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The stellar adiabatic exponents describe pressure and temperature responses to an isentropic density change:

In stellar ideal gas equations, the stellar gas constant $\mathcal{R}=k_B/m_u$ converts a dimensionless mean molecular weight $\mu$ into the specific thermal relation $P=\mathcal{R}\rho T/\mu$. Here $m_u$ is the reference atomic mass unit, commonly approximated by the proton mass in simplified hydrogen-helium models. It has units of energy per mass per temperature. It is the molar gas constant divided by the molar mass corresponding to $m_u$, rather than the molar gas constant used with a dimensional molar molecular weight.

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.

Taking the curl of axisymmetric hydrostatic equilibrium with centrifugal acceleration gives $\varpi {\partial }_z{\Omega }^2=(\nabla p\times \nabla \rho {)}_{\varphi }/{\rho }^2$. For an approximately spherical ideal gas with constant composition, use $\nabla \rho =\nabla p/c_s^2-\rho \nabla S/c_p$ to obtain the displayed relation. Latitudinal entropy differences can support noncylindrical differential rotation. A barotropic fluid gives cylindrical rotation in this leading balance; the relation does not determine the transport stresses or circulation rate.

For a self-gravitating fluid star with surface pressure $P_s$, volume $V$, kinetic energy $T$, gravitational energy $\Omega$, and $I=\int r^{2}dm$,

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.

For adiabatic perturbations of a static perfect gas, pressure obeysThe stationary balance is $\nabla (p_0+\chi )=0$. Density and entropy need not be spatially uniform. Expanding the divergence exposes the gradient terms that distinguish this equation from the homogeneous acoustic pressure wave equation. Its divergence form yields weighted acoustic Green-function reciprocity.

Temperature labels thermal equilibrium: two internally equilibrated bodies have the same temperature when contact permitting only heat exchange produces no net heat transfer. The zeroth law of thermodynamics makes the comparison transitive. If bodies A and B separately reach thermal equilibrium with the same thermometer without appreciably disturbing either body, they agree with each other. This is what lets a gas thermometer compare a metal block with water, despite their different internal energies and heat capacities. Insulation alone makes no such comparison: it can prevent heat transfer between bodies at different temperatures. ^[ref]

The zeroth law of thermodynamics establishes equality comparisons, but does not choose a numerical scale. Thermodynamic temperature supplies an absolute scale consistent with Carnot's theorem and entropy. For a simple compressible system at thermodynamic equilibrium with fixed particle number $N$ and composition,Here $U$ is internal energy, $S$ is entropy, and $V$ is volume; the second equality assumes the derivative can be inverted. Temperature is a thermodynamic conjugate variable to entropy and belongs to the intensive and extensive thermodynamic quantities as an intensive quantity. Its partial derivative tells us how much internal energy must change per unit change of entropy under the stated constraints. More general systems require their other mechanical and material variables to be held fixed too. ^[ref]

For an infinitesimal quantity $\delta Q>0$ of heat transferred from A to B, with no work or matter exchange, the equilibrium entropy changes giveFor ordinary positive temperatures, spontaneous heat transfer from $T_A>T_B$ increases total entropy; thermal equilibrium has $T_A=T_B$. This comparison uses changes in entropy, rather than the total internal energy contained in either body. A small hot object can have less internal energy than a large cool one. See entropy maximization under thermal contact for the equilibrium criterion and negative temperature for the special case of bounded energy spectra.

Express absolute thermodynamic temperature $T$ in kelvin, and a Celsius reading $t_C$ in degree Celsius:The offset is exact: $300\mathrm{K}$ corresponds to $26.85{}^{\circ }\mathrm{C}$, and $400\mathrm{K}$ to $126.85{}^{\circ }\mathrm{C}$. Their absolute temperature ratio is $4/3$; dividing their Celsius readings gives a different number. Thus gas thermometer pressure ratios and Carnot's theorem require thermodynamic temperature. The modern SI definition of the kelvin fixes the Boltzmann constant $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$; a reference bath realizes or transfers the scale rather than defining that constant.^[ref]

Temperature is not universally the mean kinetic energy of a particle. In a classical, nonrelativistic ideal gas at thermal equilibrium, the equipartition theorem gives the mean translational kinetic energy relative to the gas's bulk motion:Here $m$ is particle mass, $\mathbf{v}$ its velocity, and $\mathbf{u}$ the mean flow velocity. In this regime, different particle masses at the same temperature have equal mean translational kinetic energies but different characteristic speeds. A moving gas can also carry bulk kinetic energy without being hotter in its rest frame. Internal molecular modes and quantum statistics require further care; the equipartition theorem has classical assumptions, whereas the equilibrium entropy definition applies more widely. ^[ref]

For two ideal gas layers at a common interfacial pressure, the lower layer has smaller density precisely when $M_{-}/T_{-}<M_{+}/T_{+}$, where $M$ is mean molar mass. That ordering is unstable to an infinitesimal interchange. A heavy condensable constituent can therefore inhibit thermal convection in a light background gas even when the lower layer is hotter. This local criterion assumes no phase change during the initial displacement.

