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Efficient convection keeps the interior at nearly uniform specific entropy. For the fully ionized monatomic gas, the heat capacity ratio is $5/3$, so the adiabatic process gives $P=K_{\rho }{\rho }^{5/3}$, with $K_{\rho }$ spatially constant. The ideal gas relation $P=\mathcal{R}\rho T/\mu$ then impliesThe constant is spatial, not temporal: loss of specific entropy during contraction changes it. This is a stellar polytrope with polytropic index $n=3/2$. We neglect radiation pressure and take the thin photosphere as an atmospheric matching layer rather than a significant fraction of the stellar mass.

For a fixed dimensionless profile of the Lane-Emden equation, the stellar homology relations have the formwhere $C_{\rho },C_P$ are positive dimensionless structure constants. More explicitly, write $\rho (r)={\rho }_{c}f(x)$ and $m(r)=Mh(x)$ with $x=r/R$ and fixed dimensionless profiles. Mass conservation gives $M=4\pi {\rho }_c{R}^3{\int }_0^{1}f(x)x^{2}dx$. Integrating the hydrostatic pressure support equation, with negligible surface pressure on the interior scale, gives $P_c=(GM{\rho }_c/R){\int }_0^{1}h(x)f(x)x^{-2}dx$. Both integrals are fixed positive numbers, proving the two central scalings. Substituting these relations into $K=P_c/T_c^{5/2}$ givesThus the same polytropic index fixes $K_0$ throughout this homologous sequence, while the dimensional stellar polytrope constant changes.

At the photosphere, whose optical depth is $2/3$, identify the temperature with the effective temperature $T_e$. The ideal gas relation and the interior adiabat give ${\rho }_{\mathrm{ph}}=\mu K{T}_e^{3/2}/\mathcal{R}$. With the prescribed opacity and stellar surface boundary condition,Eliminating $K$ therefore provesUsing the Stefan–Boltzmann law, with $\sigma =ac/4$, gives the luminosity

The gravitational energy of a stellar polytrope is $\Omega =-6G{M}^2/(7R)$ for $n=3/2$. The stellar virial theorem for a monatomic gas gives $2U+\Omega =0$, and hence the total energy is $E=U+\Omega =-3G{M}^2/(7R)$. With no appreciable stellar nuclear fusion or accretion, energy conservation givesCombining the two expressions for $L$ yieldsMultiplication by $-7R^{-9/2}/2$ and integration proves the contraction law with density-linear fourth-power stellar opacity:When the accumulated contraction term dominates the initial-radius term, $R\propto {\mu }^{-4/7}{M}^{2/7}(t-t_0{)}^{-2/7}$. Substitution into the central ideal gas relation giveswhich has the stated $t^{2/7}$ behavior when the time origin is negligible. This late-time approximation still requires the assumed fully convective star, fixed composition and opacity law to remain applicable.

A stellar polytrope replaces a detailed equation of state and thermal structure by $P=K{\rho }^{1+1/n}$ for $n>0$, with spatially constant $K$ and polytropic index $n$. Together with spherical hydrostatic equilibrium and mass conservation, it gives a tractable family of stellar density profiles. Introduce the Lane-Emden variables for a stellar polytropeCombining $dP/dr=-Gm\rho /r^2$ with $dm/dr=4\pi r^2\rho$ eliminates $m$ to give $r^{-2}d[r^2{\rho }^{-1}(dP/dr)]/dr=-4\pi G\rho$. Substitution of the pressure law and the dimensionless variables gives the Lane-Emden equationA regular finite-radius model for $0\le n<5$ ends at the first zero ${\xi }_1$. Its radius and mass areThese formulas supply central-to-mean density ratios and stellar homology scalings without solving a full evolutionary model. Eliminating ${\rho }_c$, for $0<n<5$ and $n\ne 3$, gives the polytropic mass-radius relation

An efficient-convection monatomic gas has $n=3/2$. The same index describes a cold, nonrelativistic electron degeneracy pressure law; at fixed composition it gives $R\propto M^{-1/3}$ for a white dwarf. Ultrarelativistic electron degeneracy pressure has $n=3$, for which the mass is independent of ${\rho }_c$ and scales as $(K/G{)}^{3/2}$. This is the origin of the Chandrasekhar limit. A radiation-dominated, approximately constant-entropy interior also approaches $P\propto {\rho }^{4/3}$ and $n=3$. The polytrope of index zero is an incompressible comparison model, useful for illustrating the role of central concentration.

