Kelvin equation
The Kelvin equation describes the change in vapour pressure due to a curved liquid–vapor interface, such as the surface of a droplet. The vapor pressure at a convex curved surface is higher than that at a flat surface. The Kelvin equation is dependent upon thermodynamic principles and does not allude to special properties of materials. It is also used for determination of pore size distribution of a porous medium using adsorption porosimetry. The equation is named in honor of William Thomson, also known as Lord Kelvin. Formulation. The original form of the Kelvin equation, published in 1871 , is: formula_1 where: This may be written in the following form, known as the Ostwald–Freundlich equation: formula_12 where formula_13 is the actual vapour pressure, formula_14 is the saturated vapour pressure when the surface is flat, formula_15 is the liquid/vapor surface tension, formula_16 is the molar volume of the liquid, formula_17 is the universal gas constant, formula_18 is the radius of the droplet, and formula_19 is temperature. Equilibrium vapor pressure depends on droplet size. As formula_18 increases, formula_13 decreases towards formula_26, and the droplets grow into bulk liquid. If the vapour is cooled, then formula_19 decreases, but so does  formula_14. This means formula_29 increases as the liquid is cooled. formula_15 and formula_16 my be treated as approximately fixed, which means that the critical radius formula_18 must also decrease. The further a vapour is supercooled, the smaller the critical radius becomes. Ultimately it can become as small as a few molecules, and the liquid undergoes homogeneous nucleation and growth. The change in vapor pressure can be attributed to changes in the Laplace pressure. When the Laplace pressure rises in a droplet, the droplet tends to evaporate more easily. When applying the Kelvin equation, two cases must be distinguished: A drop  of liquid in its own vapor will result in a convex liquid surface, and a bubble of vapor in a liquid will result in a concave liquid surface. History. The form of the Kelvin equation here is not the form in which it appeared in Lord Kelvin's article of 1871. The derivation of the form that appears in this article from Kelvin's original equation was presented by Robert von Helmholtz (son of German physicist Hermann von Helmholtz) in his dissertation of 1885. In 2020, researchers found that the equation was accurate down to the 1nm scale. Derivation  using the Gibbs free energy. The formal definition of the Gibbs free energy for a parcel of volume formula_33, pressure formula_34 and temperature formula_19 is given by: where formula_37 is the internal energy and formula_38 is the entropy. The differential form of the Gibbs free energy can be given as where formula_40 is the chemical potential and formula_41 is the number of moles. Suppose we have a substance formula_42 which contains no impurities. Let's consider the formation of a single drop of formula_43 with radius formula_18 containing formula_45 molecules from its pure vapor. The change in the Gibbs free  energy due to this process is where formula_47 and formula_48 are the Gibbs energies of the drop and vapor respectively. Suppose we have formula_49 molecules in the vapor phase initially. After the formation of the drop, this number decreases to formula_50, where Let formula_52 and formula_53 represent the Gibbs free energy of a molecule in the vapor and liquid phase respectively. The change in the Gibbs free energy is then: where formula_55 is the Gibbs free energy associated with an interface with radius of curvature formula_18 and surface tension formula_57. The equation can be rearranged to give Let formula_59 and  formula_60 be the volume occupied by one molecule in the liquid phase and vapor phase respectively. If the drop is considered to be spherical, then The number of molecules in the drop is then given by The change in Gibbs energy is then The differential form of the Gibbs free energy of one molecule at constant temperature and constant number of molecules can be given by: If we assume that formula_65 then The vapor phase is also assumed to behave like an ideal gas, so where formula_68 is the Boltzmann constant. Thus, the change in the Gibbs free energy for  one molecule is where formula_70 is the saturated vapor pressure of formula_43 over a flat surface and formula_72 is the actual vapor pressure over the liquid. Solving the integral, we have The change in the Gibbs free energy following the formation of the drop is then The derivative of this equation with respect to formula_18 is The maximum value occurs when the derivative equals zero. The radius corresponding to this value is: Rearranging this equation gives the Ostwald–Freundlich form of the Kelvin equation: Apparent paradox. An equation similar to that of Kelvin can be derived for the solubility of  small particles or droplets in a liquid, by means of the connection between vapour pressure and solubility, thus the Kelvin equation also applies to solids, to slightly soluble liquids, and their solutions if the partial pressure formula_13 is replaced by the solubility of the solid (formula_80) (or a second liquid) at the given radius, formula_18, and formula_14 by the solubility at a plane surface (formula_83). Hence small particles (like small droplets) are more soluble than larger ones. The equation would then be given by: These results led to the problem of how new phases can ever arise from old ones . For example, if a container filled with water vapour at slightly below the saturation pressure is suddenly cooled, perhaps by adiabatic expansion, as in a cloud chamber, the vapour may become supersaturated with respect to liquid water. It is then in a metastable state, and we may expect condensation to take place. A reasonable molecular model of condensation would seem to be that two or three molecules of water vapour come together to form a tiny droplet, and that this nucleus of condensation then grows by accretion, as additional vapour molecules happen to hit it. The Kelvin equation, however, indicates  that a tiny droplet like this nucleus, being only a few ångströms in diameter, would have a vapour pressure many times that of the bulk liquid. As far as tiny nuclei are concerned, the vapour would not be supersaturated at all. Such nuclei should immediately re-evaporate, and the emergence of a new phase at the equilibrium pressure, or even moderately above it should be impossible. Hence, the over-saturation must be several times higher than the normal saturation value for spontaneous nucleation to occur. There are two ways of resolving this paradox. In the first place, we know the  statistical basis of the second law of thermodynamics. In any system at equilibrium, there are always fluctuations around the equilibrium condition, and if the system contains few molecules, these fluctuations may be relatively large. There is always a chance that an appropriate fluctuation may lead to the formation of a nucleus of a new phase, even though the tiny nucleus could be called thermodynamically unstable. The chance of a fluctuation is "e", where Δ"S" is the deviation of the entropy from the equilibrium value. It is unlikely, however, that new phases often arise by this fluctuation mechanism and the resultant  spontaneous nucleation. Calculations show that the chance, "e", is usually too small. It is more likely that tiny dust particles act as nuclei in supersaturated vapours or solutions. In the cloud chamber, it is the clusters of ions caused by a passing high-energy particle that acts as nucleation centers. Actually, vapours seem to be much less finicky than solutions about the sort of nuclei required. This is because a liquid will condense on almost any surface, but crystallization requires the presence of crystal faces of the proper kind. For a sessile drop residing on a solid surface, the Kelvin  equation is modified near the contact line, due to intermolecular interactions between the liquid drop and the solid surface. This extended Kelvin equation is given by where formula_86 is the disjoining pressure that accounts for the intermolecular interactions between the sessile drop and the solid and formula_87 is the Laplace pressure, accounting for the curvature-induced pressure inside the liquid drop. When the interactions are attractive in nature, the disjoining pressure, formula_86 is negative. Near the contact line, the disjoining pressure dominates over the Laplace pressure, implying that the solubility, formula_80 is less than formula_83. This implies that a new  phase can spontaneously grow on a solid surface, even under saturation conditions.