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 , an interior minimum has . 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.
Gibbs free energy of an ideal-gas mixture 2026-10-06
For a single-phase ideal gas mixture with molar amounts , common temperature , and pressure , extensivity and ideal mixing giveHere , is the mole fraction, is the molar gas constant, and is the standard pressure. Integrating and including the ideal mixing entropy proves . Reactions can change , so fractions alone do not set the extensive Gibbs free energy. Its composition Hessian is ; it is positive semidefinite by the weighted Cauchy-Schwarz inequality.
Past exam of the mathematics course of the University of Cambridge 2011 ii Paper 4 34D i Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2015 ii Paper 3 33C a i Solution Created 2026-09-24 Updated 2026-10-06
For a fixed amount of gas, the first law of thermodynamics gives . Define the Gibbs free energy , so . Equality of its mixed partial derivatives gives the Maxwell relation
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 303 3 a Solution Created 2026-10-03 Updated 2026-10-06
The statistical Hamiltonian in this question is already in thermal units, as indicated by without an additional . Define and . The mean field and its linear response areThese are functional derivatives of the connected generating functional; the subtraction defines the connected correlation function. In the Landau approximation, neglect loop corrections and evaluate the field integral at a stable saddle . It satisfies the Euler-Lagrange equationThis follows by varying the gradient term and integrating by parts, with periodic, decaying, or otherwise appropriate boundary conditions.
The two requested free energy functionals, following the source and Legendre conventions of the question, areIn the scalar-field source Legendre transform on the chosen stable branch, is chosen to produce , and . Thus the imposed-source Helmholtz free energy and the fixed-order-parameter Gibbs free energy have the appropriate opposite source derivatives. These names are used in the question's magnetic-ensemble convention; the defining sign relation is what fixes the calculation. At leading Landau approximation there is no fluctuation-determinant term in .
Differentiate the saddle equation with respect to . The response obeysThereforeEquivalently, the inverse Hessian relation for a connected two-point function states that is the inverse kernel of . The factor follows from twice differentiating the quartic term ; it is not . This tree-level connected response is obtained by varying the saddle. It does not require replacing the exact connected correlator by a product of the saddle values, which would incorrectly give zero.
For the requested single-momentum formula, assume a homogeneous source and a translationally invariant equilibrium phase, so and . SetThe Fourier transform with the printed positive sign sends to , while the Dirac delta function transforms to one. Hence the Ornstein--Zernike correlation function hasThe inverse convention is . For general inhomogeneous , has nonconstant coefficients and depends separately on its two positions; the displayed momentum-diagonal formula then does not follow. The preceding differential equation still holds in that case.
At zero source and for , the homogeneous saddle is for , and on either selected stable ordered branch for . Thus the Landau scalar correlation length isIn the ordered phase the negative bare quadratic coefficient is compensated by the positive curvature at the nonzero saddle. Retaining below the transition would instead give an unstable kernel and is not a physical correlation length. With the usual analytic thermal tuning , , both branches diverge as , so the correlation-length critical exponent isThe high-temperature amplitude is times the low-temperature amplitude for the same . This statement is within Landau theory; fluctuations can change critical behavior outside the mean-field regime.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 315 2 b ii Solution Created 2026-10-03 Updated 2026-10-06
Assume a single-phase ideal gas mixture in a closed parcel, with common temperature , total pressure , and molar species amounts . Define and volume mixing ratios , so . Ideal mixing makes partial pressure .
For a pure ideal species, , with molar volume . At fixed temperature, integration from standard pressure givesThe same logarithmic composition term follows from the ideal mixing entropy . Extensivity gives . Hence the Gibbs free energy of an ideal-gas mixture isEquivalently, write each standard chemical term as , using molar enthalpy and entropy. For zero abundance the limit defines the extensive expression continuously.
The fractions alone do not specify the extensive Gibbs free energy: the total amount or an equivalent elemental inventory must also be fixed. Moreover, in a chemically closed reacting system need not remain constant; for example, hydrogenation of carbon monoxide changes the total molecular count while conserving atoms. Minimization is therefore over species amounts subject to elemental constraints, rather than over fractions with an artificially fixed molecular count. Nonideal gases require activities or fugacities; condensates contribute separate phase terms.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 315 2 b i Solution Created 2026-10-03 Updated 2026-10-06
Let the closed atmospheric parcel exchange heat and volume work with reservoirs at fixed temperature and pressure , 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 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 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 , let , 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, , where is the volume mixing ratio and the molar gas constant. ThereforeThis is a dimensionless activity-based equilibrium constant, not a bare product of dimensional pressures.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 315 4 e Solution Created 2026-10-03 Updated 2026-10-06
The three parameter groups are local temperature, local pressure, and elemental composition. The elemental composition includes both atmospheric metallicity of a giant planet and atmospheric carbon-to-oxygen ratio; these are independent compositional controls, so a specification by and alone is incomplete. Given these inputs, thermochemical equilibrium minimizes Gibbs free energy with element conservation.
- The carbon monoxide-to-methane partition. For , cooler gas and, at fixed mole fractions, larger pressure favour the side with fewer molecules, whereas hotter gas generally favours CO. Thus relatively hot photospheres commonly show CO near and , while cooler conditions favour methane bands near . The exact boundary depends on composition and pressure; carbon monoxide–methane quenching can invalidate an equilibrium diagnosis.
