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For a fully ionized, nonrelativistic monatomic ideal gas, $\gamma =5/3$ and the virial theorem becomesSince $U$ increases with its particle-weighted mean temperature, the equilibrated self-gravitating object has negative heat capacity: adding total energy allows expansion, makes $\Omega$ less negative, and reduces $U$. Conversely, radiating energy permits contraction and raises the internal temperature. This refers to a sequence of gravitational equilibria, not to the positive local heat capacity of a parcel at fixed volume.

In an ordinary nondegenerate nuclear-burning star, a rise in nuclear heating tends to expand and cool the burning region, lowering its temperature-sensitive reaction rate. Reduced heating leads to contraction and reheating. This expansion feedback is the nuclear thermostat; it can fail in degenerate matter where pressure responds weakly to temperature. During contraction of a pre-main-sequence star, gravitational binding supplies the radiated luminosity and heats the interior. For a fixed-structure scaling, half the released gravitational energy increases $U$ and half escapes as radiation, until hydrogen burning can support a long-lived main sequence equilibrium.

Displace a small parcel upward, with negligible heat exchange, rapid pressure balance, and the same composition as its surroundings. Define the environmental gradient $\nabla =d\log T/d\log P$. An adiabatic ideal gas parcel obeys $P{\rho }^{-\gamma }=\mathrm{constant}$ and hence $T\propto P^{(\gamma -1)/\gamma }$, giving ${\nabla }_{\mathrm{ad}}=(\gamma -1)/\gamma$.

The upward displacement has $\delta \log P<0$. If $\nabla <{\nabla }_{\mathrm{ad}}$, the parcel's temperature changes more negatively than the environment's. At their common pressure it is cooler, denser and driven back down. Therefore the Schwarzschild criterion for stability isThe reversed inequality produces buoyant growth, and equality is neutrally stable in this approximation. A composition gradient requires the Ledoux criterion; pressure equilibrium alone does not ensure equal mean molecular weight.

Multiply hydrostatic equilibrium by $4\pi r^{3}dr$ and integrate from centre to surface. Integration by parts, neglecting external surface pressure, givesFor an ideal gas with a constant adiabatic index, $U=\int PdV/(\gamma -1)$. ConsequentlyA fully ionized, nonrelativistic monatomic gas has $\gamma =5/3$, provided gas rather than radiation or degeneracy dominates the thermal equation of state. It then obeys $\boxed{2U+\Omega =0}$ and $E=U+\Omega =-U=\Omega /2$.

Added nuclear heat initially raises pressure and causes expansion. The expansion makes gravitational energy less negative and can lower the temperature of the self-gravitating gas, reducing the temperature-sensitive fusion rate. Conversely, a small contraction can raise the core temperature and increase burning. This negative feedback explains stable main-sequence burning in an ordinary nondegenerate stellar core; a degenerate core need not respond the same way.

Before hydrogen ignition, a cloud loses energy by radiation and contracts. Its gravitational energy decreases; for the ideal-gas virial relation, roughly half the released binding energy heats the gas and half is radiated. Heating continues until nuclear reactions can balance the luminosity, explaining the Kelvin-Helmholtz phase of pre-main-sequence evolution.

The effective adiabatic response varies with ionization, molecular dissociation, radiation pressure and degeneracy. Partial ionization can absorb compression energy and lower the local first adiabatic exponent below $4/3$, encouraging compressibility, convection and pulsational instability. A local outer layer with $\gamma \le 4/3$ does not by itself prove global collapse: ordinary stars have ionization layers of that kind. The dynamical stability criterion concerns a suitable pressure-weighted global response and its coupling to the eigenmode; an extended star with an effectively global exponent below $4/3$ can be dynamically unstable.

The radiation pressure at a photospheric temperature near $5800$ K is only about $0.3$ Pa. For an illustrative photospheric gas density $3\times 10^{-4}\mathrm{kg}{\mathrm{m}}^{-3}$ and mean molecular weight near $1.3$, the ideal gas pressure is about $10^4$ Pa. Thus gas pressure dominates those outer layers by many orders of magnitude. Both the temperature and density are needed: low temperature alone is not a complete comparison.

For homologous stars, hydrostatic and ideal-gas estimates give $T\sim \mu m_{H}GM/(k_{B}R)$ and $\rho \sim M/R^3$. Thereforewith a structural coefficient independent of mass only under the homology assumption. As mass rises, radiation becomes increasingly important. A rough solar interior normalization of $10^{-3}$-$10^{-2}$ suggests equality after a mass increase by tens, while realistic radiation-dominated structure and composition shift the scale; stars of order $10^2{M}_{\odot }$ are a useful estimate for strong radiation support. The very small photospheric ratio must not be used as a universal normalization for their whole interiors.

