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At coexistence, the two phases have equal chemical potential. For a single component, $d\mu =-sdT+vdp$, with entropy and volume measured on the same molar or mass basis. Equating the phase derivatives along coexistence gives $\Delta vdp=\Delta sdT$. Reversible latent heat satisfies $L=T\Delta s$, henceHere $p$ and $T$ are coexistence pressure and absolute temperature, $L$ is latent heat for transforming phase 1 to phase 2, and $\Delta v=v_2-v_1$ uses the same amount of material as $L$.

For vaporization, neglect liquid molar volume compared with ideal-vapour volume $RT/p$. The ideal-vapour Clausius-Clapeyron approximation becomesUsing the normal boiling reference $T_b=373.15\mathrm{K}$, $p_b=1.01325\times 10^5\mathrm{Pa}$ and the supplied constant $L_e$ givesThis deliberately constant-latent-heat extrapolation does not reproduce the supplied triple-point pressure exactly. It predicts the liquid-vapour branch, metastable against ice below the triple-point temperature; it is not a precision sublimation-pressure prediction.

The phase diagram has solid, liquid and vapour regions meeting at the triple point. The sublimation and vaporization boundaries have positive slopes, whereas the ordinary ice–water boundary slopes negatively because ice has larger specific volume. The liquid–vapour line terminates at a critical point; the original schematic below emphasizes topology and marks the supplied triple point.

For the melting-pressure estimate, Archimedes' principle gives ${\rho }_i=0.90{\rho }_w=900\mathrm{kg}{\mathrm{m}}^{-3}$, soConvert the molar evaporation heat using water molar mass $M\simeq 0.0180\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, rounded from the tabulated value $18.0153\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref] An enthalpy cycle through vapour then givesThis uses the question's temperature-independent evaporation heat all the way to the triple point; no replacement experimental fusion heat is inserted. The melting pressure inferred from flotation and latent heats, on holding $L_f$ and $\Delta v$ constant, isusing $T_t=273.16\mathrm{K}$ and $T=270.15\mathrm{K}$. Linearizing the logarithm about the triple point gives $53.4\mathrm{MPa}$ instead of $53.7\mathrm{MPa}$. Both are estimates within the stated constant-property approximation; the negative melting slope makes a positive pressure necessary to lower the melting temperature.

For a simple compressible one-component equilibrium system, the fundamental thermodynamic relation isThe internal energy has natural variables $(S,V,N)$: its derivatives give $T$, $-p$ and chemical potential $\mu$. The natural variables of a thermodynamic potential identify the independent arguments whose conjugates appear directly in a potential's differential. A Legendre transform exchanges a variable for its conjugate, giving potentials suited to different controlled conditions:

For fixed composition and the appropriate closed-system constraints, equilibrium minimizes $U$ at fixed $S,V$, $H$ at fixed $S,p$, $F$ at fixed $T,V$, or $G$ at fixed $T,p$, with other allowable internal variations understood. Enthalpy is useful for constant-pressure heat exchange with only pressure–volume work. Helmholtz free energy is suited to fixed-temperature, fixed-volume systems, and Gibbs free energy to fixed-temperature, fixed-pressure phase or chemical equilibrium. Coexisting phases of one substance have equal chemical potentials.

Where a potential is twice continuously differentiable, equality of mixed derivatives gives Maxwell relations. At fixed $N$, the four usual forms areThe fixed subscripts and signs follow from the chosen natural variables. Derivatives through a nonsmooth phase boundary require separate phase descriptions rather than automatic use of the smooth single-phase formulas.

As an application, the Helmholtz relation turns pressure data into an isothermal entropy change. For an ideal gas, $(\partial p/\partial T{)}_V=nR/V$, givingCombining the same relation with $dU=TdS-pdV$ yields $(\partial U/\partial V{)}_T=T(\partial p/\partial T{)}_V-p$, which vanishes for an ideal gas and reveals attractive-energy effects for a van der Waals gas. Thermodynamic potentials organize equilibrium constraints, while Maxwell relations convert difficult entropy derivatives into measurable response derivatives.

A thermodynamic potential packages the equilibrium state in variables suited to the imposed constraints. For a simple fluid, $U(S,V,N)$ has differential $dU=TdS-pdV+\mu dN$. Legendre transforms give $H=U+pV$, $F=U-TS$ and $G=U+pV-TS$, with natural variables of a thermodynamic potential $(S,p,N)$, $(T,V,N)$ and $(T,p,N)$ respectively. At fixed particle number, Helmholtz free energy is minimized at fixed $T,V$; Gibbs free energy at fixed $T,p$.

