\lambda\) is far smaller than the interatomic spacing. A

\[ n = \frac{p}{k_B T} = \frac{1.013\times10^{5}}{(1.381\times10^{-23})(298)} = 2.46\times10^{25}\,\mathrm{m^{-3}}, \quad n\lambda^3 = 2.46\times10^{25}\,(1.60\times10^{-11})^3 \]

Compute the number density from the ideal-gas law and form the degeneracy parameter. A

\[ n\lambda^3 = 2.46\times10^{25}\times 4.10\times10^{-33} = 1.0\times10^{-7} \]

Combine. A

\[ \lambda \approx 16\,\mathrm{pm}, \qquad n\lambda^3 \approx 1\times10^{-7} \ll 1 \]

Reading. Each argon atom occupies \(\sim10^{7}\) thermal cells, so wavepackets are enormously non-overlapping and the classical/Gibbs counting is superb — quantum-statistical corrections are of order \(10^{-7}\).

Example 2 — Absolute molar entropy of argon gas (\(298.15\,\mathrm{K}\), \(1\,\mathrm{bar}\)) and comparison with experiment.

\[ \frac{S}{N k_B} = \ln\!\left(\frac{V}{N}\left(\frac{2\pi m k_B T}{h^2}\right)^{3/2}\right) + \frac{5}{2}, \qquad \frac{V}{N} = \frac{k_B T}{p} \]

Sackur–Tetrode with volume per particle from the ideal-gas law. A

\[ \frac{V}{N} = \frac{(1.381\times10^{-23})(298.15)}{1.000\times10^{5}} = 4.117\times10^{-26}\,\mathrm{m^3} \]

Evaluate at \(p = 1\,\mathrm{bar} = 10^5\,\mathrm{Pa}\). A

\[ \frac{2\pi m k_B T}{h^2} = \frac{1.716\times10^{-45}}{(6.626\times10^{-34})^2} = \frac{1.716\times10^{-45}}{4.390\times10^{-67}} = 3.909\times10^{21}\,\mathrm{m^{-2}} \]

Reuse \(2\pi m k_B T\) from Example 1 (essentially the same \(T\)); divide by \(h^2\). A

\[ \left(3.909\times10^{21}\right)^{3/2} = 2.444\times10^{32}\,\mathrm{m^{-3}}, \qquad \frac{V}{N}\left(\dots\right)^{3/2} = 4.117\times10^{-26}\times 2.444\times10^{32} = 1.006\times10^{7} \]

Raise to the \(3/2\) power and multiply; the product is the inverse phase-space occupancy \(1/n\lambda^3\). A

\[ \frac{S}{N k_B} = \ln(1.006\times10^{7}) + 2.5 = 16.12 + 2.50 = 18.62 \]

Take the logarithm and add \(\tfrac{5}{2}\). A

\[ S_m = R\times 18.62 = (8.314\,\mathrm{J\,mol^{-1}K^{-1}})(18.62) = 154.8\,\mathrm{J\,mol^{-1}K^{-1}} \]

Multiply by \(N_A k_B = R\) to get the molar entropy. A

\[ S_m(\text{Ar}) \approx 154.8\,\mathrm{J\,mol^{-1}K^{-1}} \]

Reading. The experimental (calorimetric) standard molar entropy of argon is \(154.8\,\mathrm{J\,mol^{-1}K^{-1}}\). Sackur–Tetrode, with no fitted parameters, reproduces it to the quoted precision — a direct thermodynamic verification that entropy is absolute and quantised through \(h\).