Radiation also transfers outward momentum. Comparing radiative acceleration $\kappa L/(4\pi r^{2}c)$ with gravity gives the Eddington luminosityNear this scale, radiatively driven winds and envelope instabilities can remove matter or impede further accretion. Opacity, composition, rotation, mass loss and formation history affect the limiting stellar mass. The Eddington condition is a luminosity-to-mass constraint, not a unique composition-independent maximum stellar mass.

Use characteristic stellar values in the supplied radiative temperature gradient: $T_c/R\sim \kappa \rho L/(R^2{T}_c^3)$. Hence $L\propto T_c^{4}R/(\kappa \rho )$. With Kramers opacity, $\kappa ={\kappa }_0\rho T_c^{-3.5}$,Hydrostatic support by an ideal gas gives $T_c\propto \mu M_s/R$, or $R\propto \mu M_s/T_c$. Substitution then yieldsThe result assumes homologous density and temperature profiles, comparable structural constants, gas-pressure support and radiative transport throughout. It is not simultaneously an electron-scattering-opacity result; replacing Kramers opacity changes the exponents. Convection, degenerate cores or a different opacity regime limit quantitative application to real evolved helium stars.

Work in the rest frame of the planar shock, with nonzero mass flux $j=\rho u$. Conservation of mass gives $[j]=0$; the tangential and normal components of balance of linear momentum give $[jv]=0$ and $[p+ju]=0$. Thus $[v]=0$. For an ideal gas, specific internal energy is $e=c_{v}T=p/[(\gamma -1)\rho ]$, and specific enthalpy is $e+p/\rho =\gamma p/[(\gamma -1)\rho ]$. Total energy flux is $j[(u^2+v^2)/2+e+p/\rho ]$. Its continuity, with $v$ already continuous, givesPut $c^2=\gamma p/\rho$ and $M=u/c$. Since $p=ju/(\gamma M^2)$, the normal component of balance of linear momentum says $u[1+1/(\gamma M^2)]$ is constant. HenceThe energy equation says $u^2[1/2+1/((\gamma -1)M^2)]$ is constant, givingSet $a=M_{-}^2$, $b=M_{+}^2$, square the first relation and equate to the second. After clearing positive denominators the equation factors asThe first factor is the unchanged state. For a genuine shock the second factor vanishes, yieldingThe physical selection principle is entropy increase in the direction of flow. For $u_{\pm }>0$, the upstream side is $x<0$ and the downstream side is $x>0$. A compressive entropy-admissible shock has $\boxed{M_{-}>1,M_{+}<1}$; interpreting the printed “ahead” and “behind” spatial labels must respect these explicit side assignments. The expansion-shock branch would violate the second law.

Substitution gives the velocity ratioSince $v$ is unchanged while the normal component decreases, $|v|/u$ increases. Thus an oblique flow bends towards the shock plane, equivalently away from its normal. If $v=0$ there is no angular bending; the flow stays normal and simply slows. The tangential continuity, not a change in the tangential speed, causes the deflection.

Bose-Einstein condensation is macroscopic occupation of a single lowest-energy one-particle state. For a homogeneous three-dimensional ideal gas of conserved massive bosons, set that energy to zero. The excited-state population isHere $g_{3/2}(z)={\sum }_{r\ge 1}{z}^r/r^{3/2}$. As $\mu \to 0^{-}$, the excited states can hold only $V\zeta (3/2)/{\lambda }_T^3$ particles. Additional particles occupy the ground state. ThusThe isolated ground state must be counted separately from the continuum density of states; otherwise one misses the condensate. A dilute atomic gas is a physical realization of the quantum-degeneracy regime, but trapping geometry and interactions modify the homogeneous ideal-gas formulas.^[ref]

In the homogeneous ideal gas, one and two dimensions do not have finite-temperature condensation in the thermodynamic limit because the low-energy population integral diverges. Finite size or a trap changes that conclusion's assumptions. Condensation and superfluidity are distinct: the ideal quadratic gas has zero Landau critical velocity, while interactions can support a sound mode and superfluid flow. Neither zero viscosity nor phase-rigid flow follows from the occupation formula alone.

Use a three-dimensional ideal gas at the same number density n for both isotopes. An unpolarized helium-3 gas has spin degeneracy two, henceHelium-4 has a single spin state, and the ideal Bose-gas saturation condition givesEliminating n, using $m_3/m_4\simeq 3/4$ and $\zeta (3/2)=2.612$, yieldsThe fermionic spin degeneracy must be included. This is an ideal-gas estimate; strong interactions in liquid helium prevent identifying it with an exact experimental transition temperature.

At $T\ll T_F$, Pauli exclusion restricts thermally changed occupations to an energy interval of order $k_{B}T$ about the Fermi energy. The affected fraction is of order $T/T_F$; their typical excitation energy is of order $k_{B}T$, so the thermal energy correction is proportional to $T^2$ and the heat capacity is linear in $T$.

