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Low-temperature solid helium-3 can have greater entropy than the liquid through disordered nuclear spins, enabling compression cooling. ^[ref]

The sign follows from the Clapeyron equation. Let $S_l,S_s$ be molar entropies and $V_l,V_s$ molar volumes of liquid and solid, respectively, on their phase coexistence curve. Equality of their chemical potentials, with $d\mu =-SdT+VdP$, givesWith $S_s>S_l$ and $V_l>V_s$, the slope is negative. Reversibly freezing one mole at fixed temperature requires heat into the sample,Here $L_{s\to l}$ uses the melting convention for latent heat of fusion. This is the opposite sign from ordinary ice melting. The inequalities specify the regime; freezing does not universally cool a substance. In an approximately isolated sample, the growing higher-entropy phase draws on the sample's thermal energy instead of an external heater. Irreversible losses and incoming heat oppose the cooling.

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

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]

Reflux is condensed liquid returned toward the heated part of a distillation apparatus. In a fractionating column, descending reflux contacts rising vapor. Returning more condensate also changes the amount available for collection, so enrichment and product recovery must be considered together.^[ref]

For a two-component mixture, let $x$ be component A's mole fraction in the liquid and $y$ its mole fraction in the coexisting vapor. At interior compositions, define the relative volatility of A with respect to B by the ratio of composition odds:Thus $\alpha >1$ means A is enriched in the vapor, while $\alpha =1$ means $y=x$. Solving this definition givesHere a constant relative volatility is a stipulated model of vapor-liquid equilibrium, not a measured property of a named chemical solution. Real relative volatility can vary with composition, temperature, and pressure. The relation specifies no absolute temperature scale: even fixing $\alpha =2$ does not determine a boiling-temperature curve or a heat requirement.

Stipulate a constant relative volatility $\alpha =2$. The vapor-liquid equilibrium model then becomesAn initial liquid mole fraction $x_0=3/10$ gives an equilibrium vapor mole fraction $y_0=6/13\approx 0.461538$. Now imagine completely condensing a small sample of that vapor and using this condensate as a fresh liquid for another equilibration. Repeating this formal operation givesThese are original model calculations of ideal-stage enrichment. They are not measured compositions, a reflux design, or successive compositions of one pot undergoing simple distillation. That pot loses the more volatile component and becomes depleted; each step here instead starts from the previous enriched condensate. The displayed sequence specifies compositions, not the recoverable amount at each step.

The algebra makes the enrichment especially transparent. With composition odds $r_n=x_n/(1-x_n)$, every formal step doubles the odds, soFor every finite $n$, $x_n<1$, although $x_n\to 1$ as $n\to \infty$. This explains why repeated ideal equilibration can improve composition without making finite-step perfect separation a consequence of the model.

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.

Take identical rigid spheres of radius $a$, particle density ${\rho }_p$ and displaced liquid density ${\rho }_f$, at uniform absolute temperature $T$. Let $z$ increase upwards from a reference layer and write $\Delta \rho ={\rho }_p-{\rho }_f>0$. The buoyancy correction gives the effective gravitational massThe actual grain mass is ${\rho }_{p}V$; the displaced liquid subtracts ${\rho }_{f}V$ from the effective gravitational mass. In a dilute, noninteracting particle suspension, the Boltzmann distribution therefore givesHere $n$ is number density and $k_B$ is the Boltzmann constant. Equivalently, the dilute osmotic pressure $\Pi =n{k}_{B}T$ supports the grains above it: $d\Pi /dz=-n{m}_{b}g$. Dividing by $n$ and integrating yields the same exponential function. No dynamic viscosity appears in this equilibrium argument. If ${\rho }_p<{\rho }_f$, the population instead increases upwards; at equal densities this gravitational model predicts uniform $n$.

A microscope counts a finite optical slab, not an infinitesimal plane. For equal viewed area $A$ and slab thickness $\delta$, an ideal uniform detection window centred at $z$ givesThus equal optical volumes preserve the exponential function count ratio exactly, even for finite $\delta$, provided the windows lie wholly inside the accessible chamber and have the same detection response. Unequal depth response or overlapping a boundary can corrupt this inference. Repeated observations estimate the mean count; closely spaced visits need not be independent because grains remain in view.

