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The 1869 The Bakerian Lecture: On the Continuity of the Gaseous and Liquid States of Matter by Thomas Andrews describes dried carbon dioxide trapped above mercury in a sealed capillary tube. A steel screw pressurized water in a strong copper tube; movable mercury transmitted the load. A second capillary tube held air as a pressure gauge. Flowing water around the visible capillary tubes controlled temperature; a thermometer read the bath. A cathetometer read column positions, converted to volume readings using weighed-mercury calibration.^[ref]

Below the critical temperature, compression produced a visible liquid–gas boundary; above it, large density changes occurred without that boundary. Three trials gave $30.92{}^{\circ }\mathrm{C}$ as his critical temperature. Purity checks detected residual air; mercury height corrections accounted for unequal hydrostatic pressures. He explicitly left air nonideality and pressure-induced tube-capacity corrections unresolved, so his air-gauge pressures were approximate. These readouts support a liquid-gas critical point; they do not measure a homogeneous unstable loop or establish universal quantitative accuracy of the later Van der Waals equation.^[ref]

A disappearing liquid meniscus by itself has more than one possible explanation: in ordinary subcritical condensation or evaporation, one thermodynamic phase can simply run out. The stronger critical-point inference combines temperature control, compression readings and the existence of a continuous route around the endpoint of the phase coexistence curve. On that route a dilute gas can become dense without crossing a two-phase boundary. This is what continuity of liquid and gas means; it does not remove their distinct equilibrium number densities below $T_c$.

In his 12 December 1910 Nobel lecture, Johannes Diderik van der Waals explained gases and liquids through the same moving molecules, with attraction and finite size modifying their pressure. He also rejected the idea that constant $a,b$ should give universal numerical agreement. His critical-volume discussion explicitly retained derivatives of a volume-dependent $b$: the familiar $v_c=3b$ follows only when those derivatives are neglected. Thus the elegant constant-parameter Van der Waals critical point is a model result whose limitations were already part of its author's account.^[ref]

In a closed-base cylindrical container, mass conservation makes the annular volume flux opposite to the core flux. With $h=R-a\ll a$ and annular viscosity $\lambda \mu$, Couette-Poiseuille flow in a thin gap gives $q_g=wh/2-h^3{P}_z/(12\lambda \mu )\simeq -aw/2$, where $P$ removes the annular hydrostatic pressure. Hence $P_z=6\lambda \mu aw/h^3[1+O(h/a)]$. The resulting column equation is $(a^2{w}_z{)}_z/a^2=\Delta \rho g/(3\mu )+2\lambda aw/h^3$. Curvature and the Couette flux change only the neglected relative order $h/a$.

Archimedes' principle states that buoyancy equals the downward gravitational weight, or weight, of the displaced fluid. In a resting fluid of constant mass density ${\rho }_f$ under uniform gravitational acceleration $g$, its magnitude isHere $V_{\mathrm{displaced}}$ is the volume from which the body excludes the fluid: the whole external volume for a fully immersed solid body, and just the immersed volume for a partly immersed one. The body's own mass does not enter this buoyancy formula. ^[ref]

A short proof outline for a fully immersed body is to imagine replacing it by exactly the fluid it displaces. The surrounding hydrostatic pressure is unchanged. That replacement fluid is at rest, so its upward pressure force must balance its downward force from gravity. The original body receives the same inward pressure forces on the same closed surface, hence the same upward buoyancy.

Equivalently, let $\Omega$ be the displaced region, $\partial \Omega$ its closed surface and ${\mathbf{e}}_z$ the upward unit direction. The divergence theorem, applied component by component to the surface integral, and hydrostatic pressure giveThis also explains why horizontal forces cancel even for an irregular shape. A spatially constant surface pressure contributes zero to the closed-surface integral. If the mass density varies with depth, replace ${\rho }_f{V}_{\mathrm{displaced}}$ by the mass ${\int }_{\Omega }{\rho }_{f}dV$ of the displaced fluid.