A flow vessel is an open control volume, even if its walls are enclosed. For each inlet, pressure forces carry flow work $pdV$ in addition to internal energy. The steady first-law balance isWith no heat exchange, shaft work, or significant kinetic and gravitational energy differences, the enthalpy fluxes are equal. Mass flow must also balance. This equality concerns mass-flow-weighted specific enthalpies, not an unweighted sum over pipes.

Use the liquefaction yield with separate warm-gas and liquid outlets. The wording specifies that the outgoing gas is at 14 K; it does not require the outgoing liquid to have that temperature. The two outlet streams need not be in mutual thermal equilibrium. Taking $H$ as the printed specific enthalpy, the steady-flow energy balance givesFor fixed outlet enthalpies the greatest yield occurs at the smallest inlet enthalpy. Among the tabulated pressures this is 30 atm. Approximating the atmospheric-pressure gas enthalpy by the tabulated zero-pressure limiting value givesThis uses a low-pressure approximation: $p_0=1$ atm is not literally the zero-pressure table entry. Linear interpolation to 1 atm would instead give $H(p_0)\simeq 8.651\times 10^4\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$ and $\alpha \simeq 0.193$. The sparse table does not determine an exact continuous optimum between entries.

Atmospheric-pressure liquid helium must be much colder than the 14 K gas outlet: its normal boiling point is about 4.224 K.^[ref] A counterflow heat exchanger can return cold gas through the incoming stream before that gas leaves the overall control volume, while the collected liquid leaves separately at the cold end. This is a possible interpretation of the whole apparatus, rather than an extra diagram specified by the question. The enthalpy calculation needs only the stated inlet and outlet properties; it does not assume one equilibrium vessel containing both phases at 14 K. In particular, $H(p_0)-H_{\ell }$ is an outlet-stream enthalpy difference, not the latent heat of vaporization at one common temperature.

Use the enthalpy $H=U+pV$. The equation of state gives $pV=RT+pb$, soAt fixed pressure, $pb$ is constant. ThereforeSince $C_V$ and $R$ are constants, this ratio is independent of both temperature and pressure. Although the gas is not mechanically ideal when $b\ne 0$, its constant-pressure/constant-volume heat-capacity difference remains $R$ in this model.

Entropy at absolute zero. The third law of thermodynamics concerns the low-temperature limit of equilibrium states. In the zero-entropy formulation, the entropy of a perfect crystal with a nondegenerate ground state tends to zero as $T\to 0$. The statistical interpretation is $S=k_B\ln \Omega$: a unique accessible ground state has $\Omega =1$. The entropy-difference formulation of the third law of thermodynamics states that entropy differences associated with regular changes of mechanical parameters tend to zero as absolute zero is approached. These statements require equilibrium and appropriate ground-state assumptions; a frozen disordered material or a degenerate ground state can carry residual entropy.

For a phase of fixed composition and pressure,A finite limiting entropy constrains this integral. Ordinary equilibrium heat capacities therefore vanish as temperature approaches zero, with familiar examples $C\propto T^3$ for three-dimensional acoustic phonons and $C\propto T$ for a normal degenerate Fermi liquid. A constant positive heat capacity extrapolated to zero would give a logarithmically divergent entropy difference and is impossible in this regime. Discontinuous transitions and anomalous limits need separate treatment rather than extending a high-temperature approximation indefinitely.

Gibbs free energy. For fixed particle numbers,Thus, when $S\to 0$, the Gibbs free energy approaches its zero-temperature value with a horizontal tangent:The equality $G(0,p)=H(0,p)$ uses the enthalpy $H=U+pV$; neither limiting value needs to be zero. The low-temperature Gibbs free-energy expansion makes the approach quantitative. For $C_p=a{T}^r$ with $r>0$ and $S(0)=0$,The second derivative is ${\partial }_T^{2}G=-C_p/T\le 0$, although it need not tend to zero. For a reaction between such equilibrium phases, ${\partial }_T\Delta G=-\Delta S\to 0$, and $\Delta G$ and $\Delta H$ approach one another. If a genuine residual entropy $S_0$ is retained instead, $G=G(0,p)-S_{0}T+\cdots$ has limiting slope $-S_0$. A horizontal tangent cannot be asserted without the zero-entropy condition.

