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Regularity at the center requires $m(0)=0$ and a finite central density ${\rho }_c$. The equation of state fixes $p_c=p({\rho }_c)$, and the regular central series beginsFor each admissible ${\rho }_c$, ordinary-differential-equation uniqueness determines $m,p,\rho$ outward until the first zero of $p$, which defines $R$. The remaining additive constant in $\Phi$ is fixed by matching to the exterior Schwarzschild time coordinate. Thus smooth stars form a one-parameter family labelled uniquely by ${\rho }_c$.

The four stellar structure equations aretogether with the perfect-gas equation of stateHere $a_{\mathrm{rad}}$ is the radiation constant, $ϵ$ is the specific stellar energy-generation rate, and $\kappa$ is the opacity.

Put $x=r/R$. Integrating the prescribed density gives the enclosed massSince $m_R=M$,Integrating hydrostatic equilibrium inward from $P(R)=0$ givesThe ideal-gas law then givesIn particular,

If $Pdx+Qdy=dF$ for a twice continuously differentiable potential $F$, then $P=F_x$ and $Q=F_y$. Equality of mixed partial derivatives gives the necessary condition for an exact differential:This condition alone does not ensure global exactness on a domain with holes.

For the Helmholtz free energy $A=U-TS$, use the stated identity for internal energy and the product rule:Thus $A_T{|}_V=-S$ and $A_V{|}_T=-p$. Equality of mixed partial derivatives yields the Maxwell relationThe temperature and volume are the independent variables here, with composition and amount of gas fixed.

Next express $S=S(T,V)$ and substitute its total differential into $dU=TdS-pdV$:Matching the coefficient of $dV$ givesThis is the thermodynamic internal pressure identity: an equation of state determines how internal energy changes under an isothermal change of volume.

For the Dieterici equation, set $c=an/(VR)$ while differentiating at fixed $V$. ThenSubstitute into the thermodynamic internal pressure identity. The term proportional to $T$ cancels $p$, leavingThe physical gas domain has $T>0$ and $V>nb$.

For the ideal gas instead, $p=nRT/V$ gives $T(\partial p/\partial T{)}_V-p=0$. Therefore $U$ is independent of volume at fixed amount of gas. Integrating the remaining condition $(\partial U/\partial T{)}_V=C_V$, with constant heat capacity at constant volume, giveswhere $U_0$ is independent of both $T$ and $V$ and fixes the reference for internal energy.

Holding $y$ fixed, the chain rule giveswhere the local change of variable from $x$ to $u$ is valid.

For the internal energy $U(S,V)$, its exact differential identifies $T=U_S$ and $p=-U_V$. The Maxwell relation from equality of mixed partial derivatives is $T_V{|}_S=-p_S{|}_V$. At fixed volume, the constant-volume heat capacity satisfiesso $T_S{|}_V=T/C_V$, assuming nonzero $C_V$ and a locally invertible relation. Applying the chain rule to $p(T,V)$ gives the isentropic temperature-volume derivativeFor the stated one-mole ideal gas, the equation of state gives $p_T{|}_V=R/V$ and $C_V=3R/2$, so the right side is $-2T/(3V)$. Along an isentropic process, $p\propto V^{-5/3}$ and $T=pV/R\propto V^{-2/3}$, giving the same left side:

A cosmological constant has fixed density and equation of state $w=p/(\rho c^2)=-1$. A separately conserved dynamical component obeysIf $w$ departs from $-1$, its history changes the Hubble parameter, luminosity distances and angular-diameter distances. Type Ia supernovae probe a relative distance-redshift relation, baryon acoustic oscillations provide standard-ruler distances and expansion information, and the cosmic microwave background fixes early-time scales and geometric combinations. Combining these breaks some degeneracies with matter density, curvature and calibration. Distance observables integrate the expansion history, so they smooth rapid variations in $w$ and cannot determine an arbitrary function exactly.

Structure-growth measurements add a dynamical test. Galaxy clustering, redshift-space distortions and weak gravitational lensing constrain the growth of matter perturbations and metric potentials. The late Integrated Sachs-Wolfe effect probes potential evolution through CMB-matter correlations. A scalar-field dark-energy model can have perturbations and an effective sound speed different from other models with a similar background expansion. Comparisons of geometry and growth can therefore distinguish some models and test gravity assumptions.