The closed-surface argument assumes that the fluid applies pressure over the entire boundary of a fully immersed body. A body sealing against a tank bottom may have no fluid beneath it; then the lower-face pressure force used in the proof is absent, and a supporting contact force must be treated separately. For partial immersion, the usual buoyancy formula assumes the exposed surface receives the common atmospheric pressure, neglects the much smaller buoyancy from the air, and neglects surface tension.

For the $0.10\mathrm{m}$ cube in the stipulated water, full immersion displaces $0.001{\mathrm{m}}^3$, so $F_b=1000\times 9.81\times 0.001=9.81\mathrm{N}$. If its uniform mass density is $600\mathrm{kg}{\mathrm{m}}^{-3}$, its mass is $m=600\times 0.001=0.600\mathrm{kg}$ and its downward force from gravity is $mg=5.886\mathrm{N}$. To hold it fully immersed at rest requires an additional downward forceOn release from rest it initially moves upward; its eventual floating equilibrium has a smaller displaced volume. A same-size cube of uniform mass density $1200\mathrm{kg}{\mathrm{m}}^{-3}$ has mass $1.200\mathrm{kg}$ and downward force from gravity $11.772\mathrm{N}$. Full immersion still gives $9.81\mathrm{N}$ of buoyancy, leaving an initial net downward force of $11.772-9.81=1.962\mathrm{N}$. Subsequent motion also involves drag.

An instrument measuring atmospheric pressure. A mercury barometer compares the hydrostatic pressure of a measured mercury column with the pressure above it. Its equation is $p_{\mathrm{atm}}=p_t+\rho gh$, with $h$ the vertical separation of the liquid levels.

A body force acts throughout a body's or fluid's volume, rather than only through its surface. Gravity is the example used in hydrostatic pressure: a gravitational field $\mathbf{g}$ gives force per unit volume $\rho \mathbf{g}$, where $\rho$ is the mass density. More generally, a body force per unit mass $\mathbf{f}$ contributes $\rho \mathbf{f}$ to momentum conservation. Forces from electromagnetism are another example. Surface pressure forces produce buoyancy on an immersed body, while its body force from gravity acts downward.

Boyle's law states that a fixed amount of substance of an ideal gas at constant absolute thermodynamic temperature has constant product of absolute pressure and total volume:Here $p$ is measured relative to a vacuum, not relative to the surrounding atmospheric pressure. Halving the volume doubles the absolute pressure; doubling the volume halves it. A real gas approximately follows Boyle's law when it is sufficiently dilute and far from a phase transition.^[ref]

For an original calculated example, suppose a sealed gas sample has $p_1=100\mathrm{kPa}$ and $V_1=100\mathrm{c}{\mathrm{m}}^3$. After isothermal compression to $V_2=50\mathrm{c}{\mathrm{m}}^3$,Both states have $C=10000\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3=10\mathrm{J}$, since $1\mathrm{kPa}\mathrm{c}{\mathrm{m}}^3=10^{-3}\mathrm{J}$. The volume ratio is dimensionless, so using the same volume units in numerator and denominator leaves the pressure in kilopascals. Although $pV$ has units of energy, its value is not the compression work: isothermal ideal-gas work involves an integral along the path.

The blue curves show calculated ideal gas states, not measurements. The left panel is a branch of a hyperbola; the right panel plots exactly the same absolute pressures against $1/V$. The dashed line uses gauge pressure with a stipulated constant atmospheric pressure of $100\mathrm{kPa}$. Its displaced intercept explains why absolute pressure in Boyle's law matters.

A short derivation starts from the ideal gas law, $pV=nRT$. Fixing the amount of substance $n$ and absolute thermodynamic temperature $T$ fixes $C=nRT$, where $R$ is the molar gas constant. The microscopic picture in ideal gas pressure from wall collisions is equally appealing: at fixed temperature, the typical particle kinetic energy stays fixed, while compression increases the number density and wall momentum transfer per unit surface area and time. A larger absolute pressure follows without requiring hotter particles.