Why helium remains a liquid. Helium atoms attract each other weakly and are exceptionally light. Their zero-point energy is large compared with the binding that would localize them in a solid. Both helium-4 and helium-3 can therefore remain liquid down to the lowest temperatures at low pressure; substantial applied pressure is needed to solidify them. This is not a violation of the third law. A quantum liquid can have a unique low-entropy ground state, and zero-point motion does not itself imply thermal entropy. The third law does not require every substance to crystallize at atmospheric pressure.

A helium-4 atom has integer spin zero and is a boson. Below the lambda point near $2.17\mathrm{K}$ at saturated vapor pressure, its liquid is superfluid. Its strong interactions mean that it is not quantitatively an ideal Bose gas, and the condensate fraction need not equal the superfluid fraction. The two-fluid model of superfluidity separates an entropy-carrying normal component of a superfluid from the superfluid component, whose entropy is zero. At sufficiently low temperature the dominant thermal excitations are acoustic phonons: their linear dispersion in three dimensions gives $U-U_0\propto V{T}^4$, $C_V\propto V{T}^3$ and $S\propto V{T}^3$. Higher-energy rotons are exponentially suppressed. Thus the normal component of a superfluid and entropy disappear as $T\to 0$, while superfluid flow and quantum motion can persist.

A helium-3 atom has spin $1/2$ and is a fermion. Well below its Fermi temperature, but above its superfluid transition, liquid helium-3 is a degenerate interacting Fermi liquid. Only quasiparticles close to the Fermi surface are thermally excited, giving $C_V\propto T$ and $S\propto T$. At temperatures of order a millikelvin, helium-3 also becomes superfluid through formation of Cooper pairs of fermions. This is not Bose-Einstein condensation of individual helium-3 atoms. Pairing reorganizes the low-energy excitation spectrum; in the fully gapped bulk B phase, the fermionic quasiparticle contribution to heat capacity is exponentially suppressed at temperatures well below the pairing gap. Gapless collective modes can still give thermal terms following a power law. The entropy nevertheless tends to zero. The normal Fermi-liquid law cannot be extrapolated through its pairing transition unchanged.^[ref]

Helium-3 also illustrates the importance of nuclear spin entropy. Over an intermediate low-temperature range, localized spins in the solid can retain approximately $k_B\ln 2$ of entropy per atom, while the degenerate liquid has much less. Reversible conversion of liquid to the higher-entropy solid then absorbs heat; adiabatic compression along the melting curve can cool the sample. This is Pomeranchuk cooling. At still lower temperature, interactions between the nuclear spins remove this nearly free-spin entropy, so a constant $k_B\ln 2$ cannot be extrapolated to absolute zero.^[ref]

Liquefaction and cooling methods. A gas must be below its liquid-gas critical point to form a distinct liquid phase by compression. Compression above the critical temperature alone cannot achieve ordinary liquid-vapour coexistence. Practical liquefaction removes the compression heat, cools the compressed gas in a heat exchanger and uses expansion to lower its temperature further.

A throttling valve operates approximately at constant specific enthalpy when kinetic-energy changes and heat leakage are negligible. The Joule-Thomson coefficient isusing consistent molar quantities, or consistent total quantities for a fixed amount. A pressure drop cools the gas if ${\mu }_{\mathrm{JT}}>0$. An ideal gas has zero Joule-Thomson coefficient; real intermolecular interactions are essential. In particular, room-temperature helium warms on throttling, so it must first be brought into a region of positive Joule-Thomson coefficient. Expansion through an engine or turbine can instead remove work and produce approximately isentropic cooling.

A regenerative liquefier sends the cold returning gas through a counterflow heat exchanger to precool the incoming compressed gas before its next expansion. Expansion engines and external precooling can supply the initial cooling; repeated operation then produces liquid. Modern helium liquefaction commonly combines work-producing expansion with a final throttle. ^[ref]

Once liquid exists, pumping away vapour lowers its saturation pressure and boiling temperature. The escaping vapour carries latent heat away, cooling the remaining liquid. At still lower temperatures, a dilution refrigerator exploits the different entropy carried by helium-3 in concentrated and dilute helium mixtures; adiabatic demagnetization and Pomeranchuk cooling provide other routes.

Approaching zero. The unattainability formulation of the third law states that regular cooling operations with finite resources cannot reach absolute zero in finitely many stages. As the entropy available to remove becomes small and thermal couplings weaken, each method loses its ability to take the final step. An isentropic cooling trajectory starting at nonzero entropy cannot reach a common zero-entropy endpoint through regular finite changes of parameters. A fully general finite-time claim also requires assumptions about available controls and heat-transfer rates. Quantum liquids and refrigeration methods can approach $0\mathrm{K}$ while remaining consistent with vanishing equilibrium entropy; they do not provide a finite-operation route to absolute zero.

