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A liquid or gas occupying connected or isolated pore space. Its pressure and compressibility affect the mechanical and seismic response of porous rock.

Pressure measures the compressive normal force per unit surface area. For a uniformly loaded surface of surface area $A$, the magnitude of the perpendicular force is $F_{\perp }=PA$; locally, $P=d{F}_{\perp }/dA$. Its SI unit is the pascal, $1\mathrm{Pa}=1\mathrm{N}{\mathrm{m}}^{-2}$. A force parallel to the surface contributes shear stress, rather than pressure. In a gas, the steady macroscopic force comes from many microscopic transfers of momentum; ideal gas pressure from wall collisions makes this connection explicit. ^[ref]

For a computed example, a uniform pressure of $100\mathrm{kPa}$ on a piston of surface area $10\mathrm{c}{\mathrm{m}}^2=10^{-3}{\mathrm{m}}^2$ produces a perpendicular force of $100\mathrm{N}$. If the opposite face is exposed to $80\mathrm{kPa}$, the net force due to the two pressures is $(100-80)\times 10^3\times 10^{-3}=20\mathrm{N}$, directed toward the lower-pressure side. The pressure on either face and their difference answer different questions; gauge pressure records a difference from a specified reference, often atmospheric pressure.

A scalar pressure describes a gas in local thermodynamic equilibrium with isotropic random particle velocities: every wall orientation has the same normal force per unit surface area. Without this isotropy, normal stresses can depend on direction and a pressure tensor is needed. Thus random motion alone is insufficient; a preferred direction matters.

In thermodynamics, pressure and volume are thermodynamic conjugate variables. For a simple compressible system, reversible expansion delivers work $\delta W=PdV$, and at fixed entropy $S$ and particle number $N$,where $U$ is internal energy. A large pressure therefore means both a large wall force per unit surface area and a large amount of expansion work per small increase in volume.

A pressure sensor converts a pressure into a recorded signal. An absolute pressure sensor uses a vacuum reference; a gauge pressure sensor uses a specified reference pressure. A closed gas syringe and an absolute pressure sensor can test Boyle's law after allowing the temperature to stabilize.

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.

In his 1895 lecture, Lord Rayleigh compared gas masses in a globe under matched pressure and temperature. Removing oxygen from air left gas about half a percent heavier than nitrogen prepared entirely from chemical sources. Several preparation routes reproduced the difference; hydrogen contamination was a concern. An eight-month storage test re-established the filling conditions: simply reweighing a sealed globe could not test a change in molar mass. William Ramsay concentrated the residual gas by circulating the mixture over hot magnesium, which consumed nitrogen. A spectral test checked remaining nitrogen. Agreement of spectra from this route and the alternative spark-and-alkali separation supported a common residual gas.^[ref]

The density difference alone identifies neither a unique chemical element nor its concentration. A heavier contaminant, a lighter reference contaminant, or unmatched filling conditions can all shift a weighing. A separation process and an optical spectrum provide additional observations. This combination makes the experiment powerful: the anomaly motivates another operation whose result can be checked independently.

For tightly wound perturbations in a locally uniform, razor-thin gas disk, continuity equation and rotating-fluid momentum equations couple the surface-density perturbation to radial and azimuthal velocities. The sheet potential is $\delta \Phi =-2\pi G\delta \Sigma /|k|$. Eliminating these perturbations gives the displayed dispersion relation. radial epicyclic frequency supplies the rotation term, the sound-speed term resists short waves, and self-gravity supplies the negative term. For an axisymmetric mode $n=0$, use ${\omega }^2$ directly; the pattern speed $\omega /n$ is undefined.

