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The classical one-particle partition function, with phase-space cell $h^3$, isFor the homogeneous trap, rescale $r=V^{1/3}(kT{)}^{1/(2n)}u$. The spatial measure contributes $V(kT{)}^{3/(2n)}$, soHere $I_n=(2m\pi /h^2{)}^{3/2}{\int }_0^{\infty }4\pi u^2{e}^{-u^{2n}}du$, as specified; the last integral is $2\pi \Gamma (3/(2n))/n$.

Using the printed convention $Z_N=Z_1^N$, the free energy isThis convention treats the particles as labeled. For indistinguishable classical particles, $Z_N=Z_1^N/N!$ adds $kT\log N!$ to $F$. It changes the entropy by $-k\log N!$ but leaves the following force, energy and heat capacities unchanged. This normalization qualification is required to distinguish the printed expression from the usual Gibbs-corrected gas free energy.

The generalized force conjugate to the external parameter $V$ is $P=-(\partial F/\partial V{)}_T$, henceIt is the same equation of state as a hard-wall ideal gas, although $V$ now dilates a soft confining potential rather than being the occupied geometric volume.

Writing $L=\log V+a\log (kT)+\log I_n$, thermodynamic differentiation givesThe mean kinetic energy is $3NkT/2$ and the trap potential contributes $3NkT/(2n)$. At fixed $P$, $V=NkT/P$, so ${\partial }_{T}L{|}_P=(a+1)/T$. ConsequentlyThe same result follows from enthalpy $E+PV=(a+1)NkT$. These identities describe the classical ideal gas in a homogeneous soft trap.

On one poloidal flux surface, the field-line angular velocity $\omega (\psi )$ is constant. In a frame rotating at that rate the relative velocity is $\mathbf{v}=\mathbf{u}-R\omega {\mathbf{e}}_{\varphi }=k\mathbf{B}/\rho$, parallel to the magnetic field. The Lorentz force is perpendicular to this motion, and the Coriolis force also does no work. Centrifugal acceleration comes from the effective potential $-R^2{\omega }^2/2$. Since entropy is constant along the surface, pressure work integrates to specific enthalpy $w$, with $dw=dp/\rho$ along the flow.

Therefore kinetic energy relative to the rotating poloidal flux surface, enthalpy and effective gravitational-plus-centrifugal potential have a constant sum:This is a corotating energy integral, not the laboratory energy alone: Maxwell stress tensor stresses can transfer energy and angular momentum in the laboratory frame. In fact the expression equals $|\mathbf{u}{|}^2/2+w+\Phi -\omega R{u}_{\varphi }$, namely the total magnetohydrodynamic Bernoulli invariant minus $\omega \ell$. Differential rotation between surfaces does not invalidate the argument, because the chosen angular velocity is fixed along the particular surface being followed.

All invariants in this part refer to the selected magnetic field line. Its squared poloidal Alfvén number isAt the Alfvén surface crossing it equals unity, so ${\rho }_a={\mu }_0{k}^2$ and $M_{Ap}^2=1/y$. Subtract the two azimuthal integrals in part (a) to findFinite azimuthal fields and velocity at $y=1$ require the Alfvén-surface regularity condition for an axisymmetric wind $\ell =\omega R_a^2$. HenceHere $v_{\varphi }$ denotes the azimuthal velocity relative to field-line rotation. For gravity of a point mass at $z=0$, $\Phi =-GM/R=-GM/(R_{a}x)$.

Insert these quantities in the corotating energy integral with $w=0$, and divide by the gravitational energy scale $GM/R_a$. This gives the cold radial magnetohydrodynamic wind integral,withBoth are dimensionless. The apparent $0/0$ at $x=y=1$ must be evaluated on the smooth solution branch, not assigned an arbitrary value.

