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Activation as a cloud condensation nucleus depends primarily on dry particle size, soluble composition and hygroscopicity, together with wetting and surface-tension effects. In Köhler theory, droplet curvature raises the equilibrium vapor pressure, while dissolved solute lowers water activity. Each particle therefore has a critical supersaturation; larger or more hygroscopic particles generally activate at a lower value. A size distribution cannot be replaced by an undifferentiated total aerosol count.^[ref]

Actual droplet number also depends on the supersaturation attained in the cloud updraught. Adding particles increases competition for condensing water vapor and can lower the peak supersaturation, leaving some particles unactivated. Updraught speed, entrainment and the aerosol size distribution therefore make the number of activated droplets a nonlinear function of the available nuclei. Suitable nuclei need sufficiently low activation thresholds, and additional aerosol does not guarantee one additional droplet per particle.

Observe each gas volume, temperature, pressure and moisture condition, and reduce both volumes to the same temperature, pressure and dry-gas basis. Equal scale readings alone do not establish equal amounts of gas.

Read the calibrated volume of each gas in a graduated tube or eudiometer, with the eye at the meniscus to avoid parallax. Allow the gas and the surrounding bath to reach thermal equilibrium, then read a thermometer close to the gas. Heating expands a fixed amount of gas, so a larger reading could reflect a higher temperature rather than more gas. The temperature used in a gas law must be an thermodynamic temperature.

Read the atmospheric pressure with a barometer. If a gas is confined over a liquid, also observe the difference between the inside and outside liquid levels. With $h$ the internal surface height above the external surface, the total internal pressure isHere ${\rho }_{\ell }$ is the confining density of the liquid and $g$ the gravitational acceleration; the sign reverses automatically when $h$ is negative. Equalizing the levels before taking the volume reading removes this hydrostatic pressure correction. A mercury barometer can itself require temperature and calibration corrections for accurate work.

If the gas is saturated with water vapor, subtract the vapor pressure of water at its measured temperature:The observed pressure includes both the test gas and water vapor; counting it all as test gas overestimates the amount. The saturation assumption needs time for equilibration. For a dry gas, no water vapor subtraction is made; another collecting liquid may require its own vapor pressure correction. A collecting liquid must also be chosen so that it neither appreciably dissolves nor reacts with the gas: water, for instance, is unsuitable for preserving the whole amount of freely soluble Ammonia.

For dilute gases obeying the ideal gas law, the volume $V_0$ at a common chosen $T_0,p_0$ isConsequently the corrected comparison isThis is also the ratio of amounts of substance in the ideal gas approximation. Check for leaks, trapped air, changing composition and incomplete transfer. At high pressure or near condensation, real gas behaviour may require compressibility factors; with $pV=ZnRT$, the inferred amount ratio contains the further factor $Z_2/Z_1$.

At a stated external pressure, the boiling point of a pure liquid is the temperature at which its saturation vapor pressure equals that pressure. A vapour bubble inside the liquid must also withstand the hydrostatic head and any capillary pressure; the simple free-surface boiling-point rule neglects those small local corrections.

Increasing external pressure raises the usual boiling point, while lowering it lowers the boiling point. The Clapeyron equation explains the direction:for ordinary vaporization with positive latent heat $L$ and a larger vapour specific volume. Consequently water can boil below its usual open-vessel temperature at altitude or under partial vacuum, and above it in a sufficiently pressurized vessel.

Composition also matters. A nonvolatile dissolved solute generally lowers the solvent's vapour pressure and produces boiling-point elevation at fixed external pressure. A volatile second component can instead raise or lower the boiling temperature according to the mixture's phase equilibrium; an azeotrope is a useful warning against applying the nonvolatile-solute rule to every mixture.