A statistically significant, robust departure from $w=-1$ or from constant dark-energy density would disfavor a pure cosmological constant within the stated framework. Consistency with $w=-1$ over a finite redshift range does not prove exact constancy: a slowly rolling field can mimic it. Supernova calibration/evolution, galaxy bias, photometric redshift errors, intrinsic galaxy alignments and foregrounds must be controlled; modified gravity can also change growth. Parameterizations such as $w(a)=w_0+w_a(1-a)$ are useful finite-dimensional tests rather than a complete classification of all dark-energy histories.

Cosmological variables and conventions. The scale factor $R(t)$ describes the relative separation of comoving observers; dots denote derivatives with respect to cosmic time $t$. The Hubble parameter is $H(t)=\dot{R}/R$, usually time dependent; “Hubble constant” conventionally means its present value. Here $\rho$ is mass-equivalent density, so $ϵ=\rho c^2$ is energy density, while $P$ is isotropic pressure. $G$ is the gravitational constant and $c$ the speed of light. The constant $K$ specifies cosmological spatial curvature: with $R$ a length, $K=k{c}^2$ and $k=+1,0,-1$ describe closed, flat and open spatial geometries. Alternative normalizations of $R$ move a constant factor into $K$. In the printed equations $\Lambda$ has units ${\mathrm{s}}^{-2}$: it is $c^2$ times a cosmological constant defined in inverse square metres. This time-unit convention for the cosmological constant prevents a missing factor of $c^2$.

Continuity and the first law. Differentiating the first Friedmann equation with constant $K$ and $\Lambda$, and then eliminating $\ddot{R}/R$ with the second, givesThus, on an expanding or contracting branch,The relation extends continuously to a regular turning point. For a physical volume $V(t)=V_{\ast }{R}^3(t)$ bounded by fixed comoving coordinates, $\dot{V}=3HV$. Multiplication by $c^{2}V$ givesThis adiabatic work in a comoving volume is the first law of thermodynamics for adiabatic evolution with fixed entropy and no additional chemical work. It expresses local energy conservation, including pressure work. It does not assert that $U$ is constant: radiation loses internal energy as it does work; negative pressure makes $U$ grow during expansion. A generic expanding spacetime has no global time-translation symmetry from which a conserved total energy follows.

Constant scalar-field energy density. Setting $\dot{\rho }=0$ in the continuity equation, with $H\ne 0$, givesFor a homogeneous canonical scalar field, in units $c=1$, $ϵ={\dot{\varphi }}^2/2+V(\varphi )$ and $P={\dot{\varphi }}^2/2-V(\varphi )$. A stationary potential-dominated field has precisely this equation of state. A general evolving scalar field need not have $w=-1$; the constant-density assumption is essential.

Expansion at the stipulated Higgs energy density. With zero explicit $\Lambda$ and zero $K$, constant ${ϵ}_H$ givesThe plus sign selects the expanding branch; there is also a contracting solution. The scalar-field energy is already included in $\rho$, so adding it again as an explicit cosmological constant would double count it. Using $1\mathrm{GeV}=1.602176634\times 10^{-10}\mathrm{J}$ gives ${ϵ}_H=2.4033\times 10^{45}\mathrm{J}{\mathrm{m}}^{-3}$, ${\rho }_H=2.6740\times 10^{28}\mathrm{kg}{\mathrm{m}}^{-3}$ and $H_H\simeq 3.87\times 10^9{\mathrm{s}}^{-1}$. ThereforeThis vacuum-dominated doubling time describes accelerated expansion in de Sitter spacetime, with $\ddot{R}/R=H_H^2>0$. The stipulated energy density exceeds the critical density in energy units used in Question 2 by $3\times 10^{54}$, so it cannot persist as the dominant component of the present universe. The calculation exhibits the cosmological constant problem: the absolute gravitational vacuum energy must be enormously smaller than such a particle-physics scale today. It does not establish that the Standard Model Higgs field actually remains constant or supplies a viable inflationary epoch.