A constant-energy curve obeys $f_x+f_{y}dy/dx=0$. At a nondegenerate magnetosonic point, the vanishing coefficient of the mass density derivative requires simultaneous vanishing of its numerator:The curve must admit a real crossing slope. Expanding about such a point gives $f_{yy}{m}^2+2f_{xy}m+f_{xx}=0$, $m=dy/dx$, so a regular saddle point of a scalar function crossing has $f_{xy}^2-f_{xx}{f}_{yy}>0$. These are the form of the slow and fast regularity conditions in a warm wind, with the enthalpy contribution retained in $f$.

In the strictly cold approximation used to obtain the displayed $f$, the slow speed is zero. There is a cold-limit degeneracy of a slow magnetosonic point: a finite-speed interior slow crossing need not survive. Indeed,so this cold equation has no ordinary sub-Alfvénic slow critical point. A small retained enthalpy is needed to locate that point, which can approach the launching boundary as the sound speed tends to zero. The fast critical point still obeys the simultaneous conditions above wherever the reduced expression is nonsingular.

For a fixed amount of simple compressible material, $dU=T,dS-p,dV$ and $G=U-TS+pV$ gives $dG=-S,dT+V,dp$. Equality of mixed partial derivatives of $G$ therefore yields the Maxwell relationIn the insulated throttling process, piston work into the gas is $p_1{V}_1$ and work out is $p_2{V}_2$. Neglecting changes in macroscopic kinetic and gravitational energy, the first law gives $U_2-U_1=p_1{V}_1-p_2{V}_2$. Thus $U_1+p_1{V}_1=U_2+p_2{V}_2$, so enthalpy is conserved. The irreversible process is not assumed isentropic.

Since $dH=T,dS+V,dp$ and $dS=(C_p/T)dT-(\partial V/\partial T{)}_{p}dp$, one has $dH=C_p,dT+[V-T(\partial V/\partial T{)}_p]dp$. Therefore the Joule-Thomson coefficient isFor an ideal gas, $V\propto T$ at fixed pressure and $\alpha =1/T$, so ${\mu }_{JT}=0$. A pressure drop has $dp<0$, so cooling, $dT<0$, requires ${\mu }_{JT}>0$ in the operating range.

Let $h_i,h_s$ denote ice and snow thickness, let $z$ increase downwards, and count external atmospheric fluxes as positive into the surface. The Maykut-Untersteiner model is a time-dependent column model, rather than a single thickness formula: heat conduction, heat storage, internal absorption of sunlight and moving freezing/melting boundaries must be solved together. In an ice-following description its energy equation can be writtenwhere $e_i$ is specific enthalpy, $S$ is the prescribed ice salinity, and $I$ is the downward penetrating shortwave flux. The material derivative includes coordinate motion if a moving numerical grid is used. A common attenuation form is $I=I_0{e}^{-\kappa z}$ within a homogeneous ice layer, with the origin taken at that layer's top. The corresponding source is $\kappa I$. Snow-covered cases have much less penetration; absorbed surface radiation must not also be counted as internal heating.

The liquid brine fraction contributes latent heat to the effective specific heat capacity: for the illustrative liquidus approximation ${\varphi }_b\simeq -\mu S/T$ with Celsius $T<0$ in the mixed-phase range $0<{\varphi }_b<1$, $c_{\mathrm{eff}}={\partial }_T{e}_i\simeq c_0+L_f\mu S/T^2$. The precise ice conductivity and heat capacity are salinity- and temperature-dependent. This is why the full transient temperature profile need not be linear. At the snow-ice interface impose continuity of temperature and conductive heat flux; at the ice bottom impose the seawater freezing temperature $T_f$.

Write $I_0=i_0(1-\alpha )F_{\mathrm{SW}}$ and define the atmospheric input absorbed at the surface byHere $\alpha$ is the surface albedo, $\sigma$ is the Stefan-Boltzmann constant, and $F_H,F_E$ are signed sensible and latent turbulent heat fluxes. With upward conductive flux $q_c=k{\partial }_{z}T$, the cold, nonmelting surface satisfies $Q_A+q_{c,s}=0$. Once the surface reaches its melting temperature, constrain it there and use excess energy for ablation:The mass-loss rate ${\dot{m}}_s$ first removes snow at thickness rate ${\dot{m}}_s/{\rho }_s$; after the snow is gone it removes ice at rate $m_i={\dot{m}}_s/{\rho }_i$. Snow thickness also changes through prescribed snowfall and any separately represented compaction or sublimation.