For a fixed amount of a simple compressible system, write the internal energy as $U(S,V)$. Its exact differential isAssuming the thermodynamic potential has continuous second partial derivatives, equality of its mixed partial derivatives yields the first Maxwell relation:For the second Maxwell relation, use the Legendre transform $H=U+pV$, the enthalpy. Its natural variables are $S,p$ sinceThus $T=(\partial H/\partial S{)}_p$ and $V=(\partial H/\partial p{)}_S$. Equality of the mixed partial derivatives now gives

This is the Brayton cycle. Assume fixed gas amount, constant heat capacities, and internally reversible adiabatic processes, as intended by the ideal engine model. Put $r_p=p_h/p_l=p_2/p_1>1$ andCombining $p{V}^{\gamma }=\text{constant}$ with the ideal gas law gives $T\propto p^{(\gamma -1)/\gamma }$ along either reversible ideal-gas adiabat. ThereforeDuring a constant-pressure process the heat supplied equals the enthalpy increase. If $C_p$ is the molar heat capacity at constant pressure, the positive heat input and output areThe insulated branches exchange no heat, so the first law of thermodynamics gives $W=Q_{\mathrm{in}}-Q_{\mathrm{out}}$. HenceAn adiabatic process alone need not be an isentropic process; the relations used here also require internal reversibility. Dissipative compression or expansion would require extra information and would not, in general, obey this efficiency formula.

With microstate energies held fixed when differentiating,Since $d\beta /dT=-1/(k_B{T}^2)$,The canonical ensemble fixes volume and particle number, and any other external Hamiltonian parameters. Thus the derivative is $C_V$; pressure-fixed fluctuations would naturally involve enthalpy in a different ensemble.

The translational and two-level energy variance follows because translation and internal states factorize for this dilute classical gas:Differentiating twice yieldsThe terms are independent translational and two-level binomial variances. At $k_{B}T\ll ϵ$, excitation freezes out exponentially and translation dominates. At $k_{B}T\gg ϵ$, the internal variance tends to $N{ϵ}^2/4$ while the translational variance grows as $T^2$; the internal heat capacity consequently tends to zero. Fractional total-energy fluctuations are of order $N^{-1/2}$ for large $N$. The low-temperature statement presumes translation is still nondegenerate; at strictly zero temperature the classical-gas model fails.

For an isolated system with equally probable accessible microstates, Boltzmann entropy is $S=k_B\ln \Omega$. The second law states that an isolated system's total entropy cannot decrease; it remains constant in a reversible process and increases in an irreversible one.

The insulated mixture exchanges no heat with its surroundings. At constant pressure, with only pressure-volume work, its enthalpy is conserved. Constant and equal specific heat capacities therefore give $T_f=(T_1+T_2)/2$. Calculate each entropy change along a hypothetical reversible temperature path:The surroundings gain no entropy through heat exchange; ideal mechanical work carries no entropy. The arithmetic-geometric mean inequality gives $\Delta S\ge 0$, with equality only for $T_1=T_2$. Mixing unequal temperatures is irreversible even though total energy is conserved.

Take the convection to maintain a dry, nearly reversible adiabat, with specific gas constant $R$. Hydrostatic equilibrium and an ideal-gas equation of state will be used separately to convert pressure to height. For an adiabatic displacement, the first law of thermodynamics in enthalpy form givesThe adiabat with linear heat capacity follows by integrating from the surface: this gives the monotone implicit temperature–pressure relationAn explicit form, using the positive real Lambert W branch, isAs $B\to 0$, $T=T_s(p/p_s{)}^{R/A}$. At fixed $p<p_s$,Larger heat capacity therefore keeps the upper atmosphere warmer at the same pressure. Equivalently $d\ln T/d\ln p=R/(A+BT)$ becomes smaller. The original normalized sketch below shows the resulting family of profiles; no particular gas constants or surface temperature are assumed there.

The height equations areIntegrating the latter givesThe physically selected root starts at $T_s$ and decreases. Its formal upper boundary isAt that height $T\to 0$ and $p=p_s(T/T_s{)}^{A/R}{e}^{B(T-T_s)/R}\to 0$. The $B=0$ limit is $z_{\ast }=A{T}_s/g$. This finite-height zero is a property of the assumed polytropic-like model, not a claim that a real atmosphere reaches absolute zero: molecular heat capacities, composition, transport and the continuum ideal-gas description cease to support that extrapolation.