The acoustic ruler and accelerated expansion. Before baryon drag becomes ineffective, pressure in the photon-baryon fluid drives acoustic waves around initial linear cosmological density perturbations. Their comoving sound horizon at the baryon drag epoch isAfter decoupling, this preferred separation survives in galaxy clustering as baryon acoustic oscillations: a peak in the two-point correlation function and oscillations in the matter power spectrum. Its value is calibrated by early-universe physics, commonly constrained by the cosmic microwave background. The SDSS luminous-red-galaxy measurement established a peak near $100h^{-1}\mathrm{Mpc}$; its use as a distance indicator depends on the ruler calibration rather than the length of an individual galaxy.^[ref]

For a ruler perpendicular to the line of sight and one parallel to it,where $D_M=(1+z)D_A$ is the transverse comoving distance and $D_A$ the angular diameter distance. These measure $D_M/r_d$ and $H{r}_d$, respectively. Comparing multiple redshifts tests the distance–redshift relation and expansion history. In particular,so sufficiently precise measurements of the shape of $H(z)$ can test $q<0$. In general relativity, accelerated expansion requires a sufficiently negative total $\rho +3P/c^2$; a positive cosmological constant supplies it.

A single acoustic angular scale does not independently determine acceleration, cosmological spatial curvature, the absolute sound horizon and the dark-energy equation of state. One combines redshift coverage with early-universe calibration and independent matter-density or distance information, comparing accelerating and nonaccelerating models. The galaxy selection function and redshift completeness must be modeled with random catalogues, survey covariance estimated with mock catalogues, and redshift-space distortions, galaxy bias and nonlinear broadening controlled. These tests distinguish a cosmological feature from a survey artifact. The evidence is the measured expansion history within an explicit cosmological framework, rather than the existence of an acoustic peak alone.

For a small longitudinal disturbance, let the equilibrium pressure and mass density be $p_0,{\rho }_0$, their perturbations be $p^{\prime },{\rho }^{\prime }$, and the fluid velocity be $v$. Linearizing conservation of mass and the inviscid momentum equation in one dimension givesWhen compression is adiabatic, the equation of state gives $p^{\prime }=c_s^2{\rho }^{\prime }$ with $c_s^2=(\partial p/\partial \rho {)}_s$. Differentiating and eliminating $v$ yieldsfor an ideal gas with heat-capacity ratio $\gamma =C_p/C_v$ and molar mass $M_{\mathrm{mol}}$. Compression raises temperature as well as density, explaining the factor $\gamma$ compared with the isothermal value $\sqrt{p_0/{\rho }_0}$.

A rightgoing plane sound wave has pressure and particle velocity in phase, $p^{\prime }={\rho }_0{c}_{s}v$, and mean intensity $⟨p^{\prime }v⟩={\hat{p}}^2/(2{\rho }_0{c}_s)={\rho }_0{c}_s{\hat{v}}^2/2$ for peak amplitudes. Particle displacement is a quarter-cycle out of phase with particle velocity: particles oscillate locally while the compression pattern transports energy. This continuum linear description requires wavelength much larger than the mean free path, small pressure perturbations and negligible viscous attenuation. Adiabatic propagation requires heat exchange across the compressed region to be slow on the oscillation time; strong thermal coupling would instead favour an isothermal response.

The zeroth law of thermodynamics states the transitivity of thermal equilibrium: if two systems each equilibrate thermally with a third, they equilibrate with one another. A reproducible thermometric property can therefore label their temperature. Its empirical numerical scale is not itself the complete thermodynamic content; reversible heat-engine relations establish an absolute temperature scale.

The first law of thermodynamics expresses energy conservation. For a closed system, $dU=\delta Q-\delta W$ when $\delta W$ is work done by the system. internal energy is a state function, while heat and work depend on the path. In reversible simple compressible motion, $\delta W=pdV$ and $dU=TdS-pdV$; an open system additionally needs the energies carried by transferred matter. An ideal gas can expand freely with $Q=W=0$ and unchanged internal energy. The first law alone does not decide whether the reverse spontaneous process occurs.

The Second law of thermodynamics introduces entropy and restricts direction. For a reversible path, $dS=\delta Q_{\mathrm{rev}}/T$; for an irreversible process the entropy balance includes nonnegative production. The entropy of an isolated system cannot decrease. In a cycle, the Clausius inequality is $\oint \delta Q/T\le 0$, using the appropriate boundary/reservoir temperatures. The Kelvin–Planck statement forbids a cyclic device whose sole effect is extracting heat from one reservoir and turning all of it into work; the Clausius statement forbids moving heat from cold to hot with no compensating effect. A refrigerator is allowed because it consumes work. A two-reservoir engine has $\eta \le 1-T_c/T_h$, with equality in the reversible limit. Work may be fully dissipated as heat without contradicting this one-way restriction.