At the bottom, the Stefan condition equates latent heat released by growth to upward conductive removal minus upward ocean heat flux $F_o$:Here $g_b$ is positive for basal freezing and negative for basal melting, and $L_b$ is the appropriate effective latent heat. These equations, the incident fluxes and snowfall determine the seasonal thickness trajectory. Annual sea-ice thermodynamic equilibrium means a periodic temperature field and $h_i(t+\mathcal{T})=h_i(t)$, soWith constant $L_b$, this is equivalently $\int (q_{c,b}-F_o-{\rho }_i{L}_b{m}_i)dt=0$. An “equilibrium thickness” must refer to the mean or a specified phase of this seasonal cycle.

A useful winter reduction neglects heat storage and solar penetration and treats each conductivity as constant. The two layers act as thermal resistances in series:This displays the dependence on snow and ice thickness, but is not a replacement for the full seasonally forced Maykut-Untersteiner model.

Increasing ocean heat flux subtracts directly from basal growth and increases bottom ablation. On a stable perennial branch it lowers annual equilibrium thickness, and sufficiently large input can eliminate a year-round ice cover. With zero oceanic input, removing this heat sink can allow much thicker thermodynamic ice; it does not remove the seasonal surface energy balance.

Additional annual snowfall has competing effects. Larger $h_s/k_s$ suppresses winter heat conduction and basal growth, favoring thinner ice. But persistent bright snow raises albedo, consumes energy when it melts and delays exposure of dark bare ice, suppressing summer ice ablation. The original column calculations show near compensation for moderate snowfall and substantial thickening when much more snow survives into summer. Thus annual equilibrium thickness need not decrease with snowfall, even though greater snow thickness at fixed winter temperatures always reduces the conductive growth rate. The balance also depends on snowfall timing and the treatment of snow-to-ice conversion. These seasonal responses are illustrated by the original model calculations.

Reducing surface albedo increases absorbed shortwave radiation. It promotes surface ablation, internal heating and, where sunlight reaches the ocean, basal ablation; the ice-albedo feedback reinforces the reduction in equilibrium ice thickness. Increasing albedo has the opposite effect. This sensitivity is strongest when sunlight is available and dark surfaces are exposed, rather than during polar night.

For the observed bottom melt, use ${\rho }_i\simeq 917\mathrm{kg}{\mathrm{m}}^{-3}$, $L_f\simeq 3.34\times 10^5\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$ and approximately 122 days. The basal melt heat flux equivalent isThe integrated latent heat is about $6.1\times 10^8\mathrm{J}{\mathrm{m}}^{-2}$. This is energy used for melting, not automatically the complete ocean-to-ice heat input: the basal balance gives $F_o={\rho }_i{L}_b(-g_b)+q_{c,b}$, with sensible heat and salinity corrections if needed.

A likely major source is solar heating of an unusually large open-water fraction. Dark water absorbs much more sunlight than snow-covered ice, and warm water transfers that stored heat to floe bottoms by mixing and lateral transport. Thin or ponded ice also transmits more sunlight. Wind-driven divergence and export expose additional water, while advection of warm Pacific or Atlantic water and mixing of subsurface heat can contribute regionally. The supplied melt amount alone cannot partition these sources. The 2007 Beaufort Sea heat-budget analysis found that locally absorbed solar heat was sufficient to account for the enhanced basal melt. The latent-heat equivalent is about $60\mathrm{W}{\mathrm{m}}^{-2}$; increased solar absorption by exposed ocean is a well-supported principal explanation.

The one-particle partition function isThe classical indistinguishable noninteracting gas has $Z_N=Z_1^N/N!$; for labeled particles omit $N!$, which changes none of the requested derivatives. With $\beta =(k_{B}T{)}^{-1}$,The allowed equation of state gives $PV=N{k}_{B}T$, so the enthalpy is $E+PV=5N{k}_{B}T/2$ and $\boxed{C_P=5N{k}_B/2}$, at fixed particle number.

