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For entropy gain from impact and thermal equilibration, let the large lake be a reservoir at an initially unspecified temperature $T_L$, in kelvin, and take the final meteorite temperature to be $T_L$. Its initial temperature is $T_i=1273.15\mathrm{K}$. The meteorite loses thermal energy $mc(T_i-T_L)$, and its kinetic energy $m{v}^2/2$ is also dissipated into the lake. ThusandBoth heat equilibration and dissipation contribute positively. For the requested estimate, assume a lake temperature $T_L=300\mathrm{K}$. This gives $\Delta S_{\mathrm{universe}}\simeq 2.93\times 10^3\mathrm{J}{\mathrm{K}}^{-1}$. Taking $273$–$300\mathrm{K}$ instead gives about $3.3$–$2.9\mathrm{kJ}{\mathrm{K}}^{-1}$; this shows the estimate's sensitivity without asserting an unprovided lake temperature. The boxed expression retains the full temperature dependence. The model uses the stated constant heat capacity, no phase change of the meteorite, and a lake large enough that its temperature change is negligible.^[ref]

The Second law of thermodynamics restricts which energy-conserving processes can occur. In the Kelvin-Planck statement of the second law, no cyclic engine can have as its sole effect the extraction of heat from one reservoir and its complete conversion into work. In the Clausius statement of the second law, no process can have as its sole effect transferring heat from a colder reservoir to a hotter one. An engine or refrigerator combined with the reverse of a reversible device shows that a violation of either statement would violate the other.

A heat engine takes heat $Q_h>0$ from a reservoir at $T_h$, rejects $Q_c>0$ to one at $T_c<T_h$, and returns its working substance to the initial state. Energy conservation gives $W=Q_h-Q_c$ and efficiency $\eta =W/Q_h=1-Q_c/Q_h$. Carnot's theorem states that no engine between these reservoirs is more efficient than a reversible one, and that all reversible engines between them have the same efficiency. Otherwise, drive a reversible engine backwards with the work of a more efficient engine; the composite cycle transfers heat from cold to hot without external work, contradicting the Clausius statement.

A reversible Carnot cycle consists of two isotherms joined by two reversible adiabats. For an ideal gas, $Q_h=N{k}_B{T}_h\log (V_2/V_1)$ and $Q_c=N{k}_B{T}_c\log (V_3/V_4)$. The adiabats give $V_2/V_1=V_3/V_4$, soCarnot's theorem makes this independent of the working substance and defines the thermodynamic absolute-temperature scale consistently with ideal-gas temperature.^[ref]

For any reversible cycle, $\oint \delta Q_{\mathrm{rev}}/T=0$. This makes the integral between two equilibrium states independent of the reversible path, defining the state function entropy:Heat itself is path dependent; entropy is not. An irreversible change can be evaluated using any imagined reversible path between its equilibrium endpoints. For example, a fixed amount of ideal gas with constant $C_V$ has $\Delta S=C_V\log (T_B/T_A)+N{k}_B\log (V_B/V_A)$.

For a general cycle the Clausius inequality is $\oint \delta Q/T_{\mathrm{boundary}}\le 0$. The entropy balance for a process is $\Delta S\ge \int \delta Q/T_{\mathrm{boundary}}$, with equality for reversible heat exchange and no entropy production. The surroundings' entropy must be included, giving $\Delta S_{\mathrm{universe}}\ge 0$. A system's own entropy can decrease by exporting entropy; an isolated system's cannot.

The engine's reservoirs have net entropy change $Q_c/T_c-Q_h/T_h\ge 0$, sowith equality for a reversible engine. Friction, heat flow through a finite temperature difference and unrestrained expansion produce entropy and reduce usable work. Thus energy conservation fixes the balance of heat and work, while the second law fixes their direction and the attainable efficiency.^[ref]

For the ideal dilute Maxwell-Boltzmann gas implicit in a single-particle partition function, indistinguishability gives $Z_N=Z_1^N/N!$. With Stirling's approximation,Differentiate at fixed $T,V$:This is a thermodynamic-limit ideal-gas result; arbitrary interacting classical gases do not have $Z_N=Z_1^N/N!$.

Equilibrium constraints and the Gibbs minimum. The open-system equilibrium contacts allow exchange of heat, volume and matter. For unconstrained contacts without external-field or interface corrections, equilibrium requires equal temperatures, pressures and each exchangeable species' chemical potentials across the contacts. These conditions follow by maximizing total entropy under conserved total energy, volume and species amounts.

The next minimization statement fixes the system's particle number. For a system in a reservoir at $T_0,p_0$, with $N$ fixed and only pressure-volume work, the reservoir entropy change givesFor candidate states at the imposed $T=T_0$, $p=p_0$, the expression in parentheses is $G$. Thus the Fixed-particle-number Gibbs minimum selects stable equilibrium at fixed $T,p,N$, with stationary first variation and nonnegative allowed second variation. If particle exchange remains unconstrained, $N$ is not fixed and this minimization statement must be replaced by one incorporating the particle reservoir; it is not a justification for minimizing $G$ over arbitrary $N$.

The simple-compressible-system first law givesFor a homogeneous extensive single-component phase, the Single-component Gibbs Euler relation follows from $U=TS-pV+{\mu }_{\mathrm{ch}}N$, yielding $G={\mu }_{\mathrm{ch}}N$. For a mixture, the corresponding identities are $G={\sum }_i{\mu }_i{N}_i$ and $dG=-SdT+Vdp+{\sum }_i{\mu }_{i}d{N}_i$. Surface effects or other nonextensive contributions require additional terms.^[ref]

Reaction equilibrium. Let $d\xi$ be an allowed reaction extent for the forward process, so $d{N}_A=-d\xi$ and $d{N}_B=2d\xi$. At fixed $T,p$,For an interior equilibrium allowing both reaction directions, the coefficient vanishes:At fixed $T,V$, minimizing $F$ gives the same condition. The reaction formula-unit conservation gives $M=N_A+N_B/2$, not the total number $N_A+N_B$ of gas particles. This conservation law replaces fixed total particle number for the reacting mixture.

Using the stipulated single-particle functions,The number-based dissociation equilibrium ratio is thereforeThis number-based quantity has a factor of volume; the corresponding concentration ratio $n_B^2/n_A=K_N/V$ depends on temperature alone in the stated ideal model. A standard activity-based equilibrium constant would additionally specify reference states. No bond-energy factor is added to the functions given in the paper.

Define the dissociated fraction of original A formula units by $\alpha =N_B/(2M)$. Then $N_A=M(1-\alpha )$ and $N_B=2M\alpha$, soThe dissociation fraction versus conserved density follows because the left side increases monotonically from zero to infinity on $0<\alpha <1$. At fixed $V,T$, increasing $M$ decreases $q$, hence decreases dissociation. Equivalently, at fixed $M,T$ compression raises $M/V$ and favours A. Association produces fewer independent translating particles; at high density the translational-entropy advantage of the two-particle side is smaller, so equilibrium shifts toward the one-particle side.