Finally, distinguish the equilibrium boiling point from the observed onset of bubbles. A smooth clean vessel or a shortage of nucleation sites can permit superheating; agitation or suitable existing sites can make boiling begin nearer equilibrium. Heating power mainly controls the rate of sustained boiling under otherwise fixed conditions, rather than setting an arbitrary new equilibrium boiling point. A thermometer must be placed to measure the intended liquid or saturated vapour region, rather than a locally overheated wall.

For a convex droplet, the Kelvin equation gives $\ln (p_{\mathrm{eq}}(r)/p_{\mathrm{sat}})=2\gamma {\overline{V}}_l/(rRT)$. Smaller droplets therefore require a higher equilibrium vapor pressure. At a common ambient vapor pressure, sufficiently small droplets evaporate while larger ones can grow, transferring material through the vapor. Ostwald ripening decreases the droplet count and increases a characteristic surviving radius. Direct collision and coalescence can also reduce the count but are a distinct mechanism. In a closed container the final state depends on total liquid/vapor inventory and volume; an initially undersaturated, liquid-poor system can lose all droplets rather than indefinitely grow a large one.

wetting liquid in narrow pores and gaps between fibers can have a concave meniscus. Its signed curvature is negative, so the Kelvin equation permits equilibrium at a vapor pressure below that above a flat liquid surface. capillary condensation in sufficiently small, wettable gaps explains moisture accumulation at high relative humidity. Increased humidity brings the vapor closer to flat-surface saturation and permits condensation in larger gaps. Surface adsorption can precede a bulk-like meniscus; large open spaces do not automatically fill at every humidity. The interpretation assumes a wetting contact geometry and does not require the entire cloth to be below the ordinary planar dew point.

The coarse texture comes from large ice crystals. Form many small crystals by rapid cooling with agitation and avoid extended storage near partial melting or repeated warming/cooling cycles. Curvature raises the chemical potential of smaller ice crystals, so material can transfer from smaller to larger crystals during recrystallization; partial melting followed by slow refreezing accelerates loss of a fine texture. A more viscous unfrozen phase and suitable stabilizing ingredients slow molecular transport and coarsening. Create small crystals quickly and suppress their subsequent growth with stable cold storage and slow transport. This applies interfacial and phase-change physics; it is not a claim that reducing droplet vapor pressure alone completely models a multicomponent frozen mixture.

Use the normal boiling point as the reference, $T_B=373.15\mathrm{K}$ and $P_B=101325\mathrm{Pa}$. At $T=313.15\mathrm{K}$,This is approximately 0.0795 atmosphere. It uses the supplied constant enthalpy of vaporization approximation across a 60 K interval; it should not be presented as an exact tabulated vapor pressure. Absolute temperature, rather than Celsius values, belongs in the exponential.

Let $P_{\mathrm{sat}}$ denote the flat-interface saturation pressure at the same temperature, $P_V$ the surrounding equilibrium vapor pressure, and ${\overline{V}}_l$ the incompressible liquid molar volume. Equating liquid and ideal-vapor chemical potentials to their common reference at $P_{\mathrm{sat}}$ givesThus the pressure contribution to the attractive normal force on a wetted circular area isFor capillary pressures large compared with the vapor-pressure difference, and a bridge wetting essentially the plate area,Here $F$ is the magnitude of the external separating force balancing attraction, and $P_V<P_{\mathrm{sat}}$. A finite contact angle adds a contact-line force $2\pi a\gamma \sin \theta$ at each plate's rim, counted once in the force on that plate, while the exact pressure also retains both curvatures. The rim term is normally smaller than the narrow-gap pressure term by order $d/a$ for appreciable wetting; it cannot be neglected uniformly near $\theta =90^{\circ }$. For complete wetting the leading equivalent expression is $F\simeq 2\pi a^2\gamma /d$. The corresponding equilibrium humidity is $P_V/P_{\mathrm{sat}}\simeq \exp [-2\gamma {\overline{V}}_l/(RTd)]$.