For an original chosen model, set $a=0.250\mu \mathrm{m}$, ${\rho }_p=1200\mathrm{kg}{\mathrm{m}}^{-3}$, ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$, $T=298\mathrm{K}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. These are illustrative inputs, not Jean Baptiste Perrin's measurements. Using $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ givesIf equal observation effort gives expected count $400$ in the reference layer, the expected counts $20$, $40$ and $60\mu \mathrm{m}$ higher are approximately $214.3$, $114.8$ and $61.49$. Fractions describe expected values over observations, not fractional grains. Using the actual grain mass instead of its buoyancy-corrected value would make the inferred Avogadro constant six times too small for these chosen densities.

Sensible heat is transferred heat associated with a temperature change within a single thermodynamic phase. By contrast, latent heat can change the phase fraction while the temperature stays fixed during equilibrium phase coexistence. Neither name denotes an additional stored substance: both describe how transferred heat participates in a process.

For fixed mass $m$ and composition, use the heat capacity appropriate to the heating constraint. Under the quasistatic process assumptions in heat capacity, with only pressure-volume work and unchanged bulk kinetic energy and potential energy, fixed-pressure heating givesHere $c_p$ is specific heat capacity at constant pressure; fixed-volume heating instead uses specific heat capacity at constant volume $c_V$. A cooling step has negative $Q_{\mathrm{sens}}$ with the sign convention in heat. A temperature difference has the same numerical value in units of the kelvin or degree Celsius. The integral follows by adding the small transfers $\delta Q=m{c}_p(T)dT$, rather than assuming a constant heat capacity.

As a stylized ice-to-water example for a fixed-composition closed system at fixed pressure $1\mathrm{atm}$, take $m=0.5\mathrm{kg}$, an initial temperature of $-10{}^{\circ }\mathrm{C}$, melting at $0{}^{\circ }\mathrm{C}$, and a final temperature of $20{}^{\circ }\mathrm{C}$. Assume constant specific heat capacity at constant pressure values $c_{\mathrm{ice}}=2.1\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $c_{\mathrm{water}}=4.18\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$, and latent heat of fusion $L_f=334\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$. These are stipulated, rounded model values; ignore the container and heat losses, retain all the material, and assume only pressure-volume work under the quasistatic process conditions above. The three steps areOnly the first and third terms are sensible heat; the middle term is latent heat. The plateau begins after $10.5\mathrm{kJ}$ and ends after $177.5\mathrm{kJ}$ of total supplied heat. At $94\mathrm{kJ}$ total input, half the mass is liquid water and the temperature remains $0{}^{\circ }\mathrm{C}$. The equilibrium phase transition temperature depends on pressure; this example stipulates a single fixed pressure. ^[ref]

This split makes heating budgets easy to interpret: melting consumes $167/219.3\approx 76.2\%$ of the supplied heat in this model despite causing no temperature rise. Within either single thermodynamic phase, the heating curve slope is $dT/dQ=1/(m{c}_p)$; a larger heat capacity makes the slope shallower. A flat interval represents changing phase fraction, so it cannot be modeled by applying $Q=m{c}_p\Delta T$ across the phase transition. The complementary constructed example in latent heat tracks that phase fraction explicitly.

In simple distillation, pot vapor passes directly to the laboratory condenser. An initial small condensate sample can approximate vapor-liquid equilibrium with the current pot liquid. Removing enriched distillate depletes the pot's more volatile component. A collected fraction averages the changing condensate composition, rather than preserving the first-drop mole fraction.^[ref]

Starting in 1967, a collaboration at the Stanford Linear Accelerator Center and the Massachusetts Institute of Technology directed an electron beam at liquid hydrogen, providing proton targets. Jerome Isaac Friedman, Henry Way Kendall and Richard Edward Taylor participated in this experimental program.^[ref]