The third law of thermodynamics constrains low-temperature entropy. An equilibrium perfect crystal with a nondegenerate ground state approaches zero entropy as $T\to 0$, fixing the conventional absolute entropy reference. Degenerate ground states or frozen nonequilibrium disorder require care when discussing residual entropy. Under regular equilibrium conditions the heat capacity must decrease sufficiently for $\int C/TdT$ to remain finite; a constant positive classical heat capacity cannot be extrapolated to zero temperature. The unattainability formulation states that absolute zero cannot be reached by a finite sequence of admissible cooling operations. These statements concern the limiting thermodynamic regime, rather than predicting that every substance stays in one classical phase during cooling.

The laws work together: conservation sets a device's energy budget, entropy production restricts the attainable budget, and constitutive equations give the properties of its working material. For fixed temperature and volume, Helmholtz free energy decreases toward equilibrium; at fixed temperature and pressure the corresponding potential is Gibbs free energy. Stable equilibrium, phase coexistence and Maxwell relations follow from these potentials with appropriate constraints. Calorimetry, mechanical-work measurements, thermometry and reversible-path comparisons test different parts of this framework. None of the laws by itself supplies a particular gas equation of state or microscopic heat capacity.

Molecules. In a dilute classical gas, collisions establish a Maxwell-Boltzmann distribution. The normalized speed distribution and its first two moments areMomentum transfer to a wall gives $p=nm\overline{v^2}/3=n{k}_{B}T$. Translational internal energy is $U=3N{k}_{B}T/2$; rotational and vibrational degrees of freedom add energy when thermally populated, without changing the dilute ideal-gas pressure law. Classical equipartition gives $k_{B}T/2$ per quadratic contribution, but quantum level spacing can freeze out modes.

Only the molecules moving towards a surface contribute to its one-sided number flux:For an effusive molecular energy flux, faster molecules are preferentially sampled, giving mean translational energy $2k_{B}T$ per crossing molecule and $F_E=2k_{B}TJ$. This differs from multiplying $J$ by the bulk mean $3k_{B}T/2$. Collisions and finite molecular interactions produce transport and departures from the ideal-gas equation of state.

Photons. All photons travel at $c$ in vacuum, so their thermal distribution is over frequency, not over a range of speeds. The two polarizations and Bose-Einstein occupation with zero chemical potential giveHere the spectral quantity is $u_{\nu }$ and the total energy density is $u$. Momentum flux gives $p=u/3$, while isotropic angular averaging gives the one-sided energy flux $F_E=cu/4={\sigma }_{\mathrm{SB}}{T}^4$ through a small aperture of a black-body cavity. The photon number flux is similarly $c{n}_{\gamma }/4$.

Both pressures arise from momentum transfer, but their equations of state differ: a classical molecular gas has $pV=N{k}_{B}T$ with fixed molecule number, whereas a photon gas has $pV=U/3$, $U=a_{R}V{T}^4$ and an equilibrium photon number that changes with temperature. Photon creation and absorption by the walls are essential to the zero-chemical-potential equilibrium description.

An elastic reflection transfers normal momentum $2mv\cos \theta$ to the wall. Multiplying this impulse by the collision flux and integrating givesUsing the radial integral and $\alpha =m/(2k_{B}T)$ yieldsThis agrees with the ideal-gas equation of state. The calculation assumes a stationary wall and elastic normal momentum reversal, appropriate to a gas in equilibrium with it.

An equation of state relates thermodynamic variables within a material model; a phase diagram records which equilibrium phase or phases are stable under specified controls. For a pure substance at fixed temperature and pressure, stability means minimizing Gibbs free energy. Two phases coexist when their molar chemical potentials agree. The equation of state and the free-energy stability condition must therefore be used together: an algebraic relation between $p,V,T$ alone need not identify the stable branch.

A pressure-temperature phase diagram contains single-phase regions separated by phase coexistence curves. For a one-component system the Gibbs phase rule gives $F=3-P$: a single phase has two independent controls, two-phase coexistence lies on a curve, and three phases coexist at a triple point. The liquid-gas coexistence curve ends at a liquid-gas critical point; beyond it a continuous path connects liquidlike and gaslike states without crossing a first-order coexistence boundary. A fixed $p,T$ point on a coexistence curve does not specify the proportions of the two phases: an additional extensive constraint, such as total volume, fixes the phase fraction.