The Maykut-Untersteiner model solves a seasonally forced, one-dimensional sea ice thermodynamics problem with a snow layer above saline ice. Let $z$ increase downward from the upper surface, with snow–ice interface $z=h_s$ and ice bottom $z=h_s+h_i$. Write $q=-k{T}_z$ for conductive heat flux, positive downward, and $I(z)$ for downward shortwave radiation penetrating the material. Local energy conservation givesFor $I=I_0{e}^{-\kappa z}$, the radiation term is $+\kappa I$: absorption heats the interior. Here $c_{\mathrm{eff}}$ is the derivative of specific enthalpy with temperature at prescribed ice salinity, and includes the latent heat of changing brine fraction. Using only pure-ice specific heat capacity would omit this storage. A usual approximation has ${\rho }_i{c}_{\mathrm{eff}}={\rho }_i{c}_{i,0}+a{S}_i/T_C^2$ and $k_i=k_{i,0}+b{S}_i/T_C$, with $T_C<0$ in degrees Celsius and positive, unit-dependent constants $a,b$. Thus warmer saline ice has larger effective heat capacity and lower thermal conductivity. Layerwise constant $k_i$ gives the usual finite-layer heat equation; full differentiation retains its spatial variation. Snow uses its own density and thermal conductivity, and its stronger extinction can make shortwave penetration negligible beneath it.

Define the net atmospheric input to the upper surface bywhere $\alpha$ is albedo, $H_{\uparrow }$ is turbulent sensible heat loss and $E_{\uparrow }$ is turbulent latent heat loss. Their signs can reverse. The outgoing longwave term uses the Stefan–Boltzmann law. Subtract the penetrating shortwave $I_0$, because it heats the interior rather than the infinitesimal surface. If $M_0\ge 0$ is surface ablation speed and ${\rho }_0$ the density of the material currently exposed, the upper boundary obeysBelow the melting point, $M_0=0$ and the flux balance determines $T_0$. Once it reaches $T_m$, the positive excess produces surface ablation; snow melts first, and only afterwards does that ablation remove ice. At the snow–ice interface, absent interfacial phase change,At the base, $T_i=T_f(S_w)$ and the Stefan condition isHere $G_b$ is signed bottom growth speed, and $F_o>0$ is ocean heat flux toward the ice. An upward conductive loss $-q_b>0$ freezes ice; an excess $F_o$ melts it. For a prescribed upward turbulent ocean heat flux, the model needs no separate ocean temperature solution. A water-side closure could write $F_o={\rho }_w{c}_{p,w}{K}_w{T}_{w,z}$ at the boundary in this downward coordinate. Snowfall is an independent mass input: with snowfall mass flux $P_s$, ${\dot{h}}_s=P_s/{\rho }_s-M_s$ in a constant-density approximation. The ice thickness obeys ${\dot{h}}_i=G_b-M_i$. Compaction or refreezing, if added, requires corresponding mass and energy terms rather than counting snowfall directly as ice growth. These boundary conditions and interior equations determine temperature, surface melt and basal growth together.

Under periodically repeated annual forcing, annual sea-ice thermodynamic equilibrium is a periodic annual state, not constant thickness:Winter accretion then balances net annual ablation. “Equilibrium thickness” normally denotes a mean or a specified point in this annual cycle. Different forcing can instead eliminate a perennial-ice solution.

Increasing ocean heat flux reduces bottom growth directly in the Stefan condition, lowering equilibrium thickness and eventually removing perennial ice. The stabilizing winter response is visible in the approximate conductive fluxwhen ice thins, conductive loss and winter regrowth increase. More snow lowers this flux and inhibits winter growth. However, snow also raises albedo and delays summer exposure of dark ice. The competing snow insulation and summer ice protection effects mean an annual snow sensitivity cannot be deduced from winter insulation alone. Under the original forcing, the Maykut-Untersteiner model gives an almost flat annual-thickness response below about $0.7\mathrm{m}$ annual snow depth, followed by strong thickening as snow persists into summer. This particular transition is not a universal threshold for other forcing or snowfall timing. Increasing summer albedo reduces absorbed solar energy, reduces surface and internal melting, and increases equilibrium thickness; sufficiently low albedo removes perennial ice. The original thermodynamic sensitivity calculations establish these three responses.