The gravitational energy of a stellar polytrope, $\Omega =-3G{M}^2/[(5-n)R]$, and the stellar virial theorem make polytropes useful for contraction-energy estimates. They also provide structural models for perturbation and stability calculations. A polytrope is a structural approximation, not a complete stellar evolution calculation. One must still specify stellar nuclear fusion, opacity, energy transport, composition and boundary conditions to determine the luminosity and history. Moreover, the structural exponent $1+1/n$ need not equal the adiabatic exponent relevant to a rapid perturbation. Real stars can have different effective indices in their cores and envelopes, as well as ionization zones or varying electron degeneracy pressure.

Use signed velocities in the shock frame; the Mach number depends on their squares. For a genuine nonradiative perfect gas shock wave, the Rankine-Hugoniot conditions for a perfect gas give a common mass flux $j={\rho }_L{u}_L={\rho }_R{u}_R$, a common momentum flux $p+\rho u^2$, and a common specific stagnation enthalpy $\gamma p/[(\gamma -1)\rho ]+u^2/2$. Here $e$ in the given equation of state is specific internal energy, despite the source's informal “energy density” wording.

Put $r={\rho }_R/{\rho }_L$, $P=p_R/p_L$ and $M^2={\rho }_L{u}_L^2/(\gamma p_L)$. Mass and momentum give $u_R=u_L/r$ and $P=1+\gamma M^2(1-1/r)$. Substituting these into the energy relation and clearing denominators givesThe $r=1$ factor is the no-jump solution. On the nontrivial shock wave branch,In particular $r-1=2(M^2-1)/[(\gamma -1)M^2+2]$. With $\gamma >1$, compression is therefore equivalent to $M_L^2>1$. The other Mach number isPositive pressure makes its denominator positive, and $M_R^2-1$ has the opposite sign to $M_L^2-1$. Thus a nontrivial physical shock wave has a supersonic side and a subsonic side; the entropy-admissible flow goes from the former to the latter.

The shock compression ratio also yields $u_L{u}_R=(\gamma -1)u_L^2/(\gamma +1)+2\gamma p_L/[(\gamma +1){\rho }_L]$. Rearranging produces the first velocity identity in the question. The pressure relation gives the second identity in the equivalent formThe numerator here contains mass density, as the original PDF shows; the converted TeX incorrectly replaces that mass density by pressure.

For reflection, the once-shocked gas is the common left state in both applications. Before rebound, its velocity in the shock frame is $a=u_s-U_{+}<0$ and the gas ahead has velocity $-U_{+}$; after rebound, it is $b=u_s+U_{-}>0$ and the gas next to the wall has velocity $U_{-}$. In both cases the right velocity minus the left velocity equals $-u_s$. Applying the first identity therefore gives the same quadratic equation,with roots $a$ and $b$. Their opposite signs distinguish the two roots, so $\boxed{(u_s-U_{+})(u_s+U_{-})=-\gamma p_s/{\rho }_s}$.

Write $q_0=p_0/p_s$, $q_1=p_1/p_s$. The pressure identity applied to each shock wave givesSquaring the root-product relation and eliminating the velocities yieldsand hence the normal shock reflection at a rigid wall relationTaking $p_0/p_s\to 0$ gives $\boxed{p_1/p_s\to (3\gamma -1)/(\gamma -1)}$; for a monatomic gas this limiting amplification is six.

