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The coefficient $K$ in a polytropic equation of state $p=K{\rho }^{\Gamma }$. It is spatially constant for one simple polytrope, but need not stay constant between different equilibrium models. For a fixed-composition ideal gas with $\Gamma$ equal to its adiabatic index, $K$ parametrizes specific entropy: $s=c_v\ln K+\text{constant}$ with a suitable dimensional reference inside the logarithm.

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 $H_P\Delta \log P$.

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance $n$ of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law givesHere $R$ is the molar gas constant; specifying $nRT$ fixes an energy scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:The integral of $1/V$ supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature, $\Delta U=0$; the first law of thermodynamics, $\Delta U=Q_{\mathrm{in}}-W_{\mathrm{out}}$, gives $Q_{\mathrm{in},\mathrm{rev}}=0.693147\mathrm{kJ}$ to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Now start from the same initial state, but abruptly lower the external pressure to the constant value $p_{\mathrm{ext}}=p_0/2=0.5\mathrm{MPa}$. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature $T$, mechanical equilibrium and the ideal gas law give $V_f=nRT/p_{\mathrm{ext}}=2V_0$. The work delivered to the constant load isThis work follows directly from load force times piston displacement, using $dV=Adx$ for piston area $A$. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither $p(V)=nRT/V$ at the bath temperature nor $p(V)=p_{\mathrm{ext}}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:The thermal reservoir stays at $T$ and loses heat $Q_{\mathrm{in}}=W_{\mathrm{out}}$, so $\Delta S_{\mathrm{bath}}=-W_{\mathrm{out}}/T$. Assume the ideal load stores work without an entropy change. The combined entropy production is thereforeNo numerical entropy unit is implied by $nRT=1\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost work opportunity is $T\Delta S_{\mathrm{total}}=0.193147\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume $V_0$ inside a rigid, insulated vessel of total volume $2V_0$. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so $W_{\mathrm{out}}=Q_{\mathrm{in}}=0$ and $\Delta U=0$. After equilibration the temperature is again $T$, the pressure is $p_0/2$, and the gas entropy has increased by $nR\ln 2$. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.

For a hydrostatic monatomic ideal gas with negligible external pressure, $2U+{\Omega }_f=0$, where gravitational potential energy is negative. The final total energy is therefore ${\Omega }_f/2$, giving the displayed radiated energy when no mass or other energy carrier escapes. The common $-{\Omega }_f/2$ result additionally takes the initial total energy to be negligible. Initial rest does not remove initial self-gravitational binding energy.

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.

In the ideal solution and ideal gas vapor approximation, each component's partial pressure equals its liquid mole fraction times its pure-component vapor pressure at the same temperature:These pressure formulas neglect the change in the pure-liquid reference between its saturation pressure and $P$; appreciable real gas behavior also requires a more general treatment. The ideal solution definition in terms of chemical potentials does not require either approximation.^[ref]

A short derivation explains the reference choice. At vapor-liquid equilibrium, each component has the same chemical potential in the liquid and vapor. For an ideal gas mixture, chemical potential of an ideal gas giveswhere $p^{\circ }$ is a reference pressure. Approximate ${\mu }_i^{\ast }(T,P)$ by ${\mu }_i^{\ast }(T,p_i^{\ast })$ and subtract the pure-component equation from the mixture equation:Taking the exponential function gives Raoult's law. Summing partial pressures and normalizing gives $P$ and $y_i$; this last step uses the ideal gas vapor assumption separately from the liquid model.

The law is named after François-Marie Raoult, who measured vapor pressures and freezing behavior of chemical solutions.^[ref]

A real gas departs from the ideal gas law because interactions and molecular volume matter. Nonideal equations of state, including the Dieterici equation, model such departures.

At fixed particle number and volume, a classical monatomic ideal gas has $U=3N{k}_{B}T/2$ and $C_V=3N{k}_B/2$. The canonical energy fluctuation identity $\mathrm{Var}(E)=k_B{T}^2{C}_V$ therefore gives ${\sigma }_U/U=\sqrt{2/(3N)}$. Absolute fluctuations grow as $\sqrt{N}$, while relative fluctuations decrease as $N^{-1/2}$. A particle reservoir changes the fluctuation problem because energy and particle-number fluctuations are then correlated.

For a fixed amount of ideal gas with constant heat capacities, the first law and $\delta Q=0$ givewhere $\gamma =C_p/C_V$.

The Rüchardt experiment infers heat capacity ratio from the period of an oscillation of a piston supported by trapped gas. Use a vertical precision cylinder with a close-fitting movable piston, known moving mass $M$, face area $A$, trapped equilibrium volume $V_0$, and local atmospheric pressure $p_{\mathrm{atm}}$. With negligible static friction, force balance gives the absolute gas pressure $p_0=p_{\mathrm{atm}}+Mg/A$, where $g$ is gravitational acceleration. Displace the piston slightly and release it: compression raises pressure, and expansion lowers it, producing a restoring force. The oscillation is a mechanical probe of thermal response, without separately measuring two heat inputs. ^[ref]

Let $x$ be upward displacement from equilibrium, so $V=V_0+Ax$. For constant heat capacity ratio and a reversible adiabatic process, the net upward force, including the ambient load and weight, isThe linear approximation requires $|Ax|/V_0\ll 1$. Applying Newton's second law gives simple harmonic motion withHere ${\omega }_0$ is the undamped angular frequency and $\mathcal{T}$ the undamped oscillation period, rather than temperature. Record repeated cycles, measure $M,A,V_0,p_{\mathrm{atm}}$ independently, and include all connected trapped volume. An illustrative model with $M=0.020\mathrm{kg}$, $A=2.0\times 10^{-4}{\mathrm{m}}^2$, $V_0=0.010{\mathrm{m}}^3$, $p_{\mathrm{atm}}=100000\mathrm{Pa}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\gamma =1.4$ gives $p_0=100981\mathrm{Pa}$ and $\mathcal{T}\simeq 1.182\mathrm{s}$. These stipulated values are not measurements.