The 1969 paper Observed Behavior of Highly Inelastic Electron-Proton Scattering reports inclusive electron spectra at $6^{\circ }$ and $10^{\circ }$ with incident energies from $7$ to $17\mathrm{GeV}$. Its scaling comparison estimated ${\nu }_{\mathrm{lab}}{W}_2$ under a transverse-virtual photon assumption, finding approximate agreement as a function of $\omega =1/x$ for $Q^2>1{\mathrm{GeV}}^2$ and $\omega >4$. These are the paper's conditions, not a claim of exact Bjorken scaling at every momentum transfer.^[ref]

A solvation shell is the region of solvent constituents surrounding a solute, with local arrangements influenced by their interactions. The first solvation shell contains the closest neighbors under a stated distance or interaction criterion. A coordination number counts those neighbors; it requires such a criterion and can fluctuate. Neither a drawn ring nor a few illustrated molecules determines a measured coordination number.

In a liquid, a solvation shell is an evolving neighborhood: solvent molecules reorient and exchange with the surroundings. Different ions can produce different local dynamics. Spectroscopic measurement of hydration dynamics provides an experimental way to distinguish some of those environments without pretending that a shell is a fixed cage.

The energy per unit mass required to change a saturated liquid to vapour at a stated pressure and coexistence temperature. In a steam-condensation calorimetry experiment, sensible cooling of the condensate must be subtracted from the measured receiver heat gain.

A state of matter is a broad category of material behavior. A solid resists a sustained change of shape, a liquid flows while remaining difficult to compress, and a gas expands to fill its container. These distinctions are useful first descriptions, but they do not enumerate all thermodynamic phases. Plasma adds collective behavior of charged particles, while liquid crystals show that flow and structural order need not disappear together.

Distinguish three questions: what broad state of matter does the material resemble, which thermodynamic phase is present, and what are its temperature, pressure and other state variables? Two samples at different temperatures can be in the same liquid thermodynamic phase. Conversely, crystal polymorphism gives different solid thermodynamic phases of the same substance, distinguished by their crystal structures. A phase transition between them need not change the broad state-of-matter category.

For an imagined substance with crystalline forms A and B, melting A produces a solid–liquid phase transition, while converting A to B produces a solid–solid phase transition. A sample containing A and B has two thermodynamic phases even though all its material is solid. Cutting one uniform crystal into two pieces does not create a second thermodynamic phase: count distinct equilibrium forms, rather than pieces or grains. This distinction matters when applying the phase rule.

A phase diagram locates stable thermodynamic phases using temperature and pressure. Its liquid–gas phase coexistence curve ends at a liquid-gas critical point, around which a path connects liquid and gas without a first-order phase transition. Broad categories can therefore remain useful without a sharp boundary.^[ref]

Unwanted sticking that prevents intended relative motion. In surface micromachining, a receding liquid meniscus can pull a compliant released beam onto the substrate; adhesion can then keep it there. Avoiding a liquid-vapor drying interface addresses one route to stiction, but does not remove contact adhesion during use.

Superfluidity is a quantum-fluid response with phase stiffness and flow without viscous dissipation in an appropriate low-speed regime. A superfluid supports a phase-gradient current; defects and excitations can limit that flow. Bose-Einstein condensation concerns macroscopic one-particle occupation and is a distinct criterion: the Gaussian phase stiffness theory can have divergent absolute-phase fluctuations while retaining a cost for gradients.

Why does cooling a liquid make it flow more freely through a tiny gap? Helium-4 supplies an experimental entrance to this question: the helium slit-flow experiment compares the motion of liquid levels on opposite sides of the transition into superfluidity. A flow measurement probes resistance to motion; a measurement of Bose-Einstein condensation probes occupation of a quantum state. The two-fluid model of superfluidity and the Landau criterion connect these questions to entropy transport and to the energy required to create an excitation.

A liquid maintained above its equilibrium boiling temperature at the imposed pressure. A barrier to nucleation can delay its conversion into vapor, making this a metastable state.

