Kelvin equation

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The Kelvin equation describes the change in

, also known as Lord Kelvin.

Formulation

The original form of the Kelvin equation, published in 1871, is: ^[1]

where:

• ${\displaystyle p(r)}$ = vapor pressure at a curved interface of radius ${\displaystyle r}$
• ${\displaystyle P}$ = vapor pressure at flat interface (${\displaystyle r=\infty }$) = ${\displaystyle p_{eq}}$
• ${\displaystyle \gamma }$ = surface tension
• ${\displaystyle \rho _{\rm {vapor}}}$ = density of vapor
• ${\displaystyle \rho _{\rm {liquid}}}$ = density of liquid
• ${\displaystyle r_{1}}$ , ${\displaystyle r_{2}}$ = radii of curvature along the principal sections of the curved interface.

This may be written in the following form, known as the Ostwald–Freundlich equation:

where ${\displaystyle p}$ is the actual vapour pressure, ${\displaystyle p_{\rm {sat}}}$ is the when the surface is flat, ${\displaystyle \gamma }$ is the liquid/vapor surface tension, ${\displaystyle V_{\text{m}}}$ is the molar volume of the liquid, ${\displaystyle R}$ is the , ${\displaystyle r}$ is the radius of the droplet, and ${\displaystyle T}$ is temperature.

depends on droplet size.

• If the curvature is convex, ${\displaystyle r}$ is positive, then ${\displaystyle p>p_{\rm {sat}}}$
• If the curvature is concave, ${\displaystyle r}$ is negative, then ${\displaystyle p<p_{\rm {sat}}}$

As ${\displaystyle r}$ increases, ${\displaystyle p}$ decreases towards ${\displaystyle p_{sat}}$, and the droplets grow into bulk liquid.

If the vapour is cooled, then ${\displaystyle T}$ decreases, but so does ${\displaystyle p_{\rm {sat}}}$. This means ${\displaystyle p/p_{\rm {sat}}}$ increases as the liquid is cooled. ${\displaystyle \gamma }$ and ${\displaystyle V_{\text{m}}}$ may be treated as approximately fixed, which means that the critical radius ${\displaystyle r}$ 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.^[2] In 2020, researchers found that the equation was accurate down to the 1nm scale.^[3]

Derivation using the Gibbs free energy

The formal definition of the Gibbs free energy for a parcel of volume ${\displaystyle V}$, pressure ${\displaystyle P}$ and temperature ${\displaystyle T}$ is given by:

${\displaystyle G=U+pV-TS,}$

where ${\displaystyle U}$ is the internal energy and ${\displaystyle S}$ is the entropy. The differential form of the Gibbs free energy can be given as

${\displaystyle dG=-SdT+VdP+\sum _{i=1}^{k}\mu _{i}dn_{i},}$

where ${\displaystyle \mu }$ is the chemical potential and ${\displaystyle n}$ is the number of moles. Suppose we have a substance ${\displaystyle x}$ which contains no impurities. Let's consider the formation of a single drop of ${\displaystyle x}$ with radius ${\displaystyle r}$ containing ${\displaystyle n_{x}}$ molecules from its pure vapor. The change in the Gibbs free energy due to this process is

${\displaystyle \Delta G=G_{d}-G_{v},}$

where ${\displaystyle G_{d}}$ and ${\displaystyle G_{v}}$ are the Gibbs energies of the drop and vapor respectively. Suppose we have ${\displaystyle N_{i}}$ molecules in the vapor phase initially. After the formation of the drop, this number decreases to ${\displaystyle N_{f}}$, where

${\displaystyle N_{f}=N_{i}-n_{x}.}$

Let ${\displaystyle g_{v}}$ and ${\displaystyle g_{l}}$ represent the Gibbs free energy of a molecule in the vapor and liquid phase respectively. The change in the Gibbs free energy is then:

${\displaystyle \Delta G=N_{f}g_{v}+n_{x}g_{l}+4\pi r^{2}\sigma -N_{i}g_{v},}$

where ${\displaystyle 4\pi r^{2}\sigma }$ is the Gibbs free energy associated with an interface with radius of curvature ${\displaystyle r}$ and surface tension ${\displaystyle \sigma }$. The equation can be rearranged to give