The constraints are essential. A leak changes the amount of substance. Rapid compression can raise the temperature before heat transfer restores thermal equilibrium, as in the isothermal gas syringe experiment. Molecular size and interactions in a dense real gas invalidate the ideal gas approximation; the Van der Waals equation describes one such correction. Equal endpoint temperatures allow an endpoint test of Boyle's law, but do not establish that every intervening state lies on an isothermal process.

For fixed mass $m$, mass density satisfies $\rho =m/V=mp/C$, so it is proportional to absolute pressure. In a dilute bubbly flow, Boyle's law allows bubble volume and its contribution to buoyancy to increase as hydrostatic pressure decreases, provided gas mass is retained, dissolution is negligible, and the temperature stays approximately fixed. Its usefulness is that a simple ratio predicts a substantial size change without tracking individual molecules.

Buoyancy is the resultant upward force exerted by a resting fluid on an immersed body in a downward gravitational field. The fluid pushes inward, perpendicular to each part of the body's surface. Because hydrostatic pressure increases with depth, the upward push from below exceeds the downward push from above. Opposite horizontal forces cancel; buoyancy is the remaining vertical force. A sinking body still receives this upward support. ^[ref]

More precisely, pressure $p$ on a small surface of area $dA$ with outward unit normal vector $\mathbf{n}$ exertsThe minus sign says the force points into the body. On a cube, the top force points down, the bottom force points up, and the side forces cancel in pairs because opposite sides have the same pressure at each depth. This is a surface traction; gravity on the body itself is a separate body force. ^[ref]

For a computed example, stipulate resting water with constant mass density ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$ and uniform gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. A fully immersed cube has side $L=0.10\mathrm{m}$, top depth $h_t=0.20\mathrm{m}$ and bottom depth $h_b=0.30\mathrm{m}$. Each horizontal face has area $A=L^2=0.010{\mathrm{m}}^2$, and its volume is $V=L^3=0.001{\mathrm{m}}^3$. Its top and bottom gauge pressures areThe corresponding forces relative to the common surface pressure are $p_{g,t}A=19.62\mathrm{N}$ downward and $p_{g,b}A=29.43\mathrm{N}$ upward. Therefore the net buoyancy isUsing absolute pressures adds the same $p_{\mathrm{surface}}A$ to both face forces, leaving their difference unchanged. Zero gauge pressure at the surface does not mean zero absolute pressure. Increasing both depths by the same amount also leaves buoyancy unchanged in this constant-mass density model.

Archimedes' principle turns this calculation into a rule for bodies of any shape. It explains floating equilibrium and lets us infer an immersed body's volume from the difference between its force from gravity and the force needed to support it. In particular, buoyancy depends on the displaced volume and the surrounding mass density, whereas the body's own mass density decides how its force from gravity compares with that support.

The capillary length measures when gravity and surface tension are comparably important in shaping a liquid interface. For a liquid of uniform mass density $\rho$, beneath a gas whose mass density is negligible, with uniform gravitational acceleration $g>0$ and constant surface tension $\gamma >0$,A deformation of height $a$ and horizontal length scale $L$ has hydrostatic pressure variation of order $\rho ga$ and small-slope capillary pressure of order $\gamma a/L^2$. Equating these gives $L\sim {\ell }_c$: surface tension smooths small features, whereas gravity matters for broad ones. For two resting fluids, replace $\rho$ by their positive mass density difference $\Delta \rho$ in this static balance. This change alone does not give the two-fluid wave dynamics, because both fluids can have inertia. ^[ref]

With the stipulated illustrative parameters $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the capillary length is ${\ell }_c=2.71\mathrm{mm}$. These parameters define the ideal calculations below; they are not a measurement of a particular sample.

