The Kelvin Equation

Blow a soap bubble and watch a tiny one shrink while a large one grows. Small droplets and small bubbles are inherently unstable partners to large ones nearby. This happens because curvature raises the internal pressure. Less obviously, curvature also raises the vapour pressure of the liquid itself. The Kelvin equation is the quantitative bridge between curvature and volatility. This relationship matters everywhere surfaces bend. It explains why fine fog droplets evaporate faster than coarse mist. It explains why small crystals dissolve to feed larger ones (Ostwald ripening). It also explains why water condenses inside narrow pores below the bulk saturation pressure. It sets the thermodynamic stage for nucleation theory. It also underlies capillary condensation used in pore-size analysis. This article builds the Kelvin equation from the Laplace pressure and the chemical potential of a liquid under mechanical stress. You will see the full derivation. Then you will work two numerical examples. Finally you will see where the equation breaks down. You will also see how it connects to nucleation and porous-materials science.

Key takeaways

The Kelvin equation gives the vapour pressure of a curved liquid surface relative to the flat, bulk value.
Convex surfaces, like small droplets, have enhanced vapour pressure; concave surfaces, like liquid in a wetting pore, have reduced vapour pressure.
The effect scales as $1/r$ , so it is negligible for macroscopic droplets but dominant below roughly 100 nm.
It explains Ostwald ripening, capillary condensation, and the size dependence of nucleation barriers.
The equation follows directly from the Laplace pressure combined with the pressure dependence of chemical potential.
Key idea: curving a liquid surface raises the mechanical pressure inside it, and that extra pressure pushes up the chemical potential of the liquid, forcing the vapour pressure to rise to restore equilibrium.

The core insight

Curvature compresses or stretches a liquid's own internal pressure through the Laplace relation, and because chemical potential depends on pressure, that stress is transmitted directly into an exponential shift in equilibrium vapour pressure — small drops are more volatile, concave menisci less so.

The physical picture

A flat liquid surface at temperature $T$ has a well-defined equilibrium vapour pressure $p^{*}$ , the ordinary saturation vapour pressure tabulated in reference tables. Curve that surface into a small sphere of radius $r$ , however, and something new happens. The surface tension $\gamma$ pulls the surface inward (for a convex liquid surrounded by vapour), and by the Laplace equation this generates an excess pressure inside the liquid,

\Delta p = p_{\text{liquid}} - p_{\text{vapour}} = \frac{2\gamma}{r}.

That excess pressure is mechanical, but liquids are not perfectly incompressible in their thermodynamic response: raising the pressure on a liquid raises its chemical potential, $(\partial \mu / \partial p)_T = V_m$ . A liquid held at higher internal pressure therefore has a higher chemical potential than the same liquid at the flat, unstressed reference state. Since the liquid and its vapour must stay in equilibrium, the vapour must also rise to a higher chemical potential — and because $\mu_{\text{vapour}}$ depends logarithmically on pressure through the ideal-gas relation, this manifests as an exponential increase in vapour pressure. Physically: molecules on a strongly curved convex surface have fewer neighbours holding them in the liquid, so they escape into the vapour more easily. A concave surface, like the meniscus of a wetting liquid inside a narrow capillary, has the opposite curvature sign; molecules there are, in effect, more tightly held (or rather, evaporation must fight not just the flat-surface cohesion but also the tendency of the concave film to draw liquid in by capillarity), so the equilibrium vapour pressure drops below $p^{*}$ . This is precisely what allows vapour to condense inside a fine pore at pressures well below the bulk saturation pressure — capillary condensation.

Figure

Convex droplet versus concave meniscus and their vapour pressures

The core relations

\ln\left(\frac{p}{p^{*}}\right) = \frac{2\gamma V_m}{rRT}

where: $p$ is the vapour pressure over the curved surface of radius $r$ , $p^{*}$ is the flat-surface (bulk) saturation vapour pressure, $\gamma$ is the surface tension, $V_m$ is the molar volume of the liquid, $R$ is the gas constant, $T$ is temperature, and $r$ is taken positive for a convex liquid surface (droplet) and negative for a concave one (meniscus wetting a pore wall).