More precisely, with $g(E_F)$ the total density of one-particle states per unit energy, including spin,For a three-dimensional ideal gas with quadratic dispersion, $g(E_F)=3N/(2E_F)$, henceThe fixed-particle-number Sommerfeld expansion gives $U=3N{E}_F/5+({\pi }^2/4)N(k_{B}T{)}^2/E_F+\cdots$. The temperature dependence of the chemical potential must be included when obtaining this expression; holding it artificially fixed changes the intermediate energy correction.

Put $\beta =1/(k_{B}T)$ and define the thermal de Broglie wavelength of species $\alpha$ byIgnoring the internal degrees of freedom as stipulated, each molecule's translational canonical partition function is $V/{\lambda }_{\alpha }^3$. The two species do not interact, so their partition functions multiply; the treatment of indistinguishable particles requires a separate factorial for each species, not a factorial of the total particle number. ThusUsing $F=-k_{B}T\ln Z$ and Stirling's approximation gives the extensive Helmholtz free energyThese are ideal gas expressions, valid for $n_{\alpha }{\lambda }_{\alpha }^3\ll 1$, with $n_{\alpha }=N_{\alpha }/V$. The energy zero has been chosen so that separated CO and oxygen molecules have zero internal reference energy.

Substituting the chemical potentials into the equilibrium relation givesExponentiation and the positive square root give the law of mass action in number-density form:All three thermal de Broglie wavelengths have the same $T^{-1/2}$ dependence. Define the temperature-independent mass factorIf $f_{\alpha }=n_{\alpha }/n$, this exact equilibrium relation can be writtenIt already has the requested dimensional structure: $n^{3/2}{\lambda }_c^{3/2}$ has units ${\mathrm{m}}^{-3}$. Its coefficient, however, depends on the conserved elemental composition through the species fractions.

For the intended case of a gas obtained by dissociating initially pure carbon dioxide, conservation of the elemental ratio gives $n_{\mathrm{CO}}=2n_{{\mathrm{O}}_2}$ throughout. Henceand the exact relation within this ideal-mixture model isAt high temperature with $k_{B}T\gg ϵ$ and in the dilute ideal gas regime, $n{\lambda }_c^3\ll 1$, this equation is self-consistently dissociation dominated:Thus $n-n_c\simeq n$, and to leading orderUsing nominal molecular masses $28,32,44$ in common mass units gives $A\simeq 0.963$, confirming that it is of order unity. The exponential has been retained as the leading binding correction, although it tends to one as $T\to \infty$.

The printed question does not explicitly fix the elemental composition. Without the initially-pure-$\mathrm{C}{\mathrm{O}}_2$ assumption, the universal result is the boxed mass-action relation, and $A=B{f}_{\mathrm{CO}}\sqrt{f_{{\mathrm{O}}_2}}$ approaches a composition-dependent constant when both dissociation products have nonzero high-temperature fractions. It need not be of order unity for an arbitrarily unbalanced mixture. Also, $k_{B}T\gg ϵ$ alone does not justify replacing $n-n_c$ by $n$ at arbitrarily high density; the dilute classical regime is part of the model.

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

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

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

In the one-mole Van der Waals equation, $b>0$ represents excluded volume: the volume available to molecular motion is $V-b$, rather than $V$. Subtracting $b$ in the denominator therefore increases the repulsive contribution to pressure. The $a>0$ term represents intermolecular attraction, which reduces the outward momentum flux at a wall and hence appears as the negative correction $-a/V^2$. The two minus signs do not mean that both physical effects lower pressure.

For a reversible isothermal process along a homogeneous gas branch with $V_2>V_1>b$, the work done by the gas isTo find the internal energy change without guessing a caloric law, combine $dU=TdS-pdV$ with the Helmholtz free energy Maxwell relation $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$. This givesThereforeThe attractive contribution $-a/V$ becomes less negative as the gas expands: separating attracting particles requires energy. This differs from the ideal gas result $\Delta U=0$ at fixed temperature. The required heat is $Q=\Delta U+W_{\mathrm{by}}=RT\ln [(V_2-b)/(V_1-b)]$. Below the critical temperature, an equilibrium path crossing a liquid–gas coexistence region needs the equilibrium coexistence construction rather than the unstable homogeneous Van der Waals isotherm.