Use $\mathcal{R}$ for the stellar gas constant, reserving $R$ for the stellar radius. The specific internal energy of a monatomic gas obeying the ideal gas law is $e=P/[(\gamma -1)\rho ]=3P/(2\rho )$. At fixed enclosed mass, the first law of thermodynamics gives the heat available to the outgoing luminosity as $ϵ=-{\partial }_{t}e-P{\partial }_t(1/\rho )$ when there is no stellar nuclear fusion heating. ThereforeThe second coefficient includes both the mass density dependence of $e$ and pressure work. This is gravitational energy generation in a homologously contracting ideal-gas star.

Write the characteristic scales asThen $\rho ={\rho }_{0}b$, $P=P_{0}p$, $T=T_{0}p/b$, $m=Mq$ and $L_r=Ll$. Since $d/dr=R^{-1}d/dx$, direct substitution in the stellar hydrostatic equation and mass conservation givesFor radiative diffusion in a star, dividing the temperature equation by $T_0/R$ givesIn particular the fourth power here is of $\mathcal{R}$, not of the radius.

Because $q(x)$ is fixed and $M$ is constant, a fixed mass label has fixed $x$. Thus stellar homology gives ${\partial }_t\rho =-3\rho \dot{R}/R$ and ${\partial }_{t}P=-4P\dot{R}/R$. The heating rate reduces to $ϵ=-3P\dot{R}/(2\rho R)$, positive during Kelvin-Helmholtz contraction. Substitution in $d{L}_r/dr=4\pi r^2\rho ϵ$ now yieldsThe dimensionless profiles fix $D$ and $E$. In a common stellar homology family with fixed stellar composition and opacity normalization, the expression for $D$ impliesIt is independent of radius and constant in time for a particular fixed-mass star. The formula for $E$ gives $\dot{R}=-2EL{R}^2/(3G{M}^2)$, which integrates toWhen the initial-radius term is negligible,This is the Kelvin-Helmholtz contraction time law in the prescribed stellar homology model. Infinite initial radius is a limiting idealization; a finite initial radius retains the first term.

At main sequence arrival, CNO cycle heating with fixed stellar composition has luminosityThe integral is dimensionless and fixed within the stellar homology family. The constant-opacity radiative scaling still gives $L\propto M^3$. Equating the stellar nuclear fusion and transported luminosities givesFinally the Kelvin-Helmholtz contraction law evaluated at $R_{\mathrm{MS}}$ gives $t_{\mathrm{MS}}\propto M^2/(L{R}_{\mathrm{MS}})\propto M^{-34/19}$. In a coeval star cluster, the mass just arriving at the main sequence consequently satisfiesThese results comprise constant-opacity homologous contraction to CNO ignition; their mass exponents compare fixed-composition models with common dimensionless profiles.

Molecular transport across gradients. A gas can transport momentum, thermal energy and species abundance even when its average bulk velocity is zero. Molecules arriving at a point carry properties acquired near their preceding collisions. A small gradient makes the populations arriving from opposite sides slightly different, producing transport from larger to smaller values of the carried property. A continuum transport law requires a mean free path $\lambda$ small compared with the gradient length, and relaxation times short compared with the observation time.

For a dilute hard-sphere gas,where $d$ is molecular collision diameter. Both speed and collision distance matter: high speed alone does not determine a transport coefficient.

A simple kinetic estimate makes the common structure explicit. For a quantity $a(y)$ carried per particle, a molecule moving at angle $\theta$ to $+y$ brings approximately $a(y-\lambda \cos \theta )$. Expanding to first order and averaging an isotropic velocity distribution givessince $⟨\cos \theta ⟩=0$ and $⟨{\cos }^2\theta ⟩=1/3$. This mean-free-path argument estimates the coefficients; a detailed collision calculation changes their numerical prefactors and distinguishes different relaxation rates.

Momentum transport and viscosity. For a shear flow with mean velocity $U_x(y)$, the transported directed momentum per molecule is $m{U}_x$. The net flux of $x$ momentum in the $+y$ direction isHere $\eta$ is dynamic viscosity in $\mathrm{Pa}\mathrm{s}$. The associated shear stress on a material surface has the corresponding traction sign convention; the flux carries momentum from the faster-moving layer towards the slower-moving one. The kinematic viscosity is $\nu =\eta /(nm)$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$. Viscosity smooths velocity gradients, and the lost ordered mechanical energy becomes heat.

