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With gauge pressure $p_g$ referenced to atmospheric pressure $p_{\mathrm{atm}}$, absolute pressure is $p=p_g+p_{\mathrm{atm}}$. Thus Boyle's law becomesEven a constant atmospheric pressure cannot be dropped from these sums. For the calculated $100\mathrm{kPa}$ to $200\mathrm{kPa}$ example, take $p_{\mathrm{atm}}=100\mathrm{kPa}$. The initial and final gauge pressures are $0$ and $100\mathrm{kPa}$; their products with the respective volume are $0$ and $5000\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3$, which are unequal. Zero gauge pressure means balance with the surroundings, not absence of gas or absence of wall momentum transfer.

For a frictionless piston of surface area $A$ with its outer face exposed to atmospheric pressure, the additional inward force required for mechanical equilibrium is $F=(p-p_{\mathrm{atm}})A=p_{g}A$. This is why a mechanical load can reveal a pressure difference even though the equation of state requires absolute pressure. With piston friction, the load alone no longer determines $p_{g}A$; a direct pressure sensor avoids that particular inference.

For a barotropic fluid in hydrostatic equilibrium along a Killing vector, assume $\rho +p>0$. Integrating $dp/(\rho +p)=-d\log N$ gives the displayed first integral, constant on each connected fluid region. For the equation of state $p=w\rho$ with constant $w>0$, it gives $\rho N^{(1+w)/w}=C_1$.

For a pure rightgoing acoustic wave, retarded time $\theta =t-x$ and slow distance $X=Mx$ separate fast oscillations from nonlinear steepening. Expanding the equation of state gives $p=\rho +M(\gamma -1){\rho }^2/2+O(M^2)$. The leading perturbations satisfy $u_0={\rho }_0=p_0$. Adding the first-order mass conservation and momentum conservation equations cancels the unknown first corrections and yields the displayed viscous Burgers equation. Inviscid characteristics satisfy $\theta =s-(\gamma +1)F(s)X/2$, $p=F(s)$; loss of invertibility marks shock formation and the end of the smooth inviscid approximation.

In an FLRW metric with zero separate cosmological constant, $\varrho +3p\ge 0$ makes the Friedmann acceleration equation imply $\ddot{a}\le 0$. If $a_0=a(t_0)>0$ and $v_0=\dot{a}(t_0)>0$, then the scale factor is a concave function andthroughout its regular past interval. Thus a regular positive scale factor cannot persist earlier than $t_0-a_0/v_0$. If the solution continues regularly whenever $a>0$, its first past zero occurs within $a_0/v_0=H(t_0{)}^{-1}$, where $H$ is the Hubble parameter.

The continuation condition matters: without restrictions on the equation of state, an earlier curvature singularity can occur at positive $a$. For example, in fixed units $a(t)=1+\sqrt{t}$, $t>0$, with $k=0$, defines energy density $\varrho =3(\dot{a}/a{)}^2/(8\pi G)$ and pressure $p=-\varrho /3-\ddot{a}/(4\pi Ga)$. These satisfy the Friedmann equations and $\varrho +3p>0$, yet $a\to 1$ at the singular past boundary. Conversely the Milne model reaches zero without divergent curvature. A finite duration, a zero of the scale factor, and a curvature singularity are distinct conclusions.

The conductive state is the no-flow equilibrium in which heat conduction alone maintains a linear temperature profile across a layer with fixed boundary temperatures. Its hydrostatic pressure follows by integrating $P_b^{\prime }=-g{\rho }_b$, with mass density obtained from the equation of state. Under the Boussinesq approximation, deviations from this state are the variables used in the linear Rayleigh-Bénard convection problem.

For constant positive $\lambda =p_r/p_g$ and molecular gas constant $\mathcal{R}=k/({\mu }_m{m}_p)$, the gas–radiation equation of state gives $\rho =4\sigma T^3/(3c\lambda \mathcal{R})$. Vertical hydrostatic equilibrium in harmonic central gravity therefore gives $T=T_0(1-z^2/H^2)$ with $T_0={\Omega }^2{H}^2/[8(1+\lambda )\mathcal{R}]$. The mass density and pressure powers are three and four. Integrating both faces yields $\Sigma =(32/35){\rho }_{0}H$ and $\int pdz=\Sigma {\Omega }^2{H}^2/9$. With constant opacity, radiative diffusion produces a linear flux and height-independent dynamic viscosity $\mu =4c\lambda /[9\kappa (1+\lambda )]$. This is an interior formal profile; its low-density surface requires a photospheric treatment beyond diffusion.