The three curves are original qualitative sketches, not a new numerical integration or digitization. The snow marker refers to that historical forcing; the unnumbered loss boundaries and ordinate indicate trends rather than calibrated universal thresholds.

For the summer retreat discussed around the time of this examination, the proposed mechanisms act as follows.

These processes overlap: circulation causes export, export changes thickness and open-water fraction, and those changes alter solar absorption. Without specified regional budgets there is no defensible unique percentage partition. For enhanced Beaufort summer basal melt, upper-ocean solar heating and western-Arctic heat pathways are more directly relevant than invoking unmodified heat from a deep, isolated Atlantic layer. ice-mass-balance observations of extreme Beaufort melt identify increased solar absorption in open water as an important source.

Take ${\rho }_i=900\mathrm{kg}{\mathrm{m}}^{-3}$ as an explicit ice-density approximation and use the supplied $L_f=336000\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. The latent heat for $1.5\mathrm{m}$ basal ablation isJune–September contains about $122$ days, soUsing $917\mathrm{kg}{\mathrm{m}}^{-3}$ would give $43.8\mathrm{W}{\mathrm{m}}^{-2}$. This is the net flux used for melting: the gross ocean heat flux can also supply conductive losses and changes in cold content. It greatly exceeds an annual central-pack background of a few $\mathrm{W}{\mathrm{m}}^{-2}$, but is compatible with a summer upper ocean absorbing solar radiation at tens to hundreds of $\mathrm{W}{\mathrm{m}}^{-2}$. For scale, an added open-water fraction $0.3$, a shortwave input $200\mathrm{W}{\mathrm{m}}^{-2}$ and an ice–water albedo contrast $0.6$ produce $0.3\times 200\times 0.6=36\mathrm{W}{\mathrm{m}}^{-2}$ additional area-mean absorption before losses. This is an illustrative budget, not a measurement. Wind mixing and circulation can deliver stored solar or advected heat to the underside.

The ice-albedo feedback closes the warming loop: retreat exposes darker water; extra solar absorption warms the upper ocean; basal and lateral melt increase retreat. The stored heat also delays autumn freeze-up, exposing water to atmospheric heat transfer for longer. Its later release warms the atmosphere, but must not be counted as a second source in addition to the solar energy that originally entered the ocean. In winter, greater heat loss from open water and rapid growth of thin ice partly oppose the summer positive feedback.

The heat capacity at constant pressure is $C_p=(\delta Q/\partial T{)}_p=T(\partial S/\partial T{)}_p$ for reversible heat input. The enthalpy is $H=U+pV$, so $dH=TdS+Vdp$. Along an isobar this gives $\boxed{C_p=(\partial H/\partial T{)}_p}$.

Assume a single-phase ideal gas mixture in a closed parcel, with common temperature $T$, total pressure $P$, and molar species amounts $N_i$. Define $N={\sum }_i{N}_i$ and volume mixing ratios $x_i=N_i/N$, so ${\sum }_i{x}_i=1$. Ideal mixing makes partial pressure $P_i=x_{i}P$.

For a pure ideal species, $d{\mu }_i=-{\bar{S}}_{i}dT+{\bar{V}}_{i}d{P}_i$, with molar volume ${\bar{V}}_i=\mathcal{R}T/P_i$. At fixed temperature, integration from standard pressure $P^{\circ }$ givesThe same logarithmic composition term follows from the ideal mixing entropy $\Delta S_{\mathrm{mix}}=-\mathcal{R}{\sum }_i{N}_i\log x_i$. Extensivity gives $G={\sum }_i{N}_i{\mu }_i$. Hence the Gibbs free energy of an ideal-gas mixture isEquivalently, write each standard chemical term as ${\mu }_i^{\circ }=h_i^{\circ }-T{s}_i^{\circ }$, using molar enthalpy and entropy. For zero abundance the limit $x\log x\to 0$ defines the extensive expression continuously.