These are quasistatic, equilibrium expressions for an existing bridge, neglecting viscous suction, contact-angle hysteresis, disjoining forces and deformation of the plates. The wetted radius must replace $a$ if the liquid does not reach the plate edge. During actual separation a bridge can shrink, depin or break, so this local force balance is not a promise of a constant force for arbitrary travel.

Entropy at absolute zero. The third law of thermodynamics concerns the low-temperature limit of equilibrium states. In the zero-entropy formulation, the entropy of a perfect crystal with a nondegenerate ground state tends to zero as $T\to 0$. The statistical interpretation is $S=k_B\ln \Omega$: a unique accessible ground state has $\Omega =1$. The entropy-difference formulation of the third law of thermodynamics states that entropy differences associated with regular changes of mechanical parameters tend to zero as absolute zero is approached. These statements require equilibrium and appropriate ground-state assumptions; a frozen disordered material or a degenerate ground state can carry residual entropy.

For a phase of fixed composition and pressure,A finite limiting entropy constrains this integral. Ordinary equilibrium heat capacities therefore vanish as temperature approaches zero, with familiar examples $C\propto T^3$ for three-dimensional acoustic phonons and $C\propto T$ for a normal degenerate Fermi liquid. A constant positive heat capacity extrapolated to zero would give a logarithmically divergent entropy difference and is impossible in this regime. Discontinuous transitions and anomalous limits need separate treatment rather than extending a high-temperature approximation indefinitely.

Gibbs free energy. For fixed particle numbers,Thus, when $S\to 0$, the Gibbs free energy approaches its zero-temperature value with a horizontal tangent:The equality $G(0,p)=H(0,p)$ uses the enthalpy $H=U+pV$; neither limiting value needs to be zero. The low-temperature Gibbs free-energy expansion makes the approach quantitative. For $C_p=a{T}^r$ with $r>0$ and $S(0)=0$,The second derivative is ${\partial }_T^{2}G=-C_p/T\le 0$, although it need not tend to zero. For a reaction between such equilibrium phases, ${\partial }_T\Delta G=-\Delta S\to 0$, and $\Delta G$ and $\Delta H$ approach one another. If a genuine residual entropy $S_0$ is retained instead, $G=G(0,p)-S_{0}T+\cdots$ has limiting slope $-S_0$. A horizontal tangent cannot be asserted without the zero-entropy condition.

Why helium remains a liquid. Helium atoms attract each other weakly and are exceptionally light. Their zero-point energy is large compared with the binding that would localize them in a solid. Both helium-4 and helium-3 can therefore remain liquid down to the lowest temperatures at low pressure; substantial applied pressure is needed to solidify them. This is not a violation of the third law. A quantum liquid can have a unique low-entropy ground state, and zero-point motion does not itself imply thermal entropy. The third law does not require every substance to crystallize at atmospheric pressure.

A helium-4 atom has integer spin zero and is a boson. Below the lambda point near $2.17\mathrm{K}$ at saturated vapor pressure, its liquid is superfluid. Its strong interactions mean that it is not quantitatively an ideal Bose gas, and the condensate fraction need not equal the superfluid fraction. The two-fluid model of superfluidity separates an entropy-carrying normal component of a superfluid from the superfluid component, whose entropy is zero. At sufficiently low temperature the dominant thermal excitations are acoustic phonons: their linear dispersion in three dimensions gives $U-U_0\propto V{T}^4$, $C_V\propto V{T}^3$ and $S\propto V{T}^3$. Higher-energy rotons are exponentially suppressed. Thus the normal component of a superfluid and entropy disappear as $T\to 0$, while superfluid flow and quantum motion can persist.