For a uniform background velocity $U_0$ and speed of sound $c_0$, linearized conservation of mass, the Euler equation and the local equation of state $p_1=c_0^2{\rho }_1$ give this wave equation. In the coordinate $x-U_{0}t$ it is the ordinary one-dimensional wave equation. Its two characteristic curves travel at laboratory velocities $U_0+c_0$ and $U_0-c_0$; both point downstream in supersonic flow.

Consider an ideal degenerate electron gas at zero temperature. Counting two spin states per phase space cell gives number density $n_e=p_F^3/(3{\pi }^2{\hslash }^3)$. The isotropic momentum flux gives pressureHere $m_e$ is electron mass, $c$ is the speed of light, $\hslash =h/(2\pi )$ with Planck constant $h$, and special relativity gives $v(p)=pc/\sqrt{p^2+m_e^2{c}^2}$. The factor $1/3$ averages the directional momentum flux, and the spin factor is already included. This normalization makes $P_{\ast }$ a pressure, not an energy per electron.

For $x\ll 1$, expand the denominator: $I(x)\sim x^5/5$ and $P\simeq n_e{p}_F^2/(5m_e)$. For $x\gg 1$, the integrand approaches $t^3$: $I(x)\sim x^4/4$ and $P\simeq n_e{p}_{F}c/4$. At fixed composition, mass density is $\rho ={\mu }_e{m}_u{n}_e$, with ${\mu }_e$ the mean molecular weight per electron and $m_u$ the unified atomic mass unit. Since $n_e\propto x^3$, the logarithmic stiffness isPressure keeps rising; its fractional gain per fractional compression becomes smaller. Relativistic electrons approach a speed ceiling while their momenta keep increasing. Thus the high-density equation of state loses the extra compression leverage of a nonrelativistic white dwarf. The Chandrasekhar mass from central polytropic constants supplies the corresponding global mass calculation.

For a calculated local example with ${\mu }_e=2$, $x=1$ gives $\rho \simeq 1.95\times 10^9\mathrm{kg}{\mathrm{m}}^{-3}$, $I(1)=0.1537384$ and $P\simeq 7.38\times 10^{21}\mathrm{Pa}$. Its stiffness is $1.53314$, intermediate between the two limits. These are ideal equation of state values at one point, not a measured central mass density or a whole white dwarf model. Finite temperature and interactions are excluded.

Fresh water reaches maximum mass density near $4{}^{\circ }\mathrm{C}$ at atmospheric pressure. A useful local equation of state isso water becomes less dense when its temperature moves either above or below $T_m$.

The entropy density of a photon gas is its entropy per physical volume, $s=S/V$. For homogeneous equilibrium vacuum radiation with two states of electromagnetic polarization and zero photon chemical potential,The first equality is the entropy density at zero chemical potential, obtained from the thermodynamic Euler relation $U=TS-pV+{\mu }_{\gamma }{N}_{\gamma }$ with internal energy $U=uV$. The second uses the photon gas equation of state, $u=a_R{T}^4$ and $p=u/3$. If an independently constrained photon number gives a nonzero chemical potential, the general entropy density is $(u+p-{\mu }_{\gamma }{n}_{\gamma })/T$ and $u=a_R{T}^4$ need not hold. This distinction explains why the absence of a conserved number matters for entropy as well as for the spectral shape. ^[ref]

A short first law of thermodynamics check, for neighboring equilibrium states with ${\mu }_{\gamma }=0$, isThe thermodynamic Euler relation fixes the additive normalization. Equivalently, this radiation entropy tends to zero with temperature at fixed volume; no contribution from the containing walls is included. Interpreting $TdS$ as transferred heat and $pdV$ as boundary work requires a reversible thermodynamic process. For a reversible thermodynamic process that is also an adiabatic process, there is no entropy production or entropy exchange, soAn irreversible adiabatic process can produce entropy even without heat exchange, so the word adiabatic alone is insufficient to infer $T^{3}V=\text{constant}$.

For an analytic example, expand an initially equilibrated photon gas reversibly from $V_0$ to $8V_0$, keeping only the same two photon states of electromagnetic polarization. Its temperature halves, its energy density becomes $u_0/16$, its entropy density of a photon gas becomes $s_0/8$, and its total entropy stays $S_0$. The internal energy halves rather than remaining fixed: the missing half leaves as work against radiation pressure,This is a stipulated reversible radiation model, not a report of a piston experiment. Maintaining a real cavity along this path requires controlling wall heat exchange and including the wall heat capacity in any actual apparatus balance.