The characteristic thermal energy per microscopic degree-of-freedom scale is $k_{B}T$. Equipartition gives $k_{B}T/2$ per independent quadratic term under its classical assumptions.

Thermal expansion is a change in a body's dimensions as its temperature changes under a specified mechanical constraint. A heated gas under a freely moving loaded piston usually increases its volume at fixed pressure; heating the same gas in a rigid vessel instead raises its pressure. The coefficient of thermal expansion quantifies the first response. Expansion can be negative, so the name does not assert that every substance grows on heating.

For an ideal gas at fixed pressure and particle number, $V=N{k}_{B}T/P$ gives $V_2/V_1=T_2/T_1$. Heating from $300\mathrm{K}$ to $330\mathrm{K}$ therefore increases the volume by exactly $10\%$ within the ideal gas model. For a solid, specify whether the measurement concerns length or volume: for uniform expansion in all three directions, the volumetric coefficient is three times the linear thermal expansion coefficient, because $V\propto L^3$. Direction-dependent expansion requires the separate dimensional responses. ^[ref]

For a simple compressible system at fixed composition and particle number, the internal pressure is the partial derivative of internal energy with respect to volume at fixed temperature. From $dU=TdS-pdV$ and the Maxwell relation $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$, it has the measurable expression in the title. It is distinct from the mechanical pressure $p$.

For an ideal gas, $p=nRT/V$ gives ${\pi }_T=0$, so its internal energy depends only on temperature at fixed amount of gas. For the Dieterici equation $p=nRT\exp [-an/(VRT)]/(V-nb)$, direct differentiation givesThus an equation of state determines this volume sensitivity even without determining the full caloric function $U(T,V)$.

An equilibrium thermodynamic state specifies sufficient independent macroscopic properties, together with material identity and constraints, to determine the other equilibrium properties. For a uniform fixed-composition ideal gas, its amount of substance, temperature and volume determine its pressure through the ideal gas law. These properties specify a state, without specifying a thermodynamic process that prepared it.^[ref]

For a dilute classical ideal gas whose atoms carry independent nondegenerate internal levels $0,ϵ$, the canonical partition function factorizes into translation and $(1+e^{-\beta ϵ}{)}^N$. The upper-level probability is $p=(1+e^{\beta ϵ}{)}^{-1}$. Taking two derivatives of its logarithm gives the title sum of independent variances. The internal contribution freezes exponentially at low temperature and tends to $N{ϵ}^2/4$ at high temperature. The translational classical result cannot be continued through quantum degeneracy without changing the model.

An ideal gas whose particles move in an area $A$ with two independent translational momentum components. In its dilute classical limit the one-particle translational partition function is $q=A/{\Lambda }^2$, where $\Lambda$ is the thermal de Broglie wavelength, and $Z_N=q^N/N!$. The translational internal energy is $N{k}_{\mathrm{B}}T$, and the surface pressure conjugate to area is $N{k}_{\mathrm{B}}T/A$. A mobile dilute adsorbed gas is an example; a fully occupied array of fixed adsorption sites has different state counting.

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.

For a uniform ideal gas sphere, set $\rho =mn$, $p=n{k}_{B}T$ and $M=4\pi \rho R^3/3$. Building the sphere by concentric shells gives gravitational potential energy $W=-3G{M}^2/(5R)$. During a small homologous contraction, released gravitational energy balances the work against gas pressure when $3G{M}^2|dR|/(5R^2)=4\pi R^{2}p|dR|$, or $GMm/R=5k_{B}T$. Eliminating $R$ gives the displayed mass. At the same initial mass density and temperature, larger masses have stronger gravitational contraction than pressure support in this estimate. Its numerical coefficient belongs to the uniform-sphere pressure-work model; the infinite-medium Jeans instability and a pressure-confined isothermal sphere use different equilibrium and boundary assumptions.

For a dilute ideal gas vapor whose specific volume greatly exceeds that of its condensate, the Clausius-Clapeyron relation integrates to the displayed formula. If $L(T)=L_{\ast }(T_c-T)/(T_c-T_{\ast })$, the result is $\ln (p/p_{\ast })=L_{\ast }[T_c(1/T_{\ast }-1/T)-\ln (T/T_{\ast })]/[R(T_c-T_{\ast })]$. Near the liquid-gas critical point, both the ideal gas approximation and the neglect of condensed-phase volume fail; zero latent heat alone does not determine the limiting coexistence slope.

The vertical structure of a thin disk is set by the balance among vertical gravity, pressure, self-gravity, magnetic forces, and motions of gas and solids.