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For a fixed number $N$ of particles, the ideal gas equation of state isIts internal energy depends only on temperature, andWhen the heat capacity at constant volume is constant, a choice of the energy zero gives

In the gas-pressure limit $\beta \to 1$, the mixture becomes a monatomic ideal gas, soIn the radiation-pressure limit $\beta \to 0$,whereas ${\chi }_{\rho }=\beta \to 0$ and thereforeThe divergence reflects the singular constant-pressure heat capacity of pure radiation: its pressure fixes its temperature independently of density.

Combining $\Phi =-pV$ with part (i) and the expression for $⟨N⟩$ gives the ideal-gas equation of state

The one-particle translational partition function isFor a classical ideal gas,and hence

The zero-temperature internal energy isUsing part (i),Consequently the adiabatic equation of state isThe pressure is degeneracy pressure: it remains nonzero at zero temperature because of the Pauli exclusion principle. By comparison, a classical ultrarelativistic ideal gas at fixed temperature obeys $pV=N{k}_{B}T$, so its corresponding volume exponent is one and its pressure vanishes as $T\to 0$.

For $N$ particles, the ideal gas equation of state isThe internal energy of an ideal gas depends only on temperature and obeysFor constant heat capacity and a convenient choice of energy zero,

The one-dimensional mass flux and momentum flux are respectivelyIn a steady flow, the conservation laws make both the mass flux and the given energy fluxconstant. Wherever $J_m\ne 0$, their ratio is therefore constant:where $W=p/(\gamma -1)$ is the internal energy per unit volume of the ideal gas. This is the required steady-flow energy integral.

For a family in stellar homology, dimensionless radial profiles are fixed. Mass conservation and hydrostatic equilibrium then give the central scalingsThe ideal gas equation of state, with fixed mean molecular weight, consequently giveswhere fixed dimensional constants such as $G$ are suppressed in homology relations.

The proton–proton chain law $ϵ\propto \rho T^{7/2}$ gives the nuclear luminosityOn the other hand, radiative diffusion in a star givesUsing the Kramers opacity law ${\kappa }_c\propto {\rho }_c{T}_c^{-7/2}$,Thermal equilibrium requires $L_{\mathrm{nuc}}=L_{\mathrm{rad}}$. Their common mass factor cancels, leaving $R^{-13/2}\propto R^{-1/2}$, and hence

This model captures the gas-pressure support, pp-chain burning, and strongly mass-dependent luminosity of the lower main sequence, including the Sun approximately. Its fully radiative assumption is an idealization: the Sun has a convective envelope, and sufficiently low-mass red dwarfs become largely or fully convective, so real radii are not exactly constant.

Write the ideal gas law as $P=\mathcal{R}\rho T/\mu$. Since $P=g\Sigma$,The prescribed opacity and burning laws becomeThe radiative equation is consequentlyWith $y=T^7$ and $x={\Sigma }^2/2$,The energy equation similarly givesDifferentiating the first relation therefore produces the nonlinear ordinary differential equation

At the idealized zero-temperature surface, $x=0$ and $y=0$. The outward flux there is $F_s=L/(4\pi R^2)$, soAt the base, $x_0={\Sigma }_0^2/2$ and $y=T_0^7$. The core supplies no luminosity in this model, so all flux has been generated in the overlying hydrogen envelope and $F(x_0)=0$. Hence

Let the mean particle mass be $\mu m_p$. For a monatomic ideal gas with $\gamma =5/3$,A uniform sphere of mass $M=(4\pi /3){\rho }_0{r}^3$ has Newtonian gravitational energyAt the virial threshold, the pressure term $3\int PdV=3M{k}_{B}T/(\mu m_p)$ balances $|W|$. ThereforeEliminating $r$ gives the Jeans massThe numerical coefficient depends on the convention used to identify a finite cloud with a Jeans mode. For example, assigning the mass inside a sphere of radius half the standard Jeans instability wavelength givesBoth conventions have the physically invariant scaling

For a homologous star, hydrostatic equilibrium, mass conservation, and the ideal gas equation give the central scalingswhere $\mu$ is the mean molecular weight. Integrating the nuclear energy-generation law over a fixed homologous profile givesRadiative stellar structure gives independentlyEquating the two luminosities yieldsAt fixed zero-age composition, the stars are therefore homologous with

The effective temperature satisfies $L=4\pi R^2\sigma T_e^4$, so $T_e\propto M^{3/8}$. The zero-age main sequence consequently hasIt is a steep line rising toward high luminosity and high temperature on a Hertzsprung-Russell diagram.