The fractions alone do not specify the extensive Gibbs free energy: the total amount $N$ or an equivalent elemental inventory must also be fixed. Moreover, in a chemically closed reacting system $N$ 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.

In the shock frame, integrate mass conservation, momentum conservation, and total-energy conservation across a thin interval containing the stationary normal shock wave. No mass, momentum, or energy is stored in the vanishingly thin interval, so each flux has the same value on both sides. For a perfect gas, the specific enthalpy is $h=\gamma p/[(\gamma -1)\rho ]$. The resulting Rankine-Hugoniot conditions for a perfect gas areThe momentum flux includes both transported momentum and pressure force. The energy flux includes kinetic energy and enthalpy, the latter accounting for internal energy and the work needed to push gas through the interval. Entropy production can occur inside the normal shock wave even though total energy is conserved.

Use the common-density, equal-volume-heat-capacity model implicit in the stated speed law. Let the liquid fraction be $\varphi$ ahead of the melting front and $\varphi +\Delta \varphi$ behind it. Melting adds exactly the liquid needed to fill the newly created pores, so total mass conservation makes Darcy flux continuous. Consequently,Here $u$ denotes Darcy velocity and $v$ the pore-liquid velocity in the stationary rock frame. Treating the pore-space increase as storage without its simultaneous melting source would incorrectly change the Darcy flux.

Relative to the cold unmelted material, the bulk enthalpy increase behind the front is $\rho c_p\Delta T+\rho L\Delta \varphi$: all phases gain sensible heat and the melted ice consumes latent heat. The advective heat-flux difference is $\rho c_{p}U\Delta T$. Applying the Rankine-Hugoniot condition to this energy balance therefore givesso the advection-driven melting front in a porous matrix moves atThe advected latent contribution of the liquid is the same on both sides and cancels; the denominator measures sensible heating plus phase-change energy per unit bulk volume.

Put $k_a=k$ ahead of the front and $k_b=k+\Delta k$ behind. With viscosity $\mu$, Darcy's law gives the planar pressure gradients $p_{a0}^{\prime }=-\mu U/k_a$ and $p_{b0}^{\prime }=-\mu U/k_b$. Write the displacement as $\eta e^{\sigma t+i\alpha y}$ with $\alpha >0$, and take the unperturbed interface to be $x=0$ in its translating frame.

The liquid flux is divergence-free away from the interface, so the pressure perturbations are harmonic. The decaying normal modes areExpanding pressure continuity at the displaced interface and matching normal Darcy flux givesHenceThe common normal-flux perturbation is $\delta u_n=-k_b\alpha B/\mu =\alpha U\Delta k\eta /(k_a+k_b)$. The local enthalpy balance converts it into a front-speed perturbation, $\sigma \eta =\delta u_n/(1+S\Delta \varphi )$. Thus the melting-front instability due to permeability contrast hasFor a signed transverse wavenumber, replace $\alpha$ by $|\alpha |$.

For a simple compressible system with fixed particle number, the first law isDefine the enthalpy by $H=E+pV$. ThenThus $H$ has natural variables $(S,p)$ andEquality of mixed partial derivatives gives the Maxwell relation

Integrate each conservative ideal magnetohydrodynamics law through a thin pillbox around the stationary shock. Mass conservation givesThe MHD momentum-flux tensor isIts normal and tangential components givewhere $\Pi =p+B^2/(2{\mu }_0)$ is total pressure. The solenoidal condition gives $[B_x]=0$, while steady Faraday's law gives continuity of tangential electric field,Finally, material energy flux plus the normal Poynting vector component giveswhere $h$ is specific enthalpy per unit mass.

Gravity is a bounded volume force, so its integral across a shock whose thickness tends to zero vanishes. Viscosity and resistivity inside the layer may produce entropy; consequently entropy flux need not be equal on both sides, although the second law requires nonnegative net entropy production.