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]

A concrete experiment turns the phase coexistence curve into pairs of temperature and saturation pressure. A 1969 paper reports water measurements made in 1942 and 1947–1949 using a boiler and precision mercury manometer. Platinum resistance thermometers read the boiler temperature through a resistance bridge. The pressure-transfer line passed through an ice-water condenser and a dry-ice trap; helium filled connecting sections. Corrected mercury column height gave the pressure at the thermometer coils, including hydrostatic pressure contributions from water vapor, helium, mercury, and mercury vapor.^[ref]

Controls included normal/reversed-lead readings, readings at two thermometer electric currents extrapolated to zero current to remove self-heating, repeated bridge calibration, and daily triple point thermometer calibration checks. Cellar temperature measurements corrected fluid-column and gauge-block thermal expansion. One measurement series was excluded because a trapped gas bubble was suspected of corrupting the mercury head. The paper adjusted measurements to round temperatures using previously tabulated slopes; those adjusted values therefore are not independent measurements of those slopes.^[ref]

The thermodynamic inference requires liquid and gas in phase coexistence, a well-defined common temperature, and a pressure corrected to that same location. Leaks, contamination, or a temperature gradient can spoil that interpretation. A separate calorimetry measurement of latent heat, together with the two specific volumes, allows a test of the full Clapeyron equation. Converting a measured phase coexistence curve slope directly into latent heat instead assumes that equation; it is not an independent test.

Schlieren photography makes spatial changes of refractive index visible by converting deflections of light rays into contrast at an optical camera. In a gas, mass density variations can produce these deflections through refraction. A knife edge in the focused beam is a common implementation.

The strings in Boltzmann entropy of a two-box occupancy are complete microstates only within the binary toy model. They retain a coarse-grained variable for each particle and omit its exact position within a box and its momentum. An actual classical gas instead occupies continuous phase space; obtaining a dimensionless count from a phase space volume requires a specified cell convention and the appropriate treatment of particle identity. A quantum state count must respect the allowed quantum states and quantum statistics. Thus $2^N$ is not a general count of microstates for a gas obeying quantum ideal-gas statistics. ^[ref]

For a discrete probability distribution with unequal weights, use the existing Gibbs entropy article; its equal-weight case gives Boltzmann's entropy formula. Shannon entropy uses the corresponding information measure, with the logarithm base setting its units. A count alone cannot replace the specified probability distribution, as the Boltzmann distribution examples demonstrate. ^[ref]

Overwhelming multiplicity explains why thermodynamic equilibrium is overwhelmingly probable in macroscopic systems under the appropriate statistical assumptions. It does not make low-entropy thermal fluctuations logically impossible. Nor does counting alone derive irreversible macroscopic evolution: one also needs dynamics, an initial preparation, and a reason the system samples the allowed microstates on the timescale of interest. ^[ref]

Historically, Boltzmann's 1877 paper counted microscopic arrangements, called complexions, and related the natural logarithm of their number to entropy. The Sharp and Matschinsky translation appeared in Entropy 17 (2015), pages 1971–2009, DOI 10.3390/e17041971. Its notation differs from ours: Boltzmann's permutability measure $\Omega$ is already logarithmic, whereas here $\Omega$ is the count. ^[ref] Josiah Willard Gibbs's Elementary Principles in Statistical Mechanics (1902) developed the statistical ensemble framework; Chapter XIV compares several statistical definitions of entropy with thermodynamics. ^{[ref] [ref]}

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.

For a classical dilute gas with pair potential $u(r)=ϵ[1-(r/a{)}^3]$ for $r<a$ and zero otherwise, put $x=ϵ/(k_{B}T)>0$. Substitution $t=(r/a{)}^3$ in the Mayer f-function formula givesIt is positive for finite positive temperature, tends to $2\pi a^3/3$ as $T\to 0$ in the classical model, and behaves as $\pi a^3ϵ/(3k_{B}T)$ at high temperature. The low-temperature mathematical limit need not remain in the dilute classical regime for a real molecular gas.

Two equal circular rings of radius $R$ in parallel planes at $z=\pm L$ provide a boundary for an open film. In the ideal axisymmetric model, a stationary connected film has the catenoid profileThe catenoid existence threshold for equal rings explains why this equation can have two solutions, one solution, or none. The cancellation of principal curvatures holds for both solutions whenever they exist; the mean curvature equation alone does not select a stable minimal surface.