For fully ionized hydrogen and helium with $Y=1-X$,while $\kappa \propto 1+X$. At fixed mass,and the corresponding radius relation is

At the pure-hydrogen zero-age point $X=1$,whereasHenceandAs hydrogen is consumed, $X$ falls, so both $L$ and $T_e$ rise. In the usual diagram with temperature increasing leftward, the evolutionary track initially moves upward and leftward from the zero-age main sequence.

In harmonic stellar gravity, an ideal gas with a polytropic equation of state of index $n>0$ has a quadratic temperature profile and finite semi-thickness $H^2=2(n+1)\mathcal{R}{T}_0/{\Omega }^2$. This differs from the polytropic vertical structure of a self-gravitating disk.

A thin radiative envelope with constant enclosed mass, luminosity and mean molecular weight, ideal gas pressure and opacity $\kappa ={\kappa }_0{\rho }^n{T}^{-s}$ has $a=n+1$, $b=n+s+4$, and $C=16\pi a_{\mathrm{rad}}cGM/[3{\kappa }_0(\mu m_u/k_B{)}^{n}L]$. Combining radiative diffusion in a star with hydrostatic equilibrium gives ${\nabla }_{\mathrm{rad}}=P^a/(C{T}^b)$, so $P^a-P_0^a=(aC/b)(T^b-T_0^b)$ for nonzero exponents. Zero exponents give logarithmic limits. The model must be stopped or modified if the Schwarzschild criterion predicts convection.

A pressure scale height is the local distance over which pressure would change by a factor $e$ if its logarithmic gradient were constant. In hydrostatic equilibrium, $H_P=P/(\rho g)$; for an ideal gas it is $k_{B}T/(\mu m_{u}g)$. A small logarithmic pressure interval has depth approximately $H_P\Delta \log P$.

In a negligible-mass white dwarf envelope, take enclosed mass and luminosity constant, neglect radiation pressure, and use an ideal gas with the Kramers opacity law. Dividing hydrostatic pressure support equation by radiative diffusion in a star gives $PdP\propto (M/L)T^{15/2}dT$. Neglecting the surface integration constant produces the displayed pressure and mass density laws. It is a deep-envelope approximation, not a finite-temperature photospheric boundary condition.

For particles with pair potential $U$, the second virial coefficient iswith the integration measure adapted to the available relative configurations. Repulsion contributes positively and attraction negatively.

For fixed-composition monatomic ideal gas plus equilibrium blackbody radiation, let $\mathcal{R}=k/(\mu H)$ and $\beta =P_g/P$. The mass-specific heats are $c_P=\mathcal{R}(32-24\beta -3{\beta }^2)/(2{\beta }^2)$ and $c_V=\mathcal{R}(24-21\beta )/(2\beta )$. The specific-heat ratio is ${\Gamma }_1/\beta$, not generally any one of the stellar adiabatic exponents. Differentiate at constant total pressure rather than at constant $\beta$.

The gas-pressure fraction describes the relative support by an ideal gas and radiation pressure. For monatomic gas plus equilibrium radiation, it determines the adiabatic exponents of a monatomic gas-radiation mixture. Its value at the unperturbed state may be specified, but it generally changes during an isentropic parcel displacement.

Take an invented diatomic molecule with a Morse potential, setting the separated-atom energy to zero:Here $a$ is an inverse length, $r_e$ the equilibrium bond length, and ${\omega }_e=a\sqrt{2D_e/\mu }$, with reduced mass $\mu$. These stipulated parameters describe a simple model, not a measured chemical substance. For the one-dimensional Morse oscillator, energy eigenvalues above the minimum areConsequently,The harmonic estimate $A/2=0.1000\mathrm{eV}$ is slightly larger than this model's exact zero-point energy. The physical bound-level restriction matters: the quadratic expression is not an infinite ladder of vibrational quantum states.

For a separate thermal illustration using this same stipulated $D_0$, assume an ideal gas, fully active molecular rotation of the parent, negligible thermal molecular vibration, and negligible electronic excitation for all species. Each monatomic product contributes $\delta h=5RT/2$; the diatomic molecule contributes $7RT/2$. Thus the net increment is $3RT/2$, with molar gas constant $R$. At $T=300\mathrm{K}$,This last result follows from the stated thermal model, not a universal correction for every chemical bond. Frozen rotations, accessible excited states, different fragment composition, or a condensed phase change the thermal increments.

Temperature is the intensive variable thermodynamically conjugate to entropy; at equilibrium it determines the direction of heat transfer.