For a liquid interface with constant surface tension $\gamma$, the mechanical surface energy is the energy associated with its surface area $A$:The additive constant $C$ does not affect the work needed to change surface area. For several interfaces with the same surface tension, $A$ means their total surface area; the two faces of a thin soap film must both be counted. ^[ref]

With assumed $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$, creating $1\mathrm{c}{\mathrm{m}}^2=10^{-4}{\mathrm{m}}^2$ of one interface requires $7.2\mu \mathrm{J}$ of reversible work. In the sliding-rim example illustrated under surface tension, moving a $20\mathrm{mm}$ rim outwards by $1\mathrm{mm}$ creates $4.0\times 10^{-5}{\mathrm{m}}^2$ across the two faces and requires $2.88\mu \mathrm{J}$. Independently, multiplying its $2.88\mathrm{mN}$ holding force by the displacement gives the same work.

This makes the preference for a round isolated droplet precise. At fixed volume $V$, the three-dimensional isoperimetric inequality giveswith equality for a sphere. Thus a sphere minimizes surface energy among sufficiently regular closed shapes enclosing the same volume, when surface tension is uniform and other energy contributions are negligible. This is an application of the isoperimetric inequality, rather than a proof of it. Walls, wetting, and gravity can change the shape by adding constraints or other energy contributions. ^[ref]

The differential rule $d{E}_{\mathrm{s}}=\gamma dA$ integrates directly to $E_{\mathrm{s}}=\gamma A+C$ because surface tension is constant. At fixed liquid volume, decreasing surface area can drive viscous fluid flow; the released surface energy can supply viscous dissipation. This mechanical bookkeeping is useful for predicting which deformations can release energy. If surface tension depends on temperature or composition, distinguish this mechanical area-work relation from the complete interfacial thermodynamic free energy and internal energy.

Surface tension $\gamma$ is the tensile force per unit length across a line drawn in a liquid interface. For a clean interface with isotropy, it has the same value in every tangential direction: each side pulls the other along the surface, perpendicular to the line. A straight segment of length $\ell$ therefore transmits a force of magnitude $\gamma \ell$ from one interface. Here we assume constant $\gamma >0$, with fixed temperature and composition. The same coefficient measures the reversible work required to create surface area:The equality of units follows from $1\mathrm{J}=1\mathrm{N}\mathrm{m}$; the equality of the mechanical coefficients follows from the moving-rim argument below. ^{[ref] [ref]}

For a concrete example, take an ideal flat soap film spanning a rectangular wire frame with a straight sliding rim of length $\ell$. Pull the rim outwards in the plane of the film, perpendicular to the rim. Each of the film's two liquid–gas interfaces pulls the rim back towards the film with force $\gamma \ell$, so the outward holding force is $F=2\gamma \ell$. The two faces are two interfaces, each with its own surface tension. With the assumed toy value $\gamma =0.072\mathrm{N}{\mathrm{m}}^{-1}$ and $\ell =20\mathrm{mm}=0.020\mathrm{m}$, the required force is $2.88\mathrm{mN}$; one interface would contribute $1.44\mathrm{mN}$. This value of $\gamma$ is a stipulated model parameter, rather than a measured value for a particular soap solution.

Why does an interface resist being enlarged? Its molecules have a different environment from those in the bulk liquid; changing that environment has an energy cost. In the constant-surface tension model, reducing surface area lowers surface energy. This connects an easily measured rim force to the shape of a droplet, the Young–Laplace equation for its pressure, and the restoring mechanism of capillary waves. It is useful because one coefficient links forces, energy and geometry. ^[ref]

The moving-rim argument also explains the factor of two without memorizing a special formula. An outward displacement $dx$ creates surface area $\ell dx$ on each face, hence $d{A}_{\mathrm{total}}=2\ell dx$. Slow pulling without viscous dissipation requires work $Fdx=\gamma d{A}_{\mathrm{total}}$, giving $F=2\gamma \ell$. For one interface, the same argument gives $F=\gamma \ell$. A surface tension gradient instead produces Marangoni stress; the constant-$\gamma$ assumption excludes that additional effect.

A chlorine-containing liquid used as the target material in the Homestake experiment.