Heat transport. Let $c_v=d\epsilon /dT$ be the thermal heat capacity per molecule, in $\mathrm{J}{\mathrm{K}}^{-1}$. Substituting $a=\epsilon (T)$ gives Fourier's law,Thermal conductivity has units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$. For a monatomic gas, translational equipartition gives $c_v=3k_B/2$. Rotational or vibrational excitations in other gases can increase the energy carried per molecule, while collisions govern how those modes exchange energy. A temperature gradient can therefore be transported differently from directed momentum even when the dimensional estimates look similar.

Species transport. For a dilute tracer or a mixture with prescribed total density and negligible bulk flow, molecules diffuse down concentration gradients. The first-gradient number flux is Fick's first law,The diffusivity $D$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$; multiplying number flux by the species mass gives mass flux. More generally one uses diffusion flux relative to the mixture's bulk velocity and specifies mass or mole fractions. The approximation is not a claim that an arbitrary pure-gas pressure gradient can remain at rest while only a Fick flux acts. In a mixture, collision cross-sections and relative molecular speeds, including the reduced mass, determine the species diffusivities.

Dependence on density and temperature. The kinetic estimates of gas transport coefficients predict, for fixed collision diameter and molecular heat capacity,Thus dynamic viscosity and thermal conductivity are approximately density independent in the dilute continuum regime. Reducing density reduces the number of carriers but increases their mean free path in the same proportion. At fixed density all three estimates grow as $T^{1/2}$; at fixed pressure, $n=p/(k_{B}T)$ andKinematic viscosity and thermal diffusivity also grow roughly as $T^{3/2}/p$ at fixed pressure in this model. Real interaction potentials make effective collision cross-sections temperature dependent, and internal excitations change $c_v$. Dense-gas correlations and boundaries invalidate the simple cancellation. In particular, when the mean free path approaches the apparatus scale, transport becomes free molecular; it cannot be described by extending the same local, pressure-independent Fourier and viscosity coefficients indefinitely.

Heat diffusion equation. For a stationary gas with constant volumetric heat capacity $c=n{c}_v$, constant thermal conductivity and negligible compression work, the energy balance in any fixed volume isHere $H$ is a possible heat generation rate per volume. The divergence theorem and Fourier's law yieldThermal diffusivity $\alpha$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$ and is of order $\overline{v}\lambda /3$. In a moving or compressing gas, the material derivative, pressure-volume work and viscous heating must also enter the internal-energy equation. The displayed pure diffusion equation describes the approximation in which those effects are absent or negligible.

Representative solutions. With no source, a steady one-dimensional slab between temperatures $T_a,T_b$ at $x=0,L$ has $T=T_a+(T_b-T_a)x/L$ and uniform heat flux $-\kappa (T_b-T_a)/L$. Thus a constant spatial gradient transports heat without changing local temperature in time when the boundaries continually supply and remove it.

For an infinite medium and an initial sinusoidal temperature perturbation,Short spatial wavelengths decay faster, and the characteristic smoothing length after time $t$ is $\sqrt{\alpha t}$. A finite slab with both ends held at $T_b$ has a sine expansion with factors $e^{-\alpha (n\pi /L{)}^{2}t}$; its slowest relaxation time is $L^2/({\pi }^2\alpha )$.

An instantaneous heat input $Q_A$ per unit transverse area at $x=0$ in an otherwise uniform infinite medium gives the Gaussian solutionIts integral $\int c(T-T_b)dx=Q_A$ conserves the deposited heat. The peak falls as $t^{-1/2}$ while its width grows as $\sqrt{t}$; diffusion redistributes energy instead of transporting a sharp front at a single velocity.

Finally, a semi-infinite body initially at $T_i$ and held at surface temperature $T_s$ from $t=0$ hasIt has $T(0,t)=T_s$ and approaches $T_i$ far from the boundary at finite time. These solutions demonstrate steady conduction, mode decay, spreading of localized heat and boundary-driven diffusion. Their idealized behaviour at extremely short times or arbitrarily long distances must be interpreted within the continuum regime rather than as an assertion of instantaneous microscopic communication.