For a computed scale at stipulated $T=2.725\mathrm{K}$, $s\simeq 2.04122\times 10^{-14}\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{m}}^{-3}$. Combining this with the photon number density givesThis is the total equilibrium radiation entropy divided by its mean photon number, not a separately assigned entropy carried by each labeled particle. During the ideal expansion, $n_{\gamma }\propto T^3$ and $T^{3}V$ is fixed, so the mean number $N_{\gamma }=n_{\gamma }V$ also stays constant. This follows from the path; it was not imposed as the independent number constraint used to establish chemical potential.

The beautiful connection is that expansion lowers each typical photon energy and dilutes the radiation in just the proportions needed to retain its equilibrium spectral shape and total entropy. For the cosmic microwave background, homogeneous expansion of the universe has physical volume proportional to the cube of the scale factor; fixed-species cosmological entropy conservation therefore gives $T\propto a^{-1}$. This describes preservation of an already established Planck photon distribution and connects the FIRAS spectral observation in photon gas to an entropy prediction; FIRAS did not perform a direct entropy measurement. When other equilibrated species disappear and transfer energy into radiation, use the effective number of relativistic entropy degrees of freedom, $g_{\ast s}$: the conserved quantity for the combined sector is $g_{\ast s}{T}^{3}V$, under the corresponding equilibrium and no-entropy production assumptions. A changing species population, separate decoupled sectors or entropy production requires an appropriate entropy balance rather than the pure photon gas formula. ^[ref]

For one mole with temperature-independent parameters $a,b$, the equation of state is $P=RT/(V-b)-a/V^2$. A Maxwell relation gives $(\partial S/\partial V{)}_T=R/(V-b)$, so the volume dependence is $R\ln (V-b)$. Matching the ideal gas entropy at infinite volume fixes the excess at the same $V,T$ to $R\ln (1-b/V)$. The attraction contributes $-a/V$ to the Helmholtz free energy but no excess entropy when $a$ is independent of temperature. A temperature-dependent attraction parameter would change this conclusion.

An inviscid fluid with conservative body force, a materially conserved stratifying scalar $\alpha$, and an equation of state making $\nabla \rho ,\nabla p,\nabla \alpha$ coplanar conserves its Ertel potential vorticity. The stretching of absolute vorticity per unit density cancels the deformation of the scalar gradient; the remaining baroclinic triple product vanishes. In an incompressible density-stratified ideal fluid, choosing the advected density as $\alpha$ gives that cancellation directly.

Neglecting interactions and thermal pairs, the Fermi-Dirac distribution $f(p)=[\exp ((\epsilon (p)-{\mu }_{\mathrm{kin}})/(k_{B}T))+1{]}^{-1}$ with kinetic energy $\epsilon (p)=\sqrt{m_e^2{c}^4+p^2{c}^2}-m_e{c}^2$ determines the electron contribution to a stellar equation of state. The kinetic chemical potential is fixed by the electron number density, not an independently chosen pressure fraction:The electron kinetic internal energy per volume is $({\pi }^2{\hslash }^3{)}^{-1}{\int }_0^{\infty }{p}^2\epsilon (p)f(p)dp$. In the nondegenerate limit $P_e=n_e{k}_{B}T$; at zero temperature the integrals become the equation of state of a cold electron gas. These are limits of one component and must not be added together as independent electron pressures. When electron-positron thermal pair abundance matters, both charge species must be included, with charge neutrality fixing their net density.

Free expansion occurs when a gas spreads into an initially evacuated region without driving an external load. The standard idealization uses a rigid, insulated vessel: initially a partition confines the gas to volume $V_1$, while the remaining compartment contains vacuum. Open a passage and wait for thermodynamic equilibrium; the same fixed amount of substance $n$ now occupies the larger accessible volume $V_2$. The outer vessel never changes size. This simple release separates three questions: what energy is conserved, whether temperature changes, and why entropy increases. ^[ref]

Choose the gas as the closed system, neglect heat exchange with the walls, and assume negligible changes in bulk kinetic energy and potential energy between the equilibrium endpoints. Insulation gives $Q_{\mathrm{in}}=0$; no moving outer boundary or external load receives work, so $W_{\mathrm{out}}=0$. The first law of thermodynamics therefore givesThe gas can develop internal motion during the release; this does not supply external work. The intermediate state need not have one equilibrium pressure or temperature, so there is no actual quasistatic process curve $p(V)$ to integrate. In particular, substituting the equilibrium ideal gas law into $\int pdV$ would describe a different process. ^[ref]