At a first-order phase transition, first derivatives of Gibbs free energy, such as molar entropy or volume, are discontinuous. The conversion has latent heat $L=T\Delta s$. At a continuous transition the first derivatives match, while response functions associated with higher derivatives can be singular or discontinuous. At an ordinary liquid-gas critical endpoint, the two coexisting densities approach one another and the latent heat tends to zero; the material does not retain a sharp liquid-gas boundary above it.

The coexistence slope follows directly from equality of the two molar chemical potentials. Since $d\mu =-sdT+vdp$, differentiating ${\mu }_{\alpha }={\mu }_{\beta }$ along the boundary givesThis is the exact Clapeyron equation, also commonly called the Clausius-Clapeyron relation. For vaporization, both $L$ and $\Delta v$ are positive, so the saturation pressure rises with temperature. For ordinary ice melting into liquid water, $L>0$ but the molar volume decreases, giving the familiar negative melting-curve slope. The sign follows from the volume change, not from a different law.

For a dilute ideal vapour with $v_g\gg v_l$, $\Delta v\simeq RT/p$. The ideal-vapor Clausius-Clapeyron approximation becomesIf molar latent heat is approximately constant over the interval,These approximations deteriorate close to the critical point, where liquid and gas volumes become comparable, or over a range in which latent heat changes appreciably.^[ref]

The ideal gas law $pV=nRT$ is an equation of state for a dilute gas, but it has no liquid branch and cannot predict condensation. A simple model including repulsion and attraction is the Van der Waals molar equation,where $b>0$ represents excluded molar volume and $a>0$ represents attraction. At its critical point, $\partial p/\partial v={\partial }^{2}p/\partial v^2=0$, givingBelow $T_c$, the homogeneous isotherm has an interval with $\partial p/\partial v>0$. That would have negative isothermal compressibility and is mechanically unstable. Equilibrium liquid-gas coexistence replaces the loop by a constant-pressure segment. Equality of chemical potentials gives the Maxwell construction:The equal-area condition enforces the common tangent to the appropriate free energy; it is a stability construction rather than permission to follow the unstable loop physically. The average molar volume between $v_l$ and $v_g$ varies by changing the phase fraction.

The model captures a liquid-gas critical point and qualitative coexistence, but its simple constants are not a complete quantitative description of real fluids and it has no solid phase. A full phase diagram needs additional solid free energies and can contain several crystal phases. Conversely, an experimentally determined coexistence curve and latent heat constrain equations of state through the Clapeyron equation. This is the useful connection: local thermodynamic response, global free-energy stability and the geometry of phase boundaries describe the same material at different levels.

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.

An equation of state relates equilibrium state variables such as pressure, volume, temperature and material amount. At fixed composition, it reduces the number of independent thermodynamic variables. It does not by itself specify all caloric properties: information such as heat capacity or internal energy is also needed.

At a liquid-gas first-order transition, an equilibrium isotherm can contain a constant-pressure interval in which changing total volume changes the phase fractions. A metastable single-phase extrapolation can instead show a region with $(\partial p/\partial V{)}_T>0$; that region has negative isothermal compressibility and is mechanically unstable. Thermodynamic phase coexistence requires equal temperature, pressure and chemical potential, and a consistent free-energy description replaces the unstable region by coexistence. For an appropriate analytic fluid equation of state, a liquid-gas critical point occurs where $(\partial p/\partial V{)}_T=({\partial }^{2}p/\partial V^2{)}_T=0$.

These tests diagnose the phase behaviour represented by that equation of state. They do not guarantee detection of every possible transition involving an additional order parameter which the mechanical equation omits.

At every $T>0$, $V>b$,There is no constant-pressure coexistence interval, unstable loop or finite-temperature critical inflection. The isothermal compressibility is positive:Thus this equation of state predicts no liquid-gas phase transition in its physical domain. Its divergence at the excluded-volume boundary is a repulsive constraint, not a liquid-gas coexistence mechanism. This conclusion concerns the stated model and cannot rule out an ordering transition requiring additional physics or an additional thermodynamic variable. Extrapolation to absolute zero is also outside the classical gas model's reliable regime.