The Clapeyron equation for molar latent heat $L$ is $dp/dT=L/[T(V_{\mathrm{vap}}-V_{\mathrm{liq}})]$. Neglect the liquid molar volume, treat the vapor as an ideal gas, and approximate $L$ as constant over the interval. ThenUsing one atmosphere as $101325\mathrm{Pa}$ givesThis is the effective molar enthalpy of vaporization inferred from the Integrated Clausius-Clapeyron relation. If $L$ varies with temperature, the inferred value is weighted by $1/T^2$, not an unweighted arithmetic mean. Real helium is not exactly an ideal gas over the whole interval, and its liquid changes superfluid state within it, so this is an estimate under the stated approximations rather than a precise temperature-independent material constant.

For a reversible adiabatic process of an ideal gas, $\delta Q=0$ and $dU=-pdV$. At fixed composition and constant sample heat capacity at constant volume $C_V$,Mayer's relation gives $n_{\mathrm{mol}}R=C_p-C_V$, so $T{V}^{\gamma -1}$ is constant andThe finite-path power law requires approximately constant heat capacities as well as reversibility. When rotational or vibrational populations change appreciably, use $\int C_V(T)dT/T+n_{\mathrm{mol}}R\ln V=\mathrm{constant}$ instead; the local isentropic pressure exponent remains $\gamma (T)$.

For an intact heteronuclear diatomic gas, quantum excitation progressively activates molecular motion. With rotational and vibrational modes frozen, the three translational degrees of freedom give $C_{V,\mathrm{m}}=3R/2$ and $\gamma =5/3$. Two active rotations add $R$, giving $C_{V,\mathrm{m}}=5R/2$ and $\gamma =7/5$. One active harmonic vibration contributes a further $R$, through kinetic and potential terms, giving $C_{V,\mathrm{m}}=7R/2$ and $\gamma =9/7$.

For molecular moment of inertia $I={\mu }_{\mathrm{red}}{r}_e^2$, the rigid rotor levels are $E_J={\hslash }^{2}J(J+1)/(2I)$, so the rotational excitation scale isThe first gap corresponds to $2{\Theta }_{\mathrm{rot}}$, so the crossover occurs at temperatures of that order. For bond force constant $k_s$ and reduced molecular mass ${\mu }_{\mathrm{red}}$, the vibrational scale isTypically ${\Theta }_{\mathrm{vib}}\gg {\Theta }_{\mathrm{rot}}$. Real gases may condense at the low-temperature end or dissociate and excite electronic states at the high-temperature end; the three plateau values describe the ideal intact-molecule model.

For the piston apparatus, mechanical equilibrium requires $p_{0}A=p_{\mathrm{air}}A+mg$, hence $p_0=p_{\mathrm{air}}+mg/A$. Let $x$ be a small upward displacement, so $V=V_0+Ax$. The local isentropic compressibility givesConsequently the adiabatic gas-spring piston period isThis neglects friction, gas inertia, leakage and dissipation, with internal equilibration much faster than the oscillation and external thermal equilibration much slower. The small oscillation probes $\gamma$ locally; it need not keep $\gamma$ constant over a large temperature interval.

With $p_{\mathrm{air}}=101325\mathrm{Pa}$, the supplied dimensions give $p_0=101351.4\mathrm{Pa}$ and $B=2.20566\times 10^{-5}{\mathrm{s}}^2$. ThusHelium is close to the monatomic ideal gas value $5/3$; NO is close to the diatomic gas value $7/5$, with rotations active and vibration largely frozen. The numbers are approximate experimental inferences, rather than exact rational values.

Use the pressure correction to a mineral reaction for the solids, and the gas thermodynamic activity $f/P_0$ for water. These are separate contributions: do not also add a gas-volume pressure correction after including $RT\ln (f/P_0)$. For this reaction,At $7000\mathrm{bar}$ with pure solids and an ideal gas, $f=P$ and the solid-activity ratio is one. ThereforeandThe pressure unit in the solid term is bar, matching the supplied $\mathrm{J}\mathrm{ba}{\mathrm{r}}^{-1}$. The gas logarithm is dimensionless because $P/P_0=7000$. High water chemical fugacity opposes dehydration and raises the equilibrium temperature substantially.

Apply the pressure correction to a mineral reaction to the solids and use the chemical fugacity of the water product. For pure solids and water vapour treated as an ideal gas,The gas contribution is already included by the logarithm and must not be counted again as an independent volume correction. With pressures in bars,

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

Use the pressure correction to a mineral reaction for the solids, with $P_0=1\mathrm{bar}$. The water contribution must instead use its chemical fugacity. For pure solid phases and water vapour treated as an ideal gas, $f_{{\mathrm{H}}_2\mathrm{O}}=P$, givingThe last term already accounts for the gas-pressure contribution; adding another gas-volume correction would double count it. The pressure ratio inside the logarithm is dimensionless.

With pressure expressed in bars, the solid correction is $-2.28(4500-1)=-10257.72\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$. ThusUsing $4500$ rather than $4499$ for the solid pressure increment would silently shift the specified one-bar reference.