For a sinusoid with wavenumber $k>0$, the relevant horizontal scale is $L=1/k$, while its wavelength is $\lambda =2\pi /k$. Thus the Bond number for this scale and the reciprocal restoring-term ratio areEquality is at $k_c=1/{\ell }_c$, hence ${\lambda }_c=2\pi {\ell }_c=17.02\mathrm{mm}$ for these parameters. Using $L=\lambda$ instead produces a different numerical Bond number; state the length convention. The capillary length itself is not the equality wavelength. This distinction helps choose whether a measured ripple should be modeled as a surface gravity wave, a nearly pure capillary wave, or a gravity-capillary wave. ^[ref]

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.

In an impermeable cylinder, an axisymmetric turbulent plume with negligible source volume flux requires a downward return flow. With uniform upward plume velocity $w$ and ambient velocity $v$, volume conservation gives $b^{2}w+(R^2-b^2)v=0$. An entrainment coefficient $\alpha$ based on the relative speed gives $Q^{\prime }=2\alpha b(w-v)$. A specified pressure closure is essential for momentum conservation. If the departure from ambient hydrostatic pressure and wall force are neglected in the whole-section integral, the total specific momentum flux $J=b^2{w}^2+(R^2-b^2)v^2$ obeys $J^{\prime }=b^2{g}^{\prime }$. Homogeneous unmodified ambient and no buoyancy loss give $b^{2}w{g}^{\prime }=B_s$. These are leading integral-model closures; restoring the ambient dynamic pressure can change the singularity criterion.

Darcy law is a linear relation between the bulk Darcy flux and the net driving force per unit volume. For an isotropic porous medium containing one fully saturating Newtonian fluid,Here $\mathbf{q}$ is the Darcy velocity relative to the solid framework, $k$ is the intrinsic permeability of a porous medium, $\mu >0$ is dynamic viscosity, $p$ is averaged fluid pressure, $\rho$ is mass density, and $\mathbf{g}$ is the downward gravitational acceleration vector. The pressure gradient exerts force density $-\nabla p$, while gravity contributes $\rho \mathbf{g}$; the minus sign makes the discharge follow their sum. The symbols $\mathbf{u}$ and $\mathbf{q}$ are both commonly used for Darcy velocity; neither denotes pore velocity here.

With upward coordinate $z$ and positive gravitational acceleration magnitude $g$, $\mathbf{g}=-g{\mathbf{e}}_z$. Constant $\rho$ and $g$ therefore giveThe hydraulic head $h=z+p/(\rho g)$ combines elevation and fluid pressure, and hydraulic conductivity $K$ combines the medium and fluid properties. Hydrostatic pressure satisfies $\nabla p=\rho \mathbf{g}$, so it produces zero Darcy flux even though fluid pressure increases downward. Conversely, equal pressures at the two ends of a vertical column do not imply zero flow: their elevations give different hydraulic heads.

The appeal of Darcy law is that a complicated connected pore geometry enters a macroscopic calculation through a measurable coefficient. It predicts filter discharge and aquifer flow; combined with mass conservation, it gives an equation for the evolving fluid pressure. Its simple linear form requires a small pore-scale Reynolds number, slow adjustment compared with pore-scale inertial relaxation, and a valid continuum approximation. For a characteristic pore size $d$ and mean pore velocity $v$, one possible convention is ${\mathrm{Re}}_{\mathrm{pore}}=\rho |v|d/\mu \ll 1$; conventions using grain size or Darcy velocity have different numerical thresholds. Merely calling flow laminar does not guarantee negligible inertia. Unsaturated pores, several fluid phases, appreciable inertia, or a fluid whose stress is not Newtonian require further modeling. Woessner and Poeter discuss these limits in their authored hydrology text.^[ref]

Henry Darcy's 1856 book Les fontaines publiques de la ville de Dijon was published by Victor Dalmont; the University Grenoble Alpes digital collection verifies this bibliographic record and supplies a digitized copy.^[ref] Woessner and Poeter's account connects Darcy law to experiments on water flowing through sand filters, relating discharge to cross-sectional area, hydraulic head difference and sample length.^[ref] The modern intrinsic permeability of a porous medium notation above should not be read as a quotation of Darcy's original notation.

