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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.

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.