\Delta p_{\text{Laplace}} = \frac{2\gamma}{r}

where: this is the mechanical pressure excess inside a spherical liquid surface, the direct input into the Kelvin derivation.

r^{*} = \frac{2\gamma V_m}{RT \ln(p/p^{*})}

where: $r^{*}$ is the critical (Kelvin) radius of a droplet that is in metastable equilibrium with vapour at supersaturation ratio $p/p^{*}$ — the germ radius central to nucleation theory.

Derivation AFrom Laplace pressure to the Kelvin equation

Goal. Derive the relation between the curvature of a liquid surface and its equilibrium vapour pressure, starting only from the Laplace equation and the definition of chemical potential.

Step 1. Mechanical equilibrium of the curved surface. For a spherical liquid surface of radius $r$ bounded by vapour, mechanical (Laplace) equilibrium requires

p_{\ell} = p_v + \frac{2\gamma}{r},

where $p_\ell$ is the pressure inside the liquid and $p_v$ is the vapour pressure just outside it, this being the ordinary curved-surface vapour pressure we want to relate to $p^{*}$ .

Step 2. Chemical potential of the liquid under this pressure. At constant $T$ , the pressure dependence of the molar Gibbs energy (chemical potential) of the liquid is

\left(\frac{\partial \mu_\ell}{\partial p_\ell}\right)_T = V_m.

Treating the liquid as essentially incompressible over this small pressure range, integrate from the flat reference state (pressure $p^{*}$ , chemical potential $\mu_\ell^{*}$ ) to the curved state (pressure $p_\ell$ ):

\mu_\ell(p_\ell) = \mu_\ell^{*}(p^{*}) + V_m\left(p_\ell - p^{*}\right).

Step 3. Chemical potential of the vapour, treated as ideal. The vapour in equilibrium with the curved liquid has chemical potential

\mu_v(p_v) = \mu_v^{\ominus} + RT\ln\left(\frac{p_v}{p^{\ominus}}\right),

and similarly for the flat-surface reference vapour at pressure $p^{*}$ :

\mu_v^{*}(p^{*}) = \mu_v^{\ominus} + RT\ln\left(\frac{p^{*}}{p^{\ominus}}\right).

Step 4. Impose liquid–vapour equilibrium at both surfaces. At the flat reference surface, $\mu_\ell^{*} = \mu_v^{*}$ . At the curved surface, $\mu_\ell(p_\ell) = \mu_v(p_v)$ . Subtracting the two equilibrium conditions eliminates the standard-state terms:

\mu_\ell(p_\ell) - \mu_\ell^{*} = \mu_v(p_v) - \mu_v^{*}.

Substituting Steps 2 and 3:

V_m\left(p_\ell - p^{*}\right) = RT \ln\left(\frac{p_v}{p^{*}}\right).

Step 5. Insert the Laplace excess pressure. Since we are interested in the small correction due to curvature, approximate $p_\ell - p^{*} \approx p_\ell - p_v = 2\gamma/r$ (the vapour pressure itself is tiny compared with the Laplace term for the purpose of this substitution — a valid approximation because $p_v \ll p_\ell$ under typical conditions, and any residual error is second order). This gives

V_m \cdot \frac{2\gamma}{r} = RT \ln\left(\frac{p_v}{p^{*}}\right).

Step 6. Rearrange. Dividing through by $RT$ and relabeling $p_v \to p$ as "the observed vapour pressure over the curved surface" gives the boxed result:

\boxed{\ln\left(\frac{p}{p^{*}}\right) = \frac{2\gamma V_m}{rRT}}

For $r>0$ (convex droplet) the vapour pressure exceeds $p^{*}$ ; for $r<0$ (concave meniscus, liquid wetting the walls of a pore) the vapour pressure is suppressed below $p^{*}$ .

