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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.

For $p=K{\rho }^{1+1/n}$ with $K,n>0$, the specific enthalpy relative to vacuum is $Q=(n+1)K{\rho }^{1/n}=(n+1)p/\rho$. It obeys $\nabla Q=\nabla p/\rho$ and $n{D}_{t}Q=-Q\nabla \cdot \mathbf{u}$. The vacuum reference is fixed when these equations are combined with the polytropic equation of state.

Specific enthalpy is enthalpy per unit mass. For a simple compressible fluid of fixed composition, the first law of thermodynamics giveswhere $T$ is temperature, $s$ is specific entropy, $p$ is pressure, and $\rho$ is mass density. Along an isentropic flow, $dh=dp/\rho$. For a perfect gas with constant specific-heat ratio $\gamma$, $h=\gamma p/[(\gamma -1)\rho ]$.