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The first-order combination of the Kelvin equation and Raoult's law for a chemical solution droplet with $A/r\ll 1$ and $B/r^3\ll 1$. Its maximum is at $r_c=\sqrt{3B/A}$, with $S_c-1=2A/(3r_c)$. For ambient saturation ratio $1<S<S_c$, the smaller equilibrium radius is stable and the larger is unstable: the sign of condensational growth is the sign of $S-S_{\mathrm{eq}}(r)$. In particular, exceeding $r_c$ alone need not activate continued growth at subcritical atmospheric supersaturation.

A convex liquid surface has additional surface free energy. The resulting Kelvin equation for a pure spherical water droplet isHere gamma is surface tension and $M_w$ is molar mass. Smaller droplets therefore need a higher environmental vapour pressure to avoid evaporating; forming pure-water embryos by homogeneous condensation has a large nucleation barrier.

An atmospheric droplet normally forms on a cloud condensation nucleus. Dissolved salt reduces water activity and hence equilibrium vapour pressure, opposing the curvature effect. Köhler theory combines them:Using the radius form, with radius and diameter conventions in Köhler activation kept distinct, the Köhler activation radius is $r_c=\sqrt{3B/A}$, with $\ln S_c=2A/(3r_c)$. Supersaturation above this maximum activates a wet aerosol into a growing cloud droplet. Solubility, particle size and wettability determine the threshold; not every particle activates. Cooling by ascent commonly supplies the necessary modest supersaturation. ^[ref]

A convex water surface has higher saturation vapour pressure than a flat one. Equality of chemical potentials with the Laplace pressure $2\sigma /r$ gives the Kelvin equationSmall pure droplets therefore need large supersaturation; many subcritical embryos evaporate, making homogeneous droplet formation inefficient under ordinary cloud conditions. Cloud condensation nuclei provide a surface and, when soluble, lower water activity. Köhler theory combines the curvature increase with this solute decrease, giving a critical supersaturation and size for activation.

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.

For a spherical liquid droplet, the Kelvin equation isHere $V_m=0.018/957=1.881\times 10^{-5}{\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$, soThe formal answer is about $1.0\mathrm{nm}$. At molecular dimensions, bulk surface tension, incompressible-liquid and continuum curvature assumptions are only approximate. There is one liquid-vapour interface here, unlike a soap bubble's two interfaces.

$S=e/e_s^l$ is the ambient saturation ratio over a flat liquid-water surface; $S-1$ is the liquid atmospheric supersaturation. Curvature increases the surface equilibrium vapour pressure, producing the Kelvin penalty $A/r$, while dissolved solute lowers it and gives the opposing Raoult contribution $B/r^3$. Thus the dilute Köhler equilibrium curve is $S_{\mathrm{eq}}-1=A/r-B/r^3$, and the numerator is the atmospheric supersaturation available to drive growth beyond surface equilibrium.

$F_d^l$ describes the resistance of vapour diffusion through the surrounding air. $F_k^l$ describes the thermal-conduction resistance to removing latent heat released by condensation: a warmed droplet surface has a higher equilibrium vapour pressure and hence a weaker mass-transfer drive. Common continuum approximations are $F_d^l\simeq R_{v}T/(D_v{e}_s^l)$ and $F_k^l\simeq L_v[k_{\mathrm{air}}T{]}^{-1}[L_v/(R_{v}T)-1]$, showing the roles of diffusivity, conductivity, saturation pressure and latent heat. Both resistances have the supplied units $\mathrm{s}\mathrm{m}\mathrm{k}{\mathrm{g}}^{-1}$ and add for this coupled transport problem.

For a spherical droplet of constant density, $m=4\pi \rho r^3/3$. Multiplying the radius equation by $4\pi \rho r$ givesAt sufficiently large radius, curvature and solute corrections are negligible, soThis limit also requires sustained positive atmospheric supersaturation. It need not describe the earliest submicrometre stage or an unactivated haze particle.

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.

For a bulk liquid and vapor at common fixed temperature and pressure, transfer $d{N}_l$ molecules from vapor to liquid while holding their sum fixed. The Gibbs free energy change is $dG=({\mu }_l-{\mu }_v)d{N}_l$. Thermodynamic equilibrium with both phases present requires stationarity under either sign of transfer, hence $\boxed{{\mu }_l={\mu }_v}$. This is the bulk coexistence condition; an interface's curvature adds a surface-energy cost.

In the capillarity approximation, take an incompressible drop with constant liquid number density $n_l$, bulk chemical potential ${\mu }_l$ at the specified reservoir conditions, and size-independent surface tension $\Gamma$. The liquid bulk is extensive, so its Gibbs thermodynamic free energy is $N_l{\mu }_l$. Creating interface area reversibly at fixed temperature adds $\Gamma A$. Equivalently, along growth at fixed bulk parameters the stated differential integrates as $d{G}_d={\mu }_{l}d{N}_l+\Gamma dA$, with the no-drop reference set to zero. For a sphere,This approximation uses bulk properties for the interior and neglects the molecular interface thickness, curvature dependence of surface tension and higher compressibility corrections. It does not apply quantitatively to drops only a few molecules across.

The vapor contains $N-N_l$ molecules. Treat its chemical potential as approximately unchanged by forming the drop, as for a large surrounding reservoir or negligible vapor depletion. ThenThe radius is limited by the available molecules in a finite sample; the constant-reservoir expression is a local nucleation model, not a description of arbitrarily complete vapor depletion. The excess over the vapor reference has derivativeIf ${\mu }_l>{\mu }_v$, it increases for every $R>0$, leaving only the stable boundary minimum $R=0$. If ${\mu }_l<{\mu }_v$, $R=0$ is a metastable local minimum and the additional stationary radius isAt that radius $\Delta G^{\prime \prime }=-8\pi \Gamma <0$, so it is an unstable maximum. The classical nucleation theory barrier is $16\pi {\Gamma }^3/[3n_l^2({\mu }_v-{\mu }_l{)}^2]$.

A fluctuation below $R_c$ tends to shrink because reducing its surface cost lowers thermodynamic free energy. A supercritical drop tends to grow, because its bulk free-energy gain outweighs the added surface cost. Growth continues while material supply and supersaturation allow it, eventually changing the vapor state or leading to macroscopic liquid. Curved-interface pressure and chemical thermodynamic equilibrium can also be described through the Young–Laplace equation and the Kelvin equation.