At fixed applied voltage, the mechanical potential of a capacitor and its source includes $-C{V}^2/2$. Increasing the liquid overlap of a partially filled coaxial capacitor therefore pulls the liquid upward. Balancing $V^2(dC/dh)/2$ against the excess hydrostatic pressure times the annular area gives the title formula for a wide external reservoir. Its differential sensitivity is $2\Delta h/V$; it vanishes at zero bias and measures the magnitude, rather than the sign, of voltage.

A draining gravity current loses fluid through its substrate while spreading horizontally. For a two-dimensional viscous current of thickness $H(x,t)$ that loses fluid according to pressure-dependent Darcy drainage,when the hydrostatic pressure is $p=\rho gH$ and $k/h\propto p^{-\beta }$.

The Dupuit approximation treats flow in a shallow unconfined aquifer as predominantly horizontal, with hydrostatic pressure $p=\rho g(h-z)$ and horizontal pressure gradient independent of depth. Integrating Darcy's law over the saturated depth then gives a volume flux per unit width depending only on the groundwater height and its slope.

Dynamic pressure is the bulk kinetic energy per unit volume associated with fluid motion at speed $U$ in a specified reference frame:Here $\rho =m/V$ is mass density. The word "pressure" identifies its units and its use as a pressure scale; $q$ is not an extra isotropic stress to add to the local static pressure in every surface traction. A stationary fluid can have large static pressure with $q=0$. Conversely, knowing $U$ and $\rho$ fixes $q$ without fixing static pressure. The definition remains useful in viscous fluid flow and compressible flow, even when the simplest Bernoulli equation is inapplicable. ^[ref]

For stipulated mass density ${\rho }_{\infty }=1.20\mathrm{kg}{\mathrm{m}}^{-3}$ and reference speed $U_{\infty }=20\mathrm{m}{\mathrm{s}}^{-1}$, $q_{\infty }=240\mathrm{Pa}$. At $40\mathrm{m}{\mathrm{s}}^{-1}$ with the same mass density, $q_{\infty }=960\mathrm{Pa}$. This quadratic scaling makes dynamic pressure a useful normalization for the lift coefficient, drag coefficient and pressure coefficient; it predicts four times a particular force only if its corresponding coefficient remains unchanged.

The connection with stagnation pressure requires hypotheses. For steady flow, constant mass density, inviscid flow, and gravity as the only body force, the Bernoulli equation gives, along one streamline,Here $p$ is static pressure, $g$ is gravitational acceleration, and $z$ is height. Bring that flow to rest without losses at the same height and the limiting static pressure is the stagnation pressure $p_0$:There must be no intervening pump supplying work or other energy input. Constancy along one streamline does not assert equal $B$ on all streamlines; irrotational flow in a connected region supplies that additional conclusion. Changing height requires retaining $\rho gz$. In compressible flow the stopping pressure difference generally differs from $\rho U^2/2$, and viscous dissipation can reduce recovered stagnation pressure. ^[ref]

A Pitot-static tube turns this relation into a measurement: its upstream-facing opening senses stagnation pressure, while separate side openings sample static pressure. Connect these to the positive and negative ports of a differential pressure transducer or manometer. Under the hypotheses above, the measured difference $\Delta p$ givesThus the stipulated $240\mathrm{Pa}$ and $1.20\mathrm{kg}{\mathrm{m}}^{-3}$ imply $20\mathrm{m}{\mathrm{s}}^{-1}$, not an independently measured speed. Align the probe with the incoming flow and keep its passages clear: blockage and pinched tubing invalidate the comparison. At very low speed, the difference can be smaller than instrument error; at high Mach number, compressibility and possibly a shock wave require a different reduction. ^[ref]