Problems

Show that the internal energy \(U = -\partial \ln Z/\partial\beta\) gives \(U = \tfrac{3}{2}Nk_BT\), and hence \(C_V = \tfrac{3}{2}Nk_B\). What is \(\gamma = C_p/C_V\)?
Solution
With \(\ln Z = N\ln(V/\lambda^3) - \ln N!\) and \(\lambda^{-3} = (2\pi m k_B T/h^2)^{3/2} \propto \beta^{-3/2}\), only the \(\lambda\) term carries \(\beta\): \(\ln Z = \text{const} - \tfrac{3}{2}N\ln\beta\). Then \(U = -\partial_\beta \ln Z = \tfrac{3}{2}N/\beta = \tfrac{3}{2}Nk_BT\). So \(C_V = (\partial U/\partial T)_V = \tfrac{3}{2}Nk_B\). Using \(C_p = C_V + Nk_B = \tfrac{5}{2}Nk_B\), \(\gamma = \tfrac{5/2}{3/2} = 5/3 \approx 1.67\).
Two identical samples of the same ideal gas, each \(N\) particles in volume \(V\) at temperature \(T\), are separated by a partition which is then removed. Using \(S = Nk_B[\ln(V/N\lambda^3) + \tfrac{5}{2}]\), show the total entropy is unchanged, and contrast with the (wrong) result obtained if the \(1/N!\) is dropped.
Solution
Before: two systems each with \(S_1 = Nk_B[\ln(V/N\lambda^3)+\tfrac52]\), total \(2S_1\). After removing the partition: \(2N\) particles in \(2V\), so \(S_{\text{tot}} = 2N k_B[\ln(2V/2N\lambda^3)+\tfrac52] = 2Nk_B[\ln(V/N\lambda^3)+\tfrac52] = 2S_1\). \(\Delta S = 0\), correct — mixing identical gases produces no entropy. Without \(1/N!\), \(S' = Nk_B[\ln(V/\lambda^3)+\tfrac32]\); before \(=2Nk_B[\ln(V/\lambda^3)+\tfrac32]\), after \(=2Nk_B[\ln(2V/\lambda^3)+\tfrac32]\), giving a spurious \(\Delta S = 2Nk_B\ln 2 > 0\) — the Gibbs paradox.
Estimate the temperature at which \(^{87}\mathrm{Rb}\) (\(m = 1.44\times10^{-25}\,\mathrm{kg}\)) at number density \(n = 1\times10^{14}\,\mathrm{cm^{-3}}\) reaches \(n\lambda^3 = 1\), the onset of quantum degeneracy where Sackur–Tetrode fails.
Solution
\(n = 1\times10^{14}\,\mathrm{cm^{-3}} = 1\times10^{20}\,\mathrm{m^{-3}}\). Condition \(n\lambda^3 = 1 \Rightarrow \lambda = n^{-1/3} = (10^{20})^{-1/3} = 2.15\times10^{-7}\,\mathrm{m}\). From \(\lambda = h/\sqrt{2\pi m k_B T}\), \(T = \dfrac{h^2}{2\pi m k_B \lambda^2} = \dfrac{(6.626\times10^{-34})^2}{2\pi(1.44\times10^{-25})(1.381\times10^{-23})(2.15\times10^{-7})^2}\). Numerator \(=4.39\times10^{-67}\). Denominator \(=2\pi(1.44\times10^{-25})(1.381\times10^{-23})(4.62\times10^{-14}) = 5.77\times10^{-61}\). \(T = 7.6\times10^{-7}\,\mathrm{K} \approx 0.8\,\mu\mathrm{K}\) — the microkelvin scale of Bose–Einstein condensation, as expected.
For neon (\(m = 20.18\,\mathrm{u}\)) at \(273.15\,\mathrm{K}\), \(1\,\mathrm{bar}\), compute the standard molar entropy from Sackur–Tetrode.
Solution
\(m = 20.18\times1.6605\times10^{-27} = 3.351\times10^{-26}\,\mathrm{kg}\). \(V/N = k_BT/p = (1.381\times10^{-23})(273.15)/10^5 = 3.772\times10^{-26}\,\mathrm{m^3}\). \(2\pi m k_BT = 2\pi(3.351\times10^{-26})(1.381\times10^{-23})(273.15) = 7.943\times10^{-46}\). Divide by \(h^2 = 4.390\times10^{-67}\): \(1.809\times10^{21}\,\mathrm{m^{-2}}\); to the \(3/2\): \((1.809\times10^{21})^{3/2} = 7.70\times10^{31}\,\mathrm{m^{-3}}\). Product with \(V/N\): \(3.772\times10^{-26}\times7.70\times10^{31} = 2.904\times10^{6}\). \(\ln = 14.88\); \(+2.5 = 17.38\). \(S_m = 8.314\times17.38 = 144.5\,\mathrm{J\,mol^{-1}K^{-1}}\) (experimental \(146.3\,\mathrm{J\,mol^{-1}K^{-1}}\) at \(298\,\mathrm{K}\); the small difference is the temperature offset).
Show that at fixed \(T\) and \(N\), doubling the pressure decreases the entropy by \(Nk_B\ln 2\). Interpret physically.
Solution
At fixed \(T,N\), Sackur–Tetrode depends on \(p\) only through \(V/N = k_BT/p\): \(S = Nk_B[\ln(k_BT/p) + \tfrac32\ln(2\pi m k_BT/h^2) + \tfrac52]\). Only the \(\ln(1/p)\) term varies, so \(S(p) = -Nk_B\ln p + \text{const}\). Then \(\Delta S = S(2p) - S(p) = -Nk_B\ln 2\). Physically, doubling the pressure at constant \(T\) halves the volume per particle, halving the number of accessible thermal cells \(V/(N\lambda^3)\), so each particle loses \(k_B\ln 2\) of entropy — the same \(k_B\ln 2\) per particle as isothermal compression to half volume.