- Water absorption, for example around and , traces the oxygen left after CO forms. In oxygen-rich gas with , water is often prominent; carbon-rich gas can have much less water and relatively more methane or hydrogen cyanide. Condensation and high-temperature dissociation are exceptions to a simple elemental bookkeeping argument.
- Carbon dioxide, with a strong band near , is sensitive to enrichment and oxidation chemistry. In a hydrogen-dominated trace-species regime, gives at fixed , so CO2 can grow roughly quadratically with heavy-element enrichment before the trace approximation fails.
CO/CH4, H2O, and CO2 bands provide complementary temperature and elemental-abundance diagnostics, but their observed strengths also depend on the atmospheric pressure-temperature profile, exoplanet clouds and observing geometry. A spectral feature is not by itself a direct equilibrium abundance.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 315 4 f Solution Created 2026-10-03 Updated 2026-10-06
Three routes away from local thermochemical equilibrium are:
- Vertical transport and quenching. When the eddy mixing time becomes shorter than the chemical relaxation time, gas retains a deeper composition. In cool, directly imaged giant exoplanets, carbon monoxide–methane quenching can preserve CO and suppress the methane expected in cool equilibrium layers. In Jupiter, CO carried from deeper hot layers, and phosphine maintained against upper-atmosphere chemical loss, are examples of transported disequilibrium species.
- Horizontal transport and quenching. Atmospheric advection time shorter than the local chemical relaxation time carries abundances between regions of different temperature. A hot Jupiter can carry CO-rich dayside gas into cooler nightside gas that would otherwise favour methane. In the Solar System, upper-atmosphere CO produced on the illuminated side of Venus can be redistributed to its dark side by circulation; this is an example of nonlocal production and transport, rather than local dark-side thermochemical equilibrium.
- Atmospheric photochemistry. Ultraviolet photons dissociate or ionize molecules and initiate reaction networks. Models of irradiated hydrogen-rich exoplanet atmospheres can produce enhanced hydrogen cyanide and hydrocarbon precursors from methane/nitrogen chemistry; whether these products accumulate depends on ultraviolet flux and transport. The terrestrial atmosphere gives a clear Solar-System example: oxygen photodissociation and subsequent reactions maintain the ozone layer, which is not a purely thermochemical-equilibrium abundance.
Vertical mixing, horizontal advection, and photochemistry supply three mechanisms and examples in both exoplanets and Solar-System atmospheres. A photochemical steady state balances production and loss; it is different from a Gibbs free energy minimum. These examples describe mechanisms and model expectations rather than asserting unique observational attribution for every planet.
Past exam of the mathematics course of the University of Cambridge 2021 iii Paper 315 4 e Solution 2026-09-28
At fixed temperature and pressure, a closed reacting system is in thermochemical equilibrium when its Gibbs free energy is minimal subject to elemental-abundance constraints. For every independent reaction,with a positive second variation in every allowed direction.
An atmosphere can be driven into disequilibrium chemistry in an exoplanet atmosphere by vertical mixing faster than chemical conversion, which causes chemical quenching; ultraviolet atmospheric photochemistry; and atmospheric escape. Lightning, energetic particles, horizontal transport, condensation, and rainout provide further mechanisms.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 315 2 d Solution 2026-09-28
At fixed temperature and pressure, thermochemical equilibrium minimizes the Gibbs free energy subject to elemental conservation. For every independent reaction with stoichiometric coefficients ,Equivalently, forward and reverse rates satisfy detailed balance. For ideal gases,
Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 4 35A c Solution Created 2026-09-24 Updated 2026-09-25
The fundamental thermodynamic relation isBecause is extensive, the Euler theorem for homogeneous functions givesSubstitution into the Gibbs free energy therefore yieldsFor a one-component equilibrium phase, the intensive chemical potential depends on the intensive state variables and .
Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 315 4 b Solution 2026-09-28
For , the forward and reverse rates areAt thermochemical equilibrium, by detailed balance, whileafter the appropriate standard-concentration factors are included. The standard reaction Gibbs free energy isConsequentlyFor other stoichiometries, the same argument uses the corresponding activity product and its standard-state factors.
Past exam of the mathematics course of the University of Cambridge 2023 ii Paper 2 37A e Solution Created 2026-09-24 Updated 2026-09-25
If a physical tension acts on a state of extension , its mechanical energy is . The canonical ensemble therefore assigns that state the Boltzmann factorComparison with the statistical weight givesEquivalently, is the thermodynamic conjugate variable to in the Gibbs free energy, just as pressure is conjugate to volume.
Past exam of the mathematics course of the University of Cambridge 2024 ii Paper 4 35B a Solution Created 2026-09-24 Updated 2026-09-25
At a first-order transition the Gibbs free energy is continuous but a first derivative such as entropy or volume jumps; there is latent heat. At a second-order transition the first derivatives are continuous while a second derivative, such as heat capacity or susceptibility, is discontinuous or divergent, and there is no latent heat.
Past exam of the mathematics course of the University of Cambridge 2025 ii Paper 4 35C a Solution Created 2026-09-24 Updated 2026-09-25
The latent heat for converting liquid to gas at coexistence iswith all quantities taken per mole or per particle consistently. Along the phase coexistence curve, the two Gibbs free energy values agree:Sincedifferentiating the equality along the curve givesTherefore the Clausius-Clapeyron relation isAt the critical point the liquid and gas become the same phase, so their entropy discontinuity disappears and