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At fixed temperature and pressure, a closed reacting system reaches thermochemical equilibrium by minimizing Gibbs free energy over nonnegative species amounts subject to conserved elemental abundances. For an allowed stoichiometric vector $\nu$, an interior minimum has ${\sum }_i{\nu }_i{\mu }_i=0$. More generally the chemical potential vector lies in the row span of the elemental conservation matrix at positive concentrations, with the appropriate one-sided inequalities for absent species. The first law of thermodynamics gives the thermodynamic differential, while the Second law of thermodynamics supplies the minimization direction.

The instantaneous-decay approximation keeps the scale factor and physical volume fixed across the decay. Energy conservation and rapid thermalization giveMeanwhile the abundance relation gives $s_{-}={\rho }_{\varphi ,-}/(Ym)\simeq 3M_{\mathrm{Pl}}^2/(Ym{t}_d^2)$. Dividing yields the entropy-injection factorThere is a numerical error in the printed coefficient: part (c) omits the factor $1/3$. The denominator $s_{-}=3M_{\mathrm{Pl}}^2/(Ym{t}_d^2)$ supplies that factor, while the radiation energy contributes $3^{3/4}$ to the numerator. Thus the consistent coefficient is $c_1{3}^{-1/4}{c}_2^{-3/4}$, rather than $c_1{3}^{3/4}{c}_2^{-3/4}$. Both cannot follow from the stated part (b) using the same decay-time approximation.

Its increase is especially transparent without dropping the original radiation. Let $R={\rho }_{\varphi ,-}/{\rho }_{r,-}$. At unchanged $g_{\star }$, $T_{+}^4/T_{-}^4=1+R$, soRelic domination means $R\gg 1$, so the leading estimate is much larger than one. This is entropy production by decay of a dominant relic.

There is no conflict with part (a). The equilibrium gas before the decay was a separately conserved, reversibly expanding system. During decay it receives energy from the nonthermal relic, and the decay products thermalize irreversibly. Its cosmological perfect-fluid continuity equation now has an energy-injection source, so $d({\rho }_{r}V)+P_{r}dV$ does not vanish. The assumptions behind cosmological entropy conservation therefore fail during the event. Moreover the zero-chemical-potential equilibrium expression in part (a) cannot be imposed on the decoupled relic as though it were already part of that same equilibrium radiation bath. The Second law of thermodynamics permits, and here requires, the resulting entropy increase.

Let the closed atmospheric parcel exchange heat and volume work with reservoirs at fixed temperature $T$ and pressure $P$, while reactions rearrange its conserved elemental inventory. For a reversible differential between equilibrium states, the first law of thermodynamics, with chemical work included, isThe Gibbs free energy $G=U+PV-TS$ therefore obeysThe first law of thermodynamics supplies this differential but does not alone choose the direction of a reaction. The Second law of thermodynamics does: for the parcel plus its reservoirs, at fixed $T,P$ and with no non-volume work,Thus spontaneous reactions decrease Gibbs free energy, and stable thermochemical equilibrium is its constrained minimum. For an allowed reaction of extent $d\xi$, let $d{N}_i={\nu }_{i}d\xi$, with positive stoichiometric vector entries for products and negative entries for reactants. At an interior minimum,The condition must hold for every independent allowed reaction, with fixed total atoms of each element and any relevant charge constraint. At a boundary with absent species, only feasible one-sided variations are permitted. These are the hypotheses of constrained Gibbs minimization for chemical equilibrium.

For an ideal gas mixture, ${\mu }_i={\mu }_i^{\circ }(T)+\mathcal{R}T\log (x_{i}P/P^{\circ })$, where $x_i$ is the volume mixing ratio and $\mathcal{R}$ the molar gas constant. ThereforeThis is a dimensionless activity-based equilibrium constant, not a bare product of dimensional pressures.

For a perfect gas with constant specific heat capacity $c_v$, the first law of thermodynamics givesUsing the pressure-density Hugoniot relation for a perfect gas, the entropy production in a perfect-gas shock is thereforeThis is positive for $P>1$, as required by the Second law of thermodynamics for the physical compressive normal shock wave.

Let two systems exchange energy inside an otherwise isolated composite system, so $E_1+E_2=E$ is fixed. By the Second law of thermodynamics, equilibrium maximizes the total entropyAt an interior maximum,Thus $T_1=T_2$, which is entropy maximization under thermal contact and the condition for thermal equilibrium.

Stability requires this entropy maximum to be locally strict, so each ordinary subsystem has a concave entropy-energy relation. Sinceone has $\partial T/\partial E>0$, or equivalentlyThis is the entropy concavity and positive heat capacity criterion.

Now let $N_{\uparrow }=\alpha N$ and $N_{\downarrow }=(1-\alpha )N$. Because the spins are distinguishable, the number of microstates is the binomial coefficientThe Boltzmann entropy $S=k_B\log \Omega$ and the Stirling formula give, to leading order for large $N$,The total energy isThereforeSolving for the up-spin fraction gives the independent spin-one-half two-level system resultFor ordinary positive absolute temperature,with $\alpha \to 0$ as $T\to 0^{+}$ and $\alpha \to 1/2$ as $T\to +\infty$. Since the spectrum is bounded above, the population-inverted range $1/2<\alpha \le 1$ is also mathematically possible and corresponds to negative temperature; $\alpha =1/2$ is the infinite-temperature state.

Substitution givesand direct differentiation yieldsThus the energy increases with temperature throughout either finite-temperature branch, in particular throughout the required positive-temperature range.

The Clausius statement of the second law says that no cyclic device can have as its sole effect the transfer of heat from a colder reservoir to a hotter one. The Kelvin-Planck statement of the second law says that no cyclic device can have as its sole effect the extraction of heat from one reservoir and its complete conversion into work.

Suppose first that the Clausius statement fails. Use the forbidden device to return to a hot reservoir the heat $Q_C$ rejected by an ordinary heat engine. The combined device extracts $Q_H-Q_C$ from the hot reservoir and converts all of it into work, violating the Kelvin--Planck statement.

Conversely, suppose the Kelvin--Planck statement fails. Use the work produced by the forbidden one-reservoir engine to drive an ordinary refrigerator. The engine's heat withdrawal and the refrigerator's heat discharge cancel by the work-energy balance, leaving only a transfer of heat from the cold reservoir to the hot reservoir with no external work. This violates the Clausius statement. Each statement therefore implies the other, so the two forms of the Second law of thermodynamics are equivalent.