Worked examples

Example 1 — Foundational: vapour pressure enhancement over a water droplet. Calculate the fractional increase in vapour pressure of a spherical water droplet of radius $r = 10\text{ nm}$ at $298\text{ K}$ , given $\gamma = 0.0720\ \text{N m}^{-1}$ , $V_m = 1.807\times10^{-5}\ \text{m}^3\ \text{mol}^{-1}$ (molar mass $18.02\ \text{g mol}^{-1}$ divided by density $997\ \text{kg m}^{-3}$ ), and $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ . Step 1 — compute the exponent:

\frac{2\gamma V_m}{rRT} = \frac{2 \times 0.0720 \times 1.807\times10^{-5}}{(10\times10^{-9})(8.314)(298)}.

Numerator: $2 \times 0.0720 \times 1.807\times10^{-5} = 2.602\times10^{-6}\ \text{J m}^2\text{mol}^{-1}\text{(per mol, dimensionally N}\,\cdot\,\text{m}^2\text{/m)}$ . More carefully, tracking units: $\gamma V_m$ has units $\text{N m}^{-1} \times \text{m}^3\text{mol}^{-1} = \text{N m}^2 \text{mol}^{-1} = \text{J m mol}^{-1}$ . So numerator $=2.602\times10^{-6}\ \text{J m mol}^{-1}$ . Denominator: $r R T = 10\times10^{-9}\ \text{m} \times 8.314\ \text{J K}^{-1}\text{mol}^{-1}\times 298\ \text{K} = 2.478\times10^{-5}\ \text{J m mol}^{-1}$ .

\frac{2\gamma V_m}{rRT} = \frac{2.602\times10^{-6}}{2.478\times10^{-5}} = 0.1050.

Step 2 — exponentiate:

\frac{p}{p^{*}} = e^{0.1050} = 1.111.

So a 10 nm water droplet has a vapour pressure about 11% higher than flat water at the same temperature. At room temperature bulk water's $p^{*} \approx 3.17\ \text{kPa}$ , so the droplet's local vapour pressure is about $3.52\ \text{kPa}$ . Sanity check: for a much larger droplet, say $r = 1\ \mu\text{m}$ (1000 nm), the exponent shrinks by a factor of 100 to $0.00105$ , giving $p/p^{*} = 1.0010$ — a mere 0.1% enhancement, confirming the effect vanishes for everyday-sized droplets, as expected.

Example 2 — Applied: capillary condensation pressure in a mesopore. A mesoporous silica sample is characterized by nitrogen adsorption at $T = 77.3\ \text{K}$ (the normal boiling point of $\text{N}_2$ , the standard condition for BET/BJH pore analysis). Liquid nitrogen at this temperature has $\gamma = 8.85\times10^{-3}\ \text{N m}^{-1}$ and molar volume $V_m = 3.47\times10^{-5}\ \text{m}^3\ \text{mol}^{-1}$ (density $806\ \text{kg m}^{-3}$ , molar mass $28.01\ \text{g mol}^{-1}$ ). Nitrogen wets silica completely, so the meniscus in a cylindrical pore of radius $r_p = 4.0\ \text{nm}$ is hemispherical with radius of curvature $r = -r_p = -4.0\ \text{nm}$ (concave, negative by convention). Find the relative pressure $p/p^{*}$ at which capillary condensation occurs in this pore. Step 1 — exponent:

\frac{2\gamma V_m}{rRT} = \frac{2\times 8.85\times10^{-3}\times 3.47\times10^{-5}}{(-4.0\times10^{-9})(8.314)(77.3)}.

Numerator: $2\times 8.85\times10^{-3}\times3.47\times10^{-5} = 6.142\times10^{-7}\ \text{J m mol}^{-1}$ . Denominator: $(4.0\times10^{-9})(8.314)(77.3) = 2.571\times10^{-6}\ \text{J m mol}^{-1}$ , carrying the negative sign from $r$ : $-2.571\times10^{-6}$ .

\frac{2\gamma V_m}{rRT} = \frac{6.142\times10^{-7}}{-2.571\times10^{-6}} = -0.2389.

Step 2 — exponentiate:

\frac{p}{p^{*}} = e^{-0.2389} = 0.7875.