This ideal geometry uses $R=1$ and $L/R=0.60$. The two profiles have different waist radii $a$, although both have $H=0$. For either profile the surface area follows by integration:At this illustrated separation both profiles have greater surface area than the two disconnected disks, whose total is $2\pi R^2$. This comparison disproves a claim of global area minimization in a class allowing disconnected competitors; it does not determine the sign of the second variation for small variations of a connected film.

A reported experiment used coaxial wire rings of radius $4\mathrm{cm}$, drawn from a surfactant solution and slowly separated until collapse. The solution contained tetradecyltrimethylammonium bromide in deionized water, with glycerol added to vary viscosity. Fluorescein enabled fluorescence imaging under cyan LED arrays; a Phantom V641 camera recorded 4,000 or 5,600 frames per second. Extracted contours gave neck radius, attachment angle, and inferred surface area. They showed neck narrowing and eventual breakup, rather than indefinite persistence of a stationary catenoid. Comparing solutions across viscosities helped distinguish liquid dissipation from the surrounding air's inertia, but the ring displacement beyond the last stable state was not precisely controlled. These are collapse observations, not direct measurements of equilibrium $H$. ^[ref]

A proposed classroom measurement would instead hold two equal rings coaxial on a translation stage, leave both sides open to the same gas, and photograph the settled silhouette beside a scale. Fit $r(z)=a\cosh (z/a)$ and inspect fit residuals before inferring cancellation of the principal curvatures. Repeat at several separations and with the camera rotated about the axis to test rotational symmetry. Keep solution composition, ring wetting, and elapsed drainage time comparable, and shield the film from drafts. Film weight, drainage, evaporation, surfactant gradients, and rupture can invalidate the ideal soap film pressure balance. A silhouette fit alone neither measures the pressure jump nor tests all possible perturbations of a stable minimal surface.

For an ideal soap film with equal constant surface tension $\gamma$ on both faces, the effective tension is $\tau =2\gamma$. Choose a unit normal $N$ from the minus side to the plus side, and $\Delta p=p_{-}-p_{+}$. Translating the Young–Laplace equation to $S=-dN$ givesA single interface uses $\tau =\gamma$. An outward-oriented spherical soap bubble has $H=-1/R$, giving $\Delta p=4\gamma /R$. The positive-convex curvature sum in the Young–Laplace equation article is $-2H$. ^[ref]

An open film with the same resting gas on both sides has zero pressure jump. Neglecting film weight, soap film pressure balance gives $H=0$: a minimal surface. A pressurized bubble instead has nonzero mean curvature. These statements assume equilibrium, constant surface tension, and negligible normal viscous stress. Surfactant gradients or motion require the fuller normal stress balance. ^[ref]

The variational outline is short. With fixed boundary, a normal displacement $fN$ gives $\delta A=-2{\int }_{M}HfdA$ by the first variation of area formula. For a closed interface oriented outward, $\delta V={\int }_{M}fdA$, so stationarity of $\tau A-\Delta pV$ gives $-2\tau H-\Delta p=0$. This is a first variation calculation; testing for a stable minimal surface needs the second variation, and deciding a global minimum needs a comparison of competitors.

Loss of strength in suitable water-saturated granular material when rising pore fluid pressure reduces the effective contact load, often during cyclic earthquake shaking. It is distinct from the condensation of a gas.

A solvent is the component designated as the medium of a chemical solution, commonly the component present in greatest proportion. Water is a common solvent, not a necessary one. A mixture of mostly ethanol with a little water can designate ethanol as solvent and water as solute. Chemical solutions can also be gas or solid mixtures; molecular nitrogen is the principal medium of ordinary well-mixed air. When proportions are comparable, state the convention explicitly.^[ref]

The solvent is part of the final sample. Choosing it influences solubility and solvation; measuring its added volume alone does not determine the final chemical solution volume. Solvent and solute masses add in a closed account, whereas the final volume should be measured rather than inferred by adding pre-mixing volumes.