The three jumps express conservation of mass flux, normal and tangential momentum flux, and total enthalpy (energy) flux across the stationary shock.

At fixed particle number, the differential of the internal energy identifies $T=(\partial U/\partial S{)}_V$ and $p=-(\partial U/\partial V{)}_S$. Equality of continuous mixed partial derivatives gives the first Maxwell relation:For the enthalpy, $dH=dU+pdV+Vdp=TdS+Vdp$. Thus $T=(\partial H/\partial S{)}_p$, $V=(\partial H/\partial p{)}_S$, and the same argument gives the enthalpy Maxwell relationThe smoothness and independent thermodynamic coordinates are understood within a single phase.

For fixed amount of substance, take the internal energy $U=U(S,V)$. The first law of thermodynamics identifies $T=U_S{|}_V$ and $-P=U_V{|}_S$. Equality of mixed second derivatives givesFor the Helmholtz free energy, subtract the product differential $d(TS)$ from the first law of thermodynamics to obtain $dF=-SdT-PdV$. Its natural variables are $(T,V)$, so $F_T=-S$ and $F_V=-P$. Exactness therefore givesFor the Gibbs free energy, the product rule gives $dG=-SdT+VdP$. Thus $G_T=-S$, $G_P=V$, andFor the enthalpy, $dH=TdS+VdP$, giving $H_S=T$, $H_P=V$ andThese are the four Maxwell relations, each following from exactness in the appropriate pair of natural variables.

Approximate the reaction enthalpy and entropy as constant over the measured interval, so ${\Delta }_r{G}^{\circ }={\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }$. A linear least squares fit of all three free energies against temperature givesand henceThe negative entropy change is reasonable because oxygen gas is incorporated into solids. The fit residuals are $+0.129,-0.257,+0.129\mathrm{kJ}\mathrm{mo}{\mathrm{l}}_{{\mathrm{O}}_2}^{-1}$, so the data support this approximate straight-line description. Extrapolating its intercept to zero temperature is a way of estimating a constant local enthalpy, not evidence that the heat capacities vanish or that the same phases persist to zero temperature. More temperatures and repeated heating/cooling measurements would test curvature and equilibration.

Interpret the supplied free energies as standard reaction free energies and use $c^{\circ }=1\mathrm{mol}{\mathrm{L}}^{-1}$. The dimensionless equilibrium constants areUnder the ideal-solution approximation, $[C]=K_C[L]$ and $[D]=(K_D/c^{\circ })[L{]}^2$. Set $k_D=K_D/c^{\circ }$, with units ${\mathrm{M}}^{-1}$. Conservation counts two original molecules in each dimer:The physical, positive root can be written without subtractive cancellation asThe resulting concentrations are:

For example, the first row satisfies $[L]+[C]+2[D]=0.100\mathrm{M}$; using $[L]+[C]+[D]$ would incorrectly count dimers. At both concentrations almost all molecular units belong to dimers.

dimer formation is favored by sufficiently high concentration because it joins two molecules. Dilution favors the separated species, whereas $[C]/[L]=K_C$ is concentration-independent in this ideal model. More than half the molecular units are dimerized when $2[D]>[L]+[C]$, equivalently $c_0>(1+K_C{)}^2/k_D\simeq 7.36\times 10^{-7}\mathrm{M}$. The supplied free energies at one temperature do not determine a temperature trend without additional enthalpy information. At $1\mathrm{M}$, neglected nonideal activities could be significant; the table is the requested concentration-based model.

Use $c_p,c_V$ for the heat capacities of the entire $n$-mole sample. For a stable simple compressible gas, a Maxwell relation giveswhere ${\alpha }_V=V^{-1}(\partial V/\partial T{)}_p$ and ${\kappa }_T=-V^{-1}(\partial V/\partial p{)}_T>0$. This heat-capacity difference and mechanical stability explains the sign without assuming the internal energy depends only on temperature. It is strictly positive for an ordinary expanding gas; the unrestricted word “always” needs the qualification that equality is possible when thermal expansion vanishes.