A helium-3 atom has spin $1/2$ and is a fermion. Well below its Fermi temperature, but above its superfluid transition, liquid helium-3 is a degenerate interacting Fermi liquid. Only quasiparticles close to the Fermi surface are thermally excited, giving $C_V\propto T$ and $S\propto T$. At temperatures of order a millikelvin, helium-3 also becomes superfluid through formation of Cooper pairs of fermions. This is not Bose-Einstein condensation of individual helium-3 atoms. Pairing reorganizes the low-energy excitation spectrum; in the fully gapped bulk B phase, the fermionic quasiparticle contribution to heat capacity is exponentially suppressed at temperatures well below the pairing gap. Gapless collective modes can still give thermal terms following a power law. The entropy nevertheless tends to zero. The normal Fermi-liquid law cannot be extrapolated through its pairing transition unchanged.^[ref]

Helium-3 also illustrates the importance of nuclear spin entropy. Over an intermediate low-temperature range, localized spins in the solid can retain approximately $k_B\ln 2$ of entropy per atom, while the degenerate liquid has much less. Reversible conversion of liquid to the higher-entropy solid then absorbs heat; adiabatic compression along the melting curve can cool the sample. This is Pomeranchuk cooling. At still lower temperature, interactions between the nuclear spins remove this nearly free-spin entropy, so a constant $k_B\ln 2$ cannot be extrapolated to absolute zero.^[ref]

Liquefaction and cooling methods. A gas must be below its liquid-gas critical point to form a distinct liquid phase by compression. Compression above the critical temperature alone cannot achieve ordinary liquid-vapour coexistence. Practical liquefaction removes the compression heat, cools the compressed gas in a heat exchanger and uses expansion to lower its temperature further.

A throttling valve operates approximately at constant specific enthalpy when kinetic-energy changes and heat leakage are negligible. The Joule-Thomson coefficient isusing consistent molar quantities, or consistent total quantities for a fixed amount. A pressure drop cools the gas if ${\mu }_{\mathrm{JT}}>0$. An ideal gas has zero Joule-Thomson coefficient; real intermolecular interactions are essential. In particular, room-temperature helium warms on throttling, so it must first be brought into a region of positive Joule-Thomson coefficient. Expansion through an engine or turbine can instead remove work and produce approximately isentropic cooling.

A regenerative liquefier sends the cold returning gas through a counterflow heat exchanger to precool the incoming compressed gas before its next expansion. Expansion engines and external precooling can supply the initial cooling; repeated operation then produces liquid. Modern helium liquefaction commonly combines work-producing expansion with a final throttle. ^[ref]

Once liquid exists, pumping away vapour lowers its saturation pressure and boiling temperature. The escaping vapour carries latent heat away, cooling the remaining liquid. At still lower temperatures, a dilution refrigerator exploits the different entropy carried by helium-3 in concentrated and dilute helium mixtures; adiabatic demagnetization and Pomeranchuk cooling provide other routes.

Approaching zero. The unattainability formulation of the third law states that regular cooling operations with finite resources cannot reach absolute zero in finitely many stages. As the entropy available to remove becomes small and thermal couplings weaken, each method loses its ability to take the final step. An isentropic cooling trajectory starting at nonzero entropy cannot reach a common zero-entropy endpoint through regular finite changes of parameters. A fully general finite-time claim also requires assumptions about available controls and heat-transfer rates. Quantum liquids and refrigeration methods can approach $0\mathrm{K}$ while remaining consistent with vanishing equilibrium entropy; they do not provide a finite-operation route to absolute zero.

Below cloud base, an initially unsaturated closed parcel expands as ambient pressure falls. In the absence of heat exchange its temperature falls approximately at the dry adiabatic lapse rate $g/c_{p,d}$. Its vapour mixing ratio is conserved, but its saturation mixing ratio decreases strongly as it cools. Hence relative humidity rises until the lifting condensation level is reached. The density normally decreases through expansion; the environmental profile determines the parcel's buoyancy and thus whether ascent can continue.

Near saturation, slight atmospheric supersaturation permits suitable cloud condensation nuclei to activate. Köhler theory describes the competition between curvature, which raises equilibrium vapour pressure, and dissolved solute, which lowers it. Cooling generates supersaturation, while condensation onto the new droplets consumes it. Their balance normally keeps in-cloud relative humidity close to, but not identically at, saturation. The activated droplet number depends on the aerosol population and the maximum supersaturation produced by the ascent.