For a reversible adiabatic process, an ideal gas obeys $T{V}^{\gamma -1}=\text{constant}$. A monatomic gas has heat capacity ratio $\gamma =5/3$. Taking the stated expansion factor to mean $V_f=9V_i$,This assumes the usual quasistatic isentropic process. Heat insulation alone does not determine the final temperature: an ideal-gas free expansion is also adiabatic but leaves the temperature unchanged. The numerical answer uses nine times the original volume, the conventional intended meaning of the source's “nine times larger”.

For the intended reversible adiabatic process, an ideal gas obeys $T{V}^{\gamma -1}=\text{constant}$. A monatomic gas has heat capacity ratio $\gamma =5/3$, soThe required extra assumption is that “adiabatic” denotes the usual quasistatic isentropic process. Absence of heat transfer alone would not fix the final temperature: an ideal gas undergoing free expansion into vacuum is also adiabatic but leaves its temperature unchanged.

For an ideal gas of fixed molecular composition, translational energy is supplemented by rotational, vibrational and eventually electronic excitations as their level spacings become thermally accessible. Activating more internal degrees of freedom generally increases the heat capacity. In a harmonic vibrational mode the contribution rises from nearly zero when $k_{B}T\ll \hslash \omega$ to its classical limit when $k_{B}T\gg \hslash \omega$; summing modes with different frequencies gives a smooth increase over an intermediate temperature range.

A monatomic gas has no molecular rotations or vibrations, so its mass-specific $c_p$ is nearly constant until electronic excitation or ionization becomes important: $B$ is then small. A simple diatomic gas can also have a small slope when its rotations are already active but its vibrational modes remain frozen. Large flexible polyatomic molecules have many internal modes, often with relatively low excitation energies, and can give a larger slope over their activation range.

The linear law is a local approximation, not an asymptotic result. A finite set of modes approaches finite classical heat-capacity contributions, whereas $A+BT$ grows without bound. At still higher temperatures dissociation and ionization change the composition and gas constant, invalidating the fixed-molecule ideal-gas model as well. The constants have units $[A]=\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $[B]=\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-2}$.

Use the shock frame, with upstream and downstream fluid velocities $u_1=u$ and $u_2=u/4$ pointing in the same direction. These are the Strong-shock Rankine-Hugoniot conditions for a nonrelativistic monatomic gas with adiabatic index $5/3$. The two scattering media approach one another at relative speed $U=u_1-u_2=3u/4$, to first order in $u/c$. An energetic electron repeatedly crosses the shock wave, scatters from magnetic turbulence carried by the local plasma, and returns. Each scattering approximately conserves its energy in the local fluid rest frame while changing its direction; comparing successive fluid frames produces a systematic energy and momentum gain. The energy comes from the converging bulk flow, through the moving magnetic field irregularities and their associated electric fields. This is diffusive shock acceleration, a first-order Fermi acceleration process.

Let $\eta$ be the positive cosine of the crossing angle with the shock wave's normal. The Lorentz transformation between fluid frames gives, for either crossing direction with the corresponding sign of the normal reversed,For an almost isotropic population, the number crossing a unit area is weighted by its normal velocity. Thus the normalized crossing distribution is $2\eta d\eta$, rather than a uniform distribution in $\eta$, andFor an ultrarelativistic electron, $E\simeq pc$ and its speed $v\simeq c$, soMore generally $dE/dp=v$ gives $⟨\Delta p⟩/p=2U/(3v)$, while a nonrelativistic kinetic energy has $⟨\Delta K⟩/K=4U/(3v)$. A complete upstream–downstream–upstream cycle contains two crossings and has ultrarelativistic fractional energy gain $u/c$.

The initially cold electrons need an injection stage: shock wave heating and small-scale plasma processes must first produce particles with speed much larger than $u$ that can return across the front. The $v\simeq c$ result is a description of that accelerated population; a cold electron cannot simply be assigned $E\simeq pc$ at its first encounter.