In Shearwood and Sloan's implementation, piston-mounted magnets induce an electromotive force in a stationary coil through Faraday's law of induction; an oscilloscope records the signal. They observed faster oscillations at smaller trapped volume values and stronger damping at small cylinder volume values. The signal tracks velocity locally, so its amplitude is not directly piston displacement. Calibration also identified inlet dead space beyond the graduated cylinder volume. ^[ref]

Repeat at smaller initial amplitudes to test the linear approximation, and compare several $V_0$ values. Heat transfer weakens the restoring force toward the isothermal process limit, biasing the inference toward $1$. Check leakage through resting-piston drift and before-and-after trapped volume readings. For viscous damping with envelope $e^{-t/\tau }$, the measured angular frequency obeys ${\omega }_0^2={\omega }_d^2+{\tau }^{-2}$; this correction does not remove thermal leakage. Internal equilibration must be faster than an oscillation period, with heat transfer and leakage slow over a run. Large amplitude or sticking friction require a richer model. ^[ref]

In the classical dilute regime, a monatomic ideal gas has entropy

For a dilute saturated vapour with constant latent heat $L$ in a dry atmosphere, set $x=R_A/c_{pA}$, $ϵ=R_A/R_B$ and $\beta =L/(R_B{T}_s)$. Its mass path is $W=ϵe_s(T_s)e^{\beta }{E}_1(\beta )/(gx)$. Substituting $y=\beta e^{x\zeta }$ in the log-pressure column integral proves this identity. For $\beta \gg 1$, $W\sim c_{pA}{T}_s{e}_s(T_s)/(gL)$ with relative correction $-1/\beta +O({\beta }^{-2})$.

The second virial coefficient is the leading interaction correction to the ideal gas law: at fixed temperature, $B_2(T)={\lim }_{n\to 0}(Z-1)/n$, where $Z=p/(n{k}_{B}T)$ and $n$ is number density. For identical classical particles in a homogeneous three-dimensional gas, with an isotropic pair potential $u(r)$ and $r$ the centre-to-centre separation,The function $f$ is the Mayer f-function. The factor $4\pi r^{2}dr$ counts the volume of a thin spherical shell; the factor $1/2$ counts each unordered pair once. The result has units of volume. Repulsive regions, $u>0$, contribute positively, while attractive regions, $u<0$, contribute negatively. ^[ref]

This is a striking bridge from a microscopic pair potential to a macroscopic pressure reading. At sufficiently small number density, $B_2>0$ raises pressure relative to an ideal gas at the same temperature and number density, while $B_2<0$ lowers it. A single value of $B_2$, however, cannot determine the entire pair potential: many different functions have the same weighted integral. Measurements over several temperatures supply more information.

The integral must exist. For example, a bounded short-range attraction outside a hard core has no difficulty. If instead $u(r)\sim A/r^s$ as $r\to \infty$, then $1-e^{-u/(k_{B}T)}\sim u/(k_{B}T)$ and the radial tail behaves like $\int r^{2-s}dr$: it converges only for $s>3$. A bare unscreened Coulomb potential therefore falls outside this formula's ordinary short-range setting. This condition on the tail does not by itself establish the stability or convergence of the full virial expansion.

For an exactly integrable toy attraction, choose the following square-well pair potential, with diameter $\sigma >0$, attraction depth $ϵ>0$, and range multiplier $a>1$:Set $T^{\ast }=k_{B}T/ϵ$ and $b_{\mathrm{HS}}=2\pi {\sigma }^3/3$, the second virial coefficient of a hard-sphere gas. Integrating the core and attractive shell separately gives the exact toy resultThe core contributes $b_{\mathrm{HS}}$; the shell contributes $-b_{\mathrm{HS}}(a^3-1)(e^{1/T^{\ast }}-1)$. Heating weakens the relative Boltzmann factor enhancement of attractive configurations, so $B_2$ approaches the positive hard-core value as $T^{\ast }\to \infty$. For the illustrative choice $a=3/2$, the zero occurs atAt this temperature, $Z=1+B_3{n}^2+\cdots$: cancellation removes the linear correction, not every correction. At low $T^{\ast }$, the negative shell contribution becomes large, requiring a correspondingly smaller number density for a useful truncated virial expansion. The square well is a deliberately chosen model, not a fit to the carbon dioxide experiment.

The gas constant per unit mass for a specified composition. An ideal gas obeys $p=\rho \mathcal{R}T$. With dimensionless mean molecular weight $\mu$ relative to an atomic mass unit, $\mathcal{R}=k_B/(\mu m_u)$; the common astrophysical notation $R^{\ast }/\mu$ uses a modified mass-based constant rather than the molar gas constant without a mass conversion.

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 $\beta$.

A speed distribution is the probability distribution of the magnitude of velocity. When it has a probability density function $f$, the fraction of particles with speeds between $v$ and $v+dv$ is $f(v)dv$, and ${\int }_0^{\infty }f(v)dv=1$. The Maxwell speed distribution describes an equilibrium ideal gas; the effusive beam speed distribution weights that population by its rate of arrival.

Buoyancy-driven flow associated with differences in air density and the vertical separation of openings; heating can drive outflow above and replacement inflow below.