Pressure and its physical terms. Solving the molar Van der Waals equation for pressure givesThe excluded-volume parameter $b$ reduces the volume available for translational motion, raising the kinetic pressure relative to $RT/V$. The attraction parameter $a$ lowers the pressure because attractive interactions reduce the momentum transferred to the boundary. This mean-field model treats $a,b$ as temperature-independent material constants. For one mole, $V$ is the sample volume with the molar amount already fixed; $b$ has units of that volume and $a$ of pressure times volume squared.

volume dependence of entropy. The Helmholtz free energy has differential $dF=-SdT-PdV$. Equality of mixed derivatives yields the Maxwell relationFor $a=0$ this derivative is $R/(V-b)$, givingEquivalently, $S=R\ln (V-b)+f(T)$ with a common reference volume understood inside the logarithm. The equation of state fixes this volume dependence, not the otherwise unspecified function of temperature.

Excess entropy at fixed volume and temperature. The ideal gas has $(\partial S_{\mathrm{id}}/\partial V{)}_T=R/V$. ThereforeIntegrate from infinite volume, where the entropy difference is specified to vanish:It is negative because excluding configurations reduces the accessible state count. It tends to zero as $V\to \infty$ and has leading dilute limit $-Rb/V$. This is the entropy excess of a van der Waals gas.

Effect of attraction. Restoring a temperature-independent $a$ changes the pressure by $-a/V^2$, but not its derivative with respect to temperature. With the same infinite-volume entropy reference, it therefore does not change $\Delta S^{\mathrm{ex}}$ at fixed $V,T$. This can also be seen fromThe attraction affects internal energy and the equilibrium volume, so it can affect entropy indirectly when different state points are compared. If $a$ were temperature dependent, its derivative would contribute directly to the entropy; that is outside the stipulated constant-parameter model.

Coexistence isotherm. Below the Van der Waals critical point, the formal homogeneous isotherm has a local minimum and maximum. Its middle branch has $\partial P/\partial V>0$, hence negative isothermal compressibility, and is mechanically unstable. Portions of the negative-slope branches can be metastable rather than globally stable. The equilibrium isotherm replaces the loop between liquid volume $V_l$ and vapour volume $V_g$ by a constant saturation pressure $P_s$. The two volume are fixed byThe Maxwell equal-area construction expresses equality of the two phases' chemical potentials. Intermediate total volume are mixtures: changing the vapour fraction changes total volume at the same pressure, rather than following the unstable homogeneous curve. The spinodals are the turning points of the formal isotherm; they are distinct from the coexistence endpoints.

Dense-phase volume at low temperature. At a general subcritical temperature, the two coexistence equations above must be solved together. For this equation of state the area condition is explicitlyThis supplies an exact implicit determination of the liquid volume even when a simple closed form is unavailable.

For the asymptotically low-temperature limit, put $\tau =RTb/a\ll 1$. The liquid is dense, $V_l\simeq b$, while the vapour is dilute, $P_s{V}_g\simeq RT$ and $V_g\gg V_l$. The area condition then givesThus the vapour volume grows exponentially in $1/\tau$, and the saturation pressure is exponentially small. It is consequently consistent to neglect $P_s$ in the liquid equation to algebraic order in $\tau$:Choose the root near $b$, not the large-volume zero-pressure root:Expanding gives the leading requested dense-phase behavior,This is the low-temperature liquid volume of a van der Waals gas. Setting $P_s=0$ is an asymptotic approximation, not an exact coexistence condition at every temperature below $T_c$. The figure illustrates generic subcritical coexistence; its finite-temperature liquid volume is determined by the full equal-area equations. The van der Waals liquid also ceases to be a realistic description at sufficiently low physical temperature, where other phases or quantum effects can matter.

This is the one-mole, or equivalently molar volume, form of the Van der Waals equation, an approximate equation of state for a real gas:Here $p$ is the gas pressure, $T$ its absolute temperature, $R$ the molar gas constant, and $V$ the volume occupied by one mole. If $V$ denotes molar volume explicitly, its units are ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$.

The positive parameter $b$ represents excluded volume arising from the finite size and short-range repulsion of the molecules. The volume available for translational motion is reduced from $V$ to $V-b$. This is an effective excluded volume, not simply the sum of the molecules' geometric volumes. Its molar units are ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$.

The positive parameter $a$ measures intermolecular attraction in a mean-field approximation. Attractions reduce the force exerted on the walls, lowering the measured pressure by $a/V^2$ relative to the repulsion-only expression. Thus $p+a/V^2$ is the pressure corrected for this attraction. The squared inverse-volume dependence reflects a contribution proportional to the square of the molecular number density. The molar units of $a$ are $\mathrm{Pa}{\mathrm{m}}^6\mathrm{mo}{\mathrm{l}}^{-2}$.