A useful control is differential-pressure calibration. For a digital manometer, zeroing establishes the reading at equal port pressures; it does not establish the gain away from zero. The manufacturer's Series 475 Handheld Digital Manometer: Specifications - Installation and Operating Instructions by Dwyer Instruments, Inc. directs the user to vent both ports to the atmosphere before zeroing. Its stated accuracy is a fraction of full scale, not of the reading, and depends on instrument temperature. Select a suitable range and record the instrument variant and units; the manual covers several ranges.^[ref]

An independent physical check can use a liquid-column manometer: apply a stable pressure difference to both instruments and compare the digital reading with the hydrostatic pressure inferred from the vertical level difference. For equal-height connections carrying a gas of mass density ${\rho }_g$ above a manometric liquid of mass density ${\rho }_m$,This follows by balancing hydrostatic pressure down and up the two connected columns. Use vertical height, even if a tube is inclined. Known mass densities, measured height and gravitational acceleration set the comparison scale. Check several differences across the working range and reverse the port connections to check the sign. This is a proposed independent check, not a claim that the cited teaching experiment performed a full instrument calibration. Recheck zero after reconnecting the sealed tubing; a zero adjustment alone cannot detect a blocked passage or an incorrect scale.

Volume preservation by the elastic accommodation in morphoelasticity. If $\mathbf{F}=\mathbf{AG}$, elastic incompressibility imposes $\det \mathbf{A}=1$, and hence $\det \mathbf{F}=\det \mathbf{G}$. Total volume can increase through growth while the elastic response remains incompressible. A hydrostatic pressure acts as a Lagrange multiplier enforcing this constraint in the Cauchy stress tensor.

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.

Gas outside the uniform-temperature bulb of a gas thermometer contributes to its conserved particle inventory. An idealized calculation divides the connecting tube into small regions with volumes $V_j$ and local temperatures $T_j$. Neglect hydrostatic pressure offsets and thermal transpiration, assume a common pressure $P$ and a local ideal gas law in each region, and let the bulb have volume $V_b$ and temperature $T_b$. Adding the particle inventories givesThus even an ideal gas in a perfectly rigid sealed apparatus need not obey $P\propto T_b$: some of its particles remain in regions with other temperatures. For a stipulated bulb of $100\mathrm{c}{\mathrm{m}}^3$, a connecting volume of $1\mathrm{c}{\mathrm{m}}^3$ maintained at $300\mathrm{K}$, and bulb reference and test temperatures of $300$ and $400\mathrm{K}$, conservation of $N$ givesThese are toy dimensions and a calculated bias, not measured data. Reducing $N$ reduces both pressures proportionally and leaves this ratio unchanged: extrapolating away gas nonideality does not eliminate the connecting-volume error. Actual apparatus also requires the tube's thermal expansion, a measured temperature profile, and the pressure corrections omitted in this model.

Gauge pressure is pressure measured relative to the surrounding atmospheric pressure; absolute pressure is instead measured relative to an ideal vacuum. At an open surface $p_{\mathrm{surface}}=p_{\mathrm{atmosphere}}$, so the constant-mass density hydrostatic pressure law becomes $p_g={\rho }_{f}gh$. Here $p_g=0$ at the surface denotes equality with atmospheric pressure. It does not denote zero absolute pressure. ^[ref]

After gravity smoothing of a Marangoni front reaches the whole spreading pool, the film approaches uniform depth while surface transport continues. A small height gradient creates a hydrostatic pressure return flow that almost cancels the net Marangoni stress flux. Setting $q\simeq 0$ in the thin-film equations with insoluble surfactant gives $h_x\simeq -3A{C}_x/(2\rho g{h}_0)$ and fractional depth variation $O(AM/(\rho g{h}_0^2{x}_N))\ll 1$. The depressed region under the surfactant therefore becomes progressively shallower.