This is the Joule-Thomson coefficient, ${\mu }_{JT}=(\partial T/\partial P{)}_H$, describing a temperature change during an isenthalpic pressure change. Here $H$ is enthalpy, $C_P=(\partial H/\partial T{)}_P$ is the heat capacity at constant pressure, and $V$ is volume, all for the same chosen amount of substance. Starting with $dH=TdS+VdP$ and using the Gibbs free-energy Maxwell relation $(\partial S/\partial P{)}_T=-(\partial V/\partial T{)}_P$ givesSetting $dH=0$ provesIn a steady insulated throttle or porous plug, negligible kinetic- and potential-energy changes make the upstream and downstream enthalpies equal. A positive coefficient gives cooling on a pressure decrease; a negative one gives warming. For an ideal gas, $V\propto T$ at fixed $P$, so the coefficient is zero. Real-gas interactions permit cooling and an inversion curve, making the effect useful in gas refrigeration and liquefaction.

The measurement of the Joule-Thomson effect with a porous plug compares pressures and thermometers on the two sides of an insulated flow restriction. Suppressing external heat exchange and separating the readings from high-speed jets are part of making the enthalpy argument applicable; an arbitrary adiabatic expansion need not be isenthalpic.

accretion onto a massive black hole releases luminosity $L\simeq \eta \dot{M}{c}^2$. Even a small coupling of this energy to surrounding gas can matter because a galactic binding energy is only of order $M_{\mathrm{gas}}{\sigma }^2$, with stellar velocity dispersion $\sigma \ll c$. active-galactic-nucleus feedback means that the energy or momentum changes the gas supply, cooling or star formation, potentially regulating further accretion. Large energy availability alone does not establish efficient coupling.

In a radiatively efficient bright nucleus, radiation pressure and winds can push on dusty or ionized gas. Outflows can remove fuel or shock-heat it; the outcome depends on cooling, covering factor, clumping and whether the flow escapes. In a mechanically dominated state, relativistic jets inflate radio cavities in a hot halo or galaxy cluster. Their expansion, shocks, sound waves and subsequent motions can distribute energy and suppress a cooling flow. The two descriptions identify useful channels rather than a universal sharp division between all nuclei.

An approximately pressure-balanced cavity containing a relativistic fluid has enthalpyX-ray pressure estimates, cavity volume and buoyant or sound-crossing ages permit comparison with radiative cooling losses. Radio images check the jet association; X-ray fronts and ripples help locate transferred energy. Deep Chandra observations of Perseus reveal such structures and motivate mechanical heating of the cooling atmosphere.^[ref]

Projection, nonrelativistic cavity contents, pressure gradients, uncertain ages and intermittent duty cycles affect the power estimate. Injected cavity energy need not all dissipate where cooling occurs. A successful negative feedback loop requires that falling gas increases accretion, heating then reduces cooling or inflow, and the delay does not destroy regulation. This can explain reduced cooling and the shortage of excessively massive star-forming galaxies; correlations between black-hole mass and bulge properties are supporting context rather than standalone proof of causation. The key comparison is coupled heating versus local cooling and binding, including where and when energy is deposited.

Without feedback, efficiently cooling gas would turn into stars too readily. Young stars return radiation, winds and supernova energy and momentum to their surroundings. This can heat gas, disrupt dense clouds and drive outflows, with the strongest fractional effect in shallow potential wells. The outflow mass-loading factor ${\eta }_w={\dot{M}}_{\mathrm{out}}/{\dot{M}}_{\ast }$ is useful, but its dependence on escape speed differs between energy-limited and momentum-limited models. Some expelled gas later recycles; feedback also transports metals and changes subsequent cooling.

active-galactic-nucleus feedback is important in massive galaxies and hot haloes. Radiative or mechanical power from black-hole accretion can suppress the supply of cold gas. Collimated jets inflate radio/X-ray cavities, while broader winds can carry significant momentum and kinetic energy. For a relativistic cavity fluid with adiabatic index $4/3$, its enthalpy is approximately $4pV$, so $4pV/t_{\mathrm{age}}$ estimates mechanical power. Comparing it with the gas cooling luminosity tests whether that energy budget is sufficient, rather than merely showing that a radio source is present.