For an ideal gas, internal energy of an ideal gas depends only on temperature. With positive heat capacity at constant volume, $U(T_2)=U(T_1)$ implies $T_2=T_1$, and the ideal gas law gives $p_2/p_1=V_1/V_2$. Doubling the accessible volume thus halves the final pressure without changing the final temperature. This is an endpoint statement; it does not assert a spatially uniform, constant temperature during the rush. Insulation alone does not imply constant temperature: a reversible adiabatic expansion cools an ideal gas because it supplies work. ^[ref]

For a real gas, $U(T,V)$ can depend on volume as well as temperature, so unchanged internal energy need not mean unchanged temperature. For a fixed-composition simple compressible system in a single thermodynamic phase, equilibrium identities using a Maxwell relation giveHere $C_V$ is the whole sample's heat capacity at constant volume. The sign depends on the equation of state; for an ideal gas the numerator vanishes. For a finite free expansion, solve $U(T_2,V_2)=U(T_1,V_1)$, or integrate this slope along equilibrium states of equal internal energy; that mathematical curve does not describe the actual release. ^[ref]

Free expansion is useful because it tests how internal energy depends on number density while eliminating external expansion work. Its beautiful surprise is that an ideal gas can finish with the same temperature and internal energy but greater entropy: a thermometer alone cannot diagnose reversibility. See entropy change in free expansion for the calculation and temperature sensitivity of a free-expansion experiment for the experimental qualification. The continuous-flow throttling associated with the Joule-Thomson coefficient has a different balance: under its usual assumptions it preserves enthalpy, whereas the gas-only idealization here preserves internal energy. ^[ref]

For a perfect fluid in general relativity of constant equation of state $P=w\rho$, conservation gives $\rho ={\rho }_0{a}^{-3(1+w)}$. In units $8\pi G=1$, the Friedmann equation becomes ${\dot{a}}^2/2+V(a)=0$. Allowed scales have $V\le 0$, and the Friedmann acceleration equation is $\ddot{a}=-V^{\prime }(a)$. The curvature term shifts the potential vertically; positive cosmological constant bends it downward and negative cosmological constant bends it upward.

A gas thermometer infers temperature from the equation of state of a confined gas. The constant-volume version measures pressure; the constant-pressure version measures volume. For an ideal gas,Here $N$ is particle number, $n$ is amount of substance, $k_B$ is the Boltzmann constant, and $R$ is the molar gas constant. The reference and test states must be internally equilibrated, and $T_0$ must be known on the absolute thermodynamic temperature scale. A pressure ratio cancels the unknown gas inventory and bulb size in the ideal constant-volume limit. This cancellation is the method's appeal: a mechanical measurement transfers an absolute temperature scale between different materials. The Maxwell speed distribution and ideal gas pressure from wall collisions explain the microscopic origin of the response. ^[ref]

For an original ideal-model calculation, choose $T_0=300\mathrm{K}$, $P_0=120\mathrm{kPa}$ and $P=160\mathrm{kPa}$. Then $T=400\mathrm{K}$. These are stipulated values, not observations. Both pressures must be absolute, referred to vacuum. If the surrounding atmospheric pressure were $100\mathrm{kPa}$, the corresponding gauge pressures would be $20$ and $60\mathrm{kPa}$; their ratio is three, which would incorrectly predict $900\mathrm{K}$. A pressure difference from the atmosphere cannot be substituted into the ideal gas law.

A motivating calibration experiment. Schooley's account of the NBS/NIST work describes a helium-filled metal bulb, a capillary, a capacitance diaphragm gauge, and a mercury pressure standard. Circulating bath liquid passed the bulb assembly and platinum resistance thermometers. The resistance probes checked the surrounding temperature; gas pressure supplied the independent thermodynamic comparison. In an early series, workers filled the cleaned bulb with purified helium at the highest test temperature, checked pressure over three hours, then measured at lower temperatures and near a reference at $0{}^{\circ }\mathrm{C}$. Six filling pressures allowed extrapolation toward zero pressure. ^[ref]

To turn this principle into a readable measurement protocol, use the following controls. These steps are a proposed teaching interpretation of the cited apparatus, rather than additional reported NBS observations:

The corresponding NBS apparatus used a protective case with an outer helium layer held close to the bulb pressure, reducing pressure deformation while aiding thermal contact. Thermal expansion was measured on bulb material. Gas in the capillary remained partly outside the hot region, requiring a measured tube temperature profile and geometry. Pure working gas also mattered: contaminants that sorb on surfaces can change the freely circulating inventory. These are physical effects to measure or bound, rather than scatter that disappears by averaging. ^[ref]

For illustration, neglect connecting-tube gas and describe a uniform bulb with a dilute virial expansion in particle number density ${\rho }_N=N/V$:Here $Z=PV/(N{k}_{B}T)$ and $B_2$ is the second virial coefficient in the particle-density convention; a molar-density convention rescales the coefficient. The second relation follows exactly from the definition of $Z$ for the same $N$, and shows the signs of the corrections. If the bulb expands by $0.1\%$ and $Z=1$ at both states, the toy pressure ratio above gives $400.4\mathrm{K}$, rather than $400\mathrm{K}$. Improved expansion and helium nonideality models really changed the temperatures inferred from the 1989 NIST data: Harvey's 2023 reanalysis found expansion updates of roughly $1$–$3\mathrm{mK}$ and nonideality updates of roughly $-0.1$ to $-0.5\mathrm{mK}$. Those are reported corrections to historical results, not the much larger illustrative correction just calculated. ^[ref]

With equal corrected bulb volumes, negligible or corrected connecting-tube gas, fixed reference and test temperatures, and a dilute classical single gas phase,Repeat the reference/test comparison for several sealed fills, reducing $N$ between comparisons but keeping it fixed within each pair. The extrapolation varies density at fixed endpoint temperatures; it does not cool one fill toward zero temperature. Real gases have different finite-density corrections but share the classical ideal gas intercept under these assumptions. The limit does not remove leaks, bulb expansion, gauge offsets, or the gas thermometer dead-space correction. At very low pressures, thermal transpiration can make remote sensor and bulb pressures differ through a nonisothermal capillary; this must be corrected too. The NBS work measured this pressure correction using a separate comparison of narrow and wider tubes. ^[ref]

At $x=1$, the toy models give $T_{\mathrm{app}}=407.92\mathrm{K}$ and $388.12\mathrm{K}$, respectively, although their imposed true test temperature is the same. Their zero-density intercept is the reusable quantity. Measuring several low-density points permits a fit and a check for curvature; asserting a straight line from one finite-density point supplies neither. A practical fit must balance nonideality against increasing relative pressure error and the capillary corrections, and keep its extrapolation uncertainty.

For the equation of state $V=N{k}_{B}T/p-NA(T)$, integrate $(\partial G/\partial p{)}_{T,N}=V$ along an isotherm. This gives the displayed Gibbs free energy difference. Its negative temperature derivative at fixed endpoint pressures gives $\Delta S=-N{k}_B\ln (p_1/p_0)+N{A}^{\prime }(T)(p_1-p_0)$. The physically admissible branch must retain positive volume.

For a barotropic perfect fluid with constant $w$, its cosmological adiabatic sound speed satisfies $c_s^2=w$. If $w<0$, the short-wavelength pressure term has the wrong sign for restoring oscillations: a mode grows as $\exp (\sqrt{-w}k\tau )$ when expansion can be neglected. For $w=-1/3$ in a coasting background with constant conformal expansion rate $K$, the exponents are $-K\pm \sqrt{K^2+k^2/3}$. This is a gradient instability, not ordinary positive-pressure Jeans instability. A network with tension or elastic anisotropic stress needs different constitutive perturbation equations; a negative background equation of state alone does not establish its perturbation sound speed.

For a fixed-composition simple compressible system, a Maxwell relation gives $C_p-C_V=T(\partial p/\partial T{)}_V(\partial V/\partial T{)}_p$. At fixed volume, the differential equation of state gives $(\partial p/\partial T{)}_V={\alpha }_V/{\kappa }_T$, proving the displayed result. Positive absolute temperature and positive isothermal compressibility imply a nonnegative difference. It vanishes when thermal expansion does, so strict inequality requires a nonzero expansion coefficient. This stability argument includes nonideal internal-energy changes, unlike a purely ideal-gas expansion-work explanation.

A dilute particle suspension can supply both excess mass density and heat. To first order its equation of state is $\rho /{\rho }_0=1+\gamma \varphi -\theta$, where $\varphi$ is the particle volume fraction and $\theta$ is fractional thermal expansion. Particle deposition flux reduces $\varphi$, while particle heating increases $\theta$. The competition can reverse the reduced gravity and create unstable density stratification.