Capillary condensation in this 4.0 nm pore occurs at about 79% of the bulk saturation pressure of nitrogen — exactly the kind of relative-pressure depression exploited in BJH (Barrett–Joyner–Halenda) pore-size analysis, where an adsorption/desorption isotherm's condensation step is inverted through this equation to extract a distribution of pore radii. Sanity check: smaller pores should show condensation at even lower relative pressure. For $r_p = 2.0\ \text{nm}$ , the exponent doubles to $-0.4778$ , giving $p/p^{*} = 0.620$ — condensation now requires only 62% relative humidity/pressure, consistent with the trend that narrower pores fill at lower pressure.

Limiting cases & common traps

Situation	Behaviour	Note
$r \to \infty$ (flat surface)	$p \to p^{*}$	Recovers ordinary tabulated vapour pressure
$r > 0$ , convex (droplet)	$p > p^{*}$	Small droplets are more volatile; drives Ostwald ripening
$r < 0$ , concave (wetting meniscus in pore)	$p < p^{*}$	Basis of capillary condensation below saturation
$r \lesssim 1$ –2 nm	Equation becomes unreliable	Continuum $\gamma$ and $V_m$ lose meaning at near-molecular scale
Non-wetting liquid in a pore (contact angle $>90°$ )	Meniscus convex into the pore	Sign of $r$ flips; suppresses rather than promotes condensation

The chemist behind it

The equation is named for William Thomson, 1st Baron Kelvin, who in 1871 combined the Young–Laplace description of curved interfaces with the thermodynamics of chemical equilibrium to predict that vapour pressure must depend on droplet curvature. His result predated a full molecular picture of surface tension and nucleation, yet it remains the exact continuum-thermodynamic relation used today, essentially unmodified, in meteorology (cloud droplet growth), materials science (Ostwald ripening, sintering), and surface science (BJH pore-size analysis from adsorption isotherms).

Origins · 1 minWilliam Thomson, Lord Kelvin and how this was discovered

William Thomson, Lord Kelvin

1824–1907 · derived the curvature dependence of vapour pressure from Laplace pressure and chemical equilibrium, foundational to nucleation and capillary condensation theory

Ask the tutor: How does the Kelvin equation combine with classical nucleation theory to predict the critical germ size for a new phase forming from a supersaturated vapour?

Why does a small droplet have a higher vapour pressure than a flat liquid surface

Curvature compresses the liquid inside a convex droplet via the Laplace relation, raising its internal pressure above the flat-surface value. Because the chemical potential of a liquid rises with pressure, the compressed liquid has higher chemical potential than flat liquid. For the vapour to stay in equilibrium with this higher-potential liquid, the vapour pressure must also rise. Physically, molecules on a highly curved surface have fewer stabilizing neighbours, so they escape into the vapour phase more readily than molecules on a flat surface at the same temperature.

How does the Kelvin equation explain capillary condensation

Inside a narrow, wetting pore, the liquid meniscus curves concavely (negative radius by convention). The Kelvin equation then predicts a vapour pressure below the flat-surface saturation value. This means vapour can condense to liquid inside the pore even when the surrounding bulk vapour pressure is well under $p^{*}$ . Narrower pores condense at lower relative pressures. This size-dependent condensation pressure is exactly what BJH and related methods invert to extract pore-size distributions from adsorption isotherms.

Is the Kelvin equation responsible for Ostwald ripening

Yes, directly. In a suspension of droplets or crystallites of different sizes, the Kelvin equation shows smaller particles have higher equilibrium vapour pressure (or, for crystals in solution, higher solubility via the analogous Ostwald–Freundlich equation). Material therefore evaporates or dissolves preferentially from small particles and redeposits on larger ones, since the surrounding medium is supersaturated relative to the large particles but undersaturated relative to the small ones. Over time small particles vanish and large ones grow, coarsening the size distribution.

Why does the Kelvin equation fail for radii below about 1–2 nanometres

The derivation treats surface tension $\gamma$ and molar volume $V_m$ as continuum, bulk properties independent of curvature. At radii of only a few molecular diameters, the liquid near a droplet's surface no longer resembles bulk liquid, and $\gamma$ itself becomes curvature-dependent (the Tolman correction) with poorly defined boundaries between "surface" and "bulk." The equation remains qualitatively correct in direction, but quantitative predictions near 1 nm should be treated cautiously, and molecular simulation or explicit corrections become necessary.