Work in the locally planar shock frame, taking $v_u,v_d>0$ towards the surface. For a stationary normal shock wave with negligible radiative loss inside the discontinuity, the Rankine-Hugoniot conditions for a perfect gas express conservation of mass, conservation of momentum, and conservation of energy per unit area asHere $h$ is specific enthalpy, including the pressure work; using only internal energy would omit that work. The gravitational potential is continuous across an infinitesimally thin front. For cold upstream gas, $P_u$ and $h_u$ are negligible, while a monatomic gas has $\gamma =5/3$ and $h=5P/(2\rho )$.

The hydrodynamic equations apply to flow parallel to the guiding magnetic field when its stress contributions cancel across the front. An oblique magnetohydrodynamic shock requires magnetic stresses and Poynting flux; a radiation-mediated front or appreciable escaping radiation requires the corresponding radiative momentum and energy flux terms. These are additional physics, not terms hidden in $h$ here.

For the stipulated Strong-shock Rankine-Hugoniot conditions, ${\rho }_d=4{\rho }_u$ and $v_d=v_u/4$. The ideal gas law is $P_d={\rho }_d{k}_B{T}_d/\mu$, with $\mu$ a dimensional mean particle mass. The specific enthalpy of a nonrelativistic monatomic gas is $5k_{B}T/(2\mu )$. Dividing the energy flux equation in d by the common mass flux givesThereforeAs an independent check, conservation of momentum gives $P_d={\rho }_u{v}_u^2-4{\rho }_u(v_u/4{)}^2=3{\rho }_u{v}_u^2/4$, and division by ${\rho }_d=4{\rho }_u$ gives exactly the same temperature. The numerical coefficient relies on $\gamma =5/3$ and the absence of dynamically important radiative or magnetic stresses.

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.

A free-fall timescale estimate follows from $R/t^2\sim GM/R^2$:Falling to a central point mass gives the more precise factor $\pi /(2\sqrt{2})$, or $1.77\times 10^3\mathrm{s}$; a distributed star changes an order-one coefficient. For any spherical mass distribution with radius $R$,The inequality is strict for an extended star; equality is approached by a thin shell at $R$. For an monatomic gas the virial theorem is $2U+\Omega =0$, so half the gravitational potential energy released by contraction heats the star and half can be radiated. The Kelvin-Helmholtz cooling time is therefore of orderA uniform-density model gives $3G{M}^2/(10RL)\simeq 9.4\times 10^6\mathrm{yr}$. The given bound fixes a scale, rather than an exact contraction history; both estimates are far shorter than the Sun's main-sequence lifetime.

A free-fall timescale estimate follows from $R/t^2\sim GM/R^2$:Falling to a central point mass gives the more precise factor $\pi /(2\sqrt{2})$, or $1.77\times 10^3\mathrm{s}$; a distributed star changes an order-one coefficient. For any spherical mass distribution with radius $R$,The inequality is strict for an extended star; equality is approached by a thin shell at $R$. For an monatomic gas the virial theorem is $2U+\Omega =0$, so half the gravitational potential energy released by contraction heats the star and half can be radiated. The Kelvin-Helmholtz cooling time is therefore of orderA uniform-density model gives $3G{M}^2/(10RL)\simeq 9.4\times 10^6\mathrm{yr}$. The given bound fixes a scale, rather than an exact contraction history; both estimates are far shorter than the Sun's main-sequence lifetime.

Truncating the singular isothermal gas profile at radius $r_0$ gives $M=2c_s^2{r}_0/G$ and boundary pressure $p_0=c_s^4/(2\pi G{r}_0^2)$. The internal energy of a monatomic gas is $U=3M{c}_s^2/2$, while shell integration gives gravitational self-energy $\Omega =-G{M}^2/r_0=-2M{c}_s^2$. Thus $2U+\Omega =3p_0\mathcal{V}$, with $\mathcal{V}=4\pi r_0^3/3$. The confining pressure supplies the virial surface term. This singular-central-density model differs from a regular pressure-confined isothermal sphere.