For an arbitrary amount $n$ measured in moles, with total container volume $V_{\mathrm{tot}}$, the corresponding formula isAt low density, when $b\ll V$ and $a/V^2\ll RT/V$, it reduces to the ideal gas law. The model captures qualitative attraction, repulsion and liquid-gas behavior, but does not supply an exact equation of state for every material.

Illustration TODO. The source's two-volume circulation sketch on page 4 is unavailable. Its exact arrows and layout cannot be recovered. The surviving flow law, salinity/temperature definitions and printed steady polynomial allow the calculation below under an explicit freshwater sign convention; no unavailable diagram values are invented.

Use $\Delta T=T_2-T_1>0$, $\Delta S=S_2-S_1$, and let positive $q$ be the thermally driven state with cold box 1 denser than warm box 2. Take positive $F$ to freshen box 1, represented in a fixed-volume salt budget by an equivalent salt removal $F{S}_1$ from that box and addition to the other reservoir. This is the freshwater convention compatible with the printed $+BF$ and positive $B$. It can represent a compensating saline volume transfer or the usual virtual-salt-flux approximation. The exact pipe assignment is not claimed to have been read from the missing illustration.

Exchange in either circulation direction reduces a salinity difference at rate proportional to $|q|\Delta S$. Steady salt balance therefore givesThe supplied linear equation of state givesEliminating $\Delta S$ yields the two-box overturning model equationHere $A$ and $B$ both have transport units when $q,F,k$ do, while $\alpha \Delta T$ and $\beta S_1$ are dimensionless. Although the source calls $q$ a mass flow, its stated Sverdrup units are volume per time; the common reference density cancels in the Boussinesq salt/volume approximation. At $F>0$, $q=0$ cannot be a finite-salinity steady state in this model. At $F=0$, zero flow needs separate treatment instead of division by $|q|$.

The circulation weakens when freshwater freshens the cold reservoir, and the weaker circulation removes the resulting salinity contrast less effectively. This salt-advection feedback is the mechanism behind multiple steady states, not an additional externally imposed switch.

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.

No. Near the liquid-gas critical point, the mass densities of liquid and vapor approach each other, so the neglected condensed-phase specific volume becomes comparable with the vapor specific volume. The vapor also ceases to obey the dilute ideal gas law. Although the latent heat tends to zero, the specific-volume difference tends to zero too: the exact Clausius-Clapeyron relation contains their ratio. Setting only the numerator to zero therefore does not justify a vanishing physical saturation pressure slope. A realistic equation of state and coexistence data are required near the surface; the dilute colder upper atmosphere can still be described approximately by the earlier expression.

Interpret the terminal infall speed as the free-fall speed from a radius much larger than the neutron star radius $R_{\ast }=12\mathrm{km}$. In the Newtonian estimate,For fully ionized hydrogen with electrons and protons at a common temperature, there are two particles per atom and $\mu =(m_p+m_e)/2\simeq m_p/2$. The result of e then yieldsThis is the thermal energy scale $k_B{T}_d$ per particle. In the same classical monatomic gas approximation, the mean random kinetic energy per particle is $3k_B{T}_d/2\simeq 59\mathrm{MeV}$. Before the shock wave, the bulk kinetic energy per proton is $m_p{v}_u^2/2=G{M}_{\ast }{m}_p/R_{\ast }\simeq 208\mathrm{MeV}$. Distinguishing these quantities avoids an ambiguity in “particle energy”. If only the ions support the downstream pressure because electrons have not equilibrated, $\mu \simeq m_p$ instead gives $k_B{T}_i\simeq 78\mathrm{MeV}$; that is a different, two-temperature assumption.

These are estimates within the question's Newtonian, $\gamma =5/3$ model. Its self-consistency is poor here: $2G{M}_{\ast }/(R_{\ast }{c}^2)\simeq 0.443$, and $39\mathrm{MeV}$ is much larger than the electron rest energy $0.511\mathrm{MeV}$. Relativistic electrons, electron–positron pair production, radiation, unequal species temperatures, and general relativity can alter the equation of state and shock wave structure. In particular, the numerical $39\mathrm{MeV}$ must not be presented as a precision surface temperature prediction.