Above cloud base, condensation lowers the vapour content and raises the liquid-water content. The latent heat released partly offsets expansion cooling, so the moist adiabat has a smaller lapse-rate magnitude than the dry one. In a dilute approximation,A closed adiabatic parcel conserves total water $q_t=q_v+q_l$ and approximately its moist entropy; retaining condensate makes this a reversible-cloud idealization, rather than a precipitating pseudoadiabat. In a shallow cloud, the saturation-mixing-ratio decrease and slow density variation yield nearly constant condensation per height, so liquid-water content is approximately linear above base. Once activation is complete, assuming nearly fixed droplet number per volume gives increasing droplet size as condensate accumulates. Entrainment, continued activation, collision-coalescence and atmospheric precipitation can all violate that simplified description; they are excluded from the stated retrieval model.

A rising parcel cools, lowering the saturation pressure. Once its water vapor supply exceeds that value, atmospheric supersaturation develops. Homogeneous nucleation of pure water is strongly inhibited at the small atmospheric supersaturations typical of clouds; existing aerosol particles provide cloud condensation nuclei. Their dry sizes and composition, together with the parcel's ascent rate, determine how many cloud droplets activate.

Köhler theory combines the Kelvin equation, which raises equilibrium vapor pressure over a curved surface, with Raoult's law, which lowers it for a chemical solution. In a dilute diameter convention, with dry diameter $D_d$ and wet diameter $D$,Here $s=e_v/E(T)-1$ is atmospheric supersaturation, ${\sigma }_w$ is surface tension, and ${\kappa }_h$ measures aerosol hygroscopicity, not the dry-adiabatic process exponent. The maximum of the Köhler curve is the critical supersaturation. Larger or more hygroscopic particles generally have lower thresholds. Cloud-droplet activation requires overcoming that maximum; total aerosol number is not the activated number.

Given an observed aerosol number size distribution $n_j(D_d)=d{N}_j/d\ln D_d$ for composition classes $j$, first assign each class a hygroscopicity and compute its critical supersaturation. If the parcel reaches maximum atmospheric supersaturation $s_{\max }$, the equilibrium-threshold estimate isFor each fitted lognormal distribution, this is its upper-tail integral; it can be evaluated analytically with an error function or numerically from measured bins.

The value $s_{\max }$ must be calculated, not guessed from the aerosol distribution alone. Prescribe parcel upward velocity $w$, initial pressure, temperature and humidity, then integrate expansion cooling, latent heat release, conservation of total water, and diffusional cloud droplet growth. Schematically,where $a_s,b_s>0$, $G$ includes water vapor diffusion and thermal conduction, and $q_l$ is liquid mass fraction. Rising motion produces atmospheric supersaturation; growing particles consume it. Their competition selects the peak and which particles cross their activation barriers. Assume a well-mixed closed parcel, liquid water, soluble spherical particles, specified mixing state, and negligible collision-coalescence and fallout during activation. Slow kinetic growth, weakly soluble particles or entrainment require a more detailed treatment than the threshold integral.

The dilute Köhler curve follows because the Kelvin equation multiplies the planar-solution saturation pressure by the curvature correction. Combining it with Raoult's law therefore givesFor a dilute chemical solution and a small curvature correction, $B/r^3\ll 1$ and $A/r\ll 1$. A first-order Taylor expansion givesThus the simplified Köhler theory relation isThe curvature term raises the required ambient vapor pressure; the solute term lowers it. These laws combine multiplicatively before linearization, so the displayed simplified relation is an approximation, not an exact identity.

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]

Relative humidity is the ratio of the partial pressure of water vapor to its saturation vapor pressure at the same temperature, usually expressed as a percentage.

A vapor whose partial pressure exceeds its vapor pressure at the local temperature; nucleation can permit condensation.