Deep Chandra observations of Perseus resolved bubbles, a density jump and weak surface-brightness ripples consistent with repeated mechanical disturbance of the intracluster gas. Their interpretation as heating requires a dissipation mechanism and a coupling history, not just a large instantaneous power.^[ref]

Observed outflow geometry, filling factors, cavity ages and cooling corrections create substantial uncertainties. Feedback can also compress gas and locally promote star formation. Simulations must track radiative losses and the scale on which energy or momentum couples: depositing unresolved thermal energy is not automatically equivalent to driving a resolved wind. A successful model should explain stellar masses, gas fractions, metallicity and circumgalactic or intracluster thermodynamics together.

Use a steady planar sonic ablation model with absorbed intensity $I_a$, critical mass density ${\rho }_c=(A_i/Z)m_u{n}_c$, and $n_c={ϵ}_0{m}_e(2\pi c/\lambda {)}^2/e^2$. In the usual isothermal-electron sonic balance, $I_a\simeq 4{\rho }_c{c}_s^3$ and the ablation pressure is $P_A\simeq 2{\rho }_c{c}_s^2$. Eliminating $c_s$ givesFor $A_i/Z\simeq 2$ and complete absorption, the supplied laser parameters give about $1.15\times 10^{13}$ Pa, or $115$ Mbar. Mach one alone fixes neither the material mass per electron nor the enthalpy/absorption coefficient. Different sonic-flow closures change the order-one constant; incomplete absorption multiplies the estimate by its fraction to power $2/3$. The robust result is the two-thirds scaling, rather than a material-independent numerical pressure.

Let the undisturbed state ahead have $({\rho }_0,p_0,u_0)=({\rho }_0,p_0,0)$, and the uniform compressed state behind have velocity $u_1=U$. The piston boundary fixes that downstream velocity. Mass and momentum jumps giveThus ${\rho }_1/{\rho }_0=V/(V-U)$. Equality of specific stagnation enthalpy in the shock frame isSubstitute $p_1=p_0+{\rho }_{0}UV$ and $1/{\rho }_1=(V-U)/({\rho }_{0}V)$, then multiply by $2(\gamma -1)V$. The resulting difference factors as $U[-2V^2+(\gamma +1)UV+2\gamma p_0/{\rho }_0]=0$. Since the piston moves into the gas with $U>0$, division givesThis determines the constant speed of the compression shock ahead of the piston.

For steady insulated flow through a porous plug, neglect changes of bulk kinetic energy and gravitational potential energy, and allow no shaft work. Each amount of gas brings flow work $p_1{V}_1$ upstream and delivers $p_2{V}_2$ downstream. The steady energy conservation balance isThus enthalpy is conserved in this Joule-Thomson effect. For an ideal gas, $H=H(T)$, so $T_1=T_2$ and $U_1=U_2$ as well. The entropy increases:The porous-plug process is irreversible despite being slow and insulated; it is not the reversible expansion of the preceding part.

A state function is determined by the equilibrium state, rather than the path used to reach it. Its differential is exact and its change around a closed thermodynamic cycle is zero. Internal energy, entropy and enthalpy are examples; transferred heat and work are not state functions.

For slow expansion in contact with a large thermal reservoir, the fixed temperature of that reservoir maintains an isothermal process. For a fixed amount of ideal gas, internal energy and enthalpy depend only on temperature, soThe ideal gas law also gives $pV=\mathrm{constant}$. For a reversible expansion the absorbed heat equals the work done by the gas, while its entropy increases by $n_{\mathrm{mol}}R\ln (V_2/V_1)$; constant internal energy does not mean zero heat transfer.

For pure phases at the $1$ bar reference pressure, the equilibrium condition is ${\Delta }_r{G}^{\circ }={\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }=0$. ThusThe calculation uses the stipulated constant reaction enthalpy and entropy, with the fluid at its reference chemical fugacity.

At the one-bar standard state, pure phases have unit thermodynamic activities. The Gibbs energy of reaction balance is ${\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }=0$, givingThis uses the stipulated temperature-independent reaction enthalpy and entropy differences, taken per mole of the written reaction.

At the one-bar standard state, pure mineral endmembers and water vapour treated as an ideal gas all have unit thermodynamic activity. The Gibbs energy of reaction is ${\Delta }_r{G}^{\circ }={\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }$. Setting it to zero givesThe quantities are taken per mole of the reaction as written. The simplified result uses the stipulated temperature-independent enthalpy and entropy differences.