Two identical cold clouds approach at speeds $\pm V$ in their center-of-momentum frame. Reflection symmetry makes the uniformly compressed gas stationary. A shock moving outward at $S$ sees upstream speed $V+S$ and downstream speed $S$; their ratio equals the strong adiabatic compression $(\gamma +1)/(\gamma -1)$. Solving gives the displayed speed. For a monatomic gas $S=V/3$. Finite-cloud breakout ends the steady upstream supply and requires a subsequent boundary-release calculation.

Take a closed system consisting of one mole of a classical monatomic gas, idealized as an ideal gas with constant heat capacity at constant volume $C_V=3nR/2$. Stipulate amount of substance $n=1\mathrm{mol}$ and molar gas constant $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ for this calculation. Set initial temperature $T_0=300\mathrm{K}$ and volume $V_0=0.010{\mathrm{m}}^3$, and define $E_0=nR{T}_0=2.4942\mathrm{kJ}$ and $p_0=E_0/V_0=249.42\mathrm{kPa}$. The specified equilibrium endpoints areThus $B$ has volume $0.030{\mathrm{m}}^3$, temperature $600\mathrm{K}$ and pressure $166.28\mathrm{kPa}$. The fixed material inventory and ideal gas law make these complete endpoint specifications for this model. All numbers below are calculated values, rather than measurements of a particular gas.

Use a frictionless piston, ignore bulk mechanical energy changes, and allow only pressure-volume work. Each step is the limit of a reversible thermodynamic process: adjust loads by infinitesimal pressure differences, and supply heat through infinitesimal temperature differences. Heating with changing temperature requires a succession of matching thermal reservoirs; contact with a single hotter heat bath would give an irreversible finite difference. On a fixed-volume leg the piston is locked. These assumptions justify using the actual equilibrium path's pressure inAn endpoint chord is not the path; substituting a guessed straight line would compute a different thermodynamic process.^[ref]

The upper route heats at fixed volume from $A$ to $C=(V_0,2T_0,2p_0)$, then undergoes isothermal expansion at $2T_0$ to $B$. The lower route undergoes isothermal expansion at $T_0$ from $A$ to $D=(3V_0,T_0,p_0/3)$, then heats at fixed volume to $B$. The tuples use the same $(V,T,p)$ ordering as the endpoints. Vertical legs do zero expansion work, and the two curved legs have different pressures at every shared intermediate volume.

The internal energy of an ideal gas gives the same endpoint difference for both routes:On either fixed-volume heating leg, $W_{\mathrm{by}}=0$ and $Q=3E_0/2$. On each isothermal expansion, $\Delta U=0$, so the first law of thermodynamics gives $Q=W_{\mathrm{by}}$. The ideal gas law and the integral of $1/V$ then giveHere $\ln$ is the natural logarithm. Rounded to $0.001\mathrm{kJ}$, the upper route has $(Q,W_{\mathrm{by}},\Delta U)=(9.222,5.480,3.741)\mathrm{kJ}$ and the lower route has $(6.481,2.740,3.741)\mathrm{kJ}$. Subtracting separately rounded entries can differ by $0.001\mathrm{kJ}$; the unrounded balances are exact in the stipulated model.

The shaded area between the curves explains the entire difference:Following the upper route forward and the lower route backward makes a thermodynamic cycle with this positive net work and matching net heat. Its internal energy change is zero. The two open routes themselves are reversible, so their different work outputs should not be mistaken for a loss caused by entropy production: they receive different heat transfers at different temperatures.

Two other state functions provide independent endpoint checks. With enthalpy $H=5nRT/2$ in this model,To obtain the entropy change along either route, integrate $C_{V}dT/T$ on its heating leg, giving $C_V\ln 2$, and $nRdV/V$ on its isothermal expansion, giving $nR\ln 3$. Reordering the legs changes their heat totals but leaves the sum of their entropy changes unchanged. In particular, neither route's total $Q$ divided by its final temperature equals $\Delta S$, and neither total $Q$ equals $\Delta H$: both routes include an isothermal expansion, rather than a single fixed-pressure heating step.