${\displaystyle \Delta G=(N_{i}-n_{x})g_{v}+n_{x}g_{l}+4\pi r^{2}\sigma -N_{i}g_{v}=n_{x}(g_{l}-g_{v})+4\pi r^{2}\sigma .}$

Let ${\displaystyle v_{l}}$ and ${\displaystyle v_{v}}$ be the volume occupied by one molecule in the liquid phase and vapor phase respectively. If the drop is considered to be spherical, then

${\displaystyle n_{x}v_{l}={\frac {4}{3}}\pi r^{3}.}$

The number of molecules in the drop is then given by

${\displaystyle n_{x}={\frac {4\pi r^{3}}{3v_{l}}}.}$

The change in Gibbs energy is then

${\displaystyle \Delta G={\frac {4\pi r^{3}}{3v_{l}}}(g_{l}-g_{v})+4\pi r^{2}\sigma .}$

The differential form of the Gibbs free energy of one molecule at constant temperature and constant number of molecules can be given by:

${\displaystyle dg=(v_{l}-v_{v})dP.}$

If we assume that ${\displaystyle v_{v}\gg v_{l}}$ then

${\displaystyle dg\simeq -v_{v}dP.}$

The vapor phase is also assumed to behave like an ideal gas, so

${\displaystyle v_{v}={\frac {kT}{P}},}$

where ${\displaystyle k}$ is the Boltzmann constant. Thus, the change in the Gibbs free energy for one molecule is

${\displaystyle \Delta g=-kT\int \limits _{P_{sat}}^{P}{\frac {dP}{P}},}$

where ${\displaystyle P_{sat}}$ is the

of ${\displaystyle x}$ over a flat surface and ${\displaystyle P}$ is the actual vapor pressure over the liquid. Solving the integral, we have

${\displaystyle \Delta g=g_{l}-g_{v}=-kT\ln {\Bigl (}{\frac {P}{P_{sat}}}{\Bigr )}.}$

The change in the Gibbs free energy following the formation of the drop is then

${\displaystyle \Delta G=-{\frac {4}{3}}\pi r^{3}{\frac {kT}{v_{l}}}\ln {\Bigl (}{\frac {P}{P_{sat}}}{\Bigr )}+4\pi r^{2}\sigma .}$

The derivative of this equation with respect to ${\displaystyle r}$ is

${\displaystyle {\frac {\partial {\bigl (}\Delta G{\bigr )}}{\partial r}}=-4\pi r^{2}{\frac {kT}{v_{l}}}\ln {\Bigl (}{\frac {P}{P_{sat}}}{\Bigr )}+8\pi r\sigma .}$

The maximum value occurs when the derivative equals zero. The radius corresponding to this value is:

${\displaystyle r={\frac {2v_{l}\sigma }{kT\ln {\Bigl (}{\frac {P}{P_{sat}}}{\Bigr )}}}.}$

Rearranging this equation gives the Ostwald–Freundlich form of the Kelvin equation:

${\displaystyle \ln {\Bigl (}{\frac {P}{P_{sat}}}{\Bigr )}={\frac {2v_{l}\sigma }{rkT}}.}$

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 ${\displaystyle p}$ is replaced by the solubility of the solid (${\displaystyle c}$) (or a second liquid) at the given radius, ${\displaystyle r}$, and ${\displaystyle p_{\rm {sat}}}$ by the solubility at a plane surface (${\displaystyle c_{\rm {sat}}}$). Hence small particles (like small droplets) are more soluble than larger ones. The equation would then be given by:

${\displaystyle \ln {\frac {c}{c_{\rm {sat}}}}={\frac {2\gamma V_{\text{m}}}{rRT}}.}$

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

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^−ΔS/k, where ΔS is the deviation of the entropy from the equilibrium value.^[4]

It is unlikely, however, that new phases often arise by this fluctuation mechanism and the resultant spontaneous nucleation. Calculations show that the chance, e^−ΔS/k, 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^[5]

${\displaystyle \ln {\frac {c}{c_{\rm {sat}}}}={\frac {V_{\text{m}}}{RT}}\left({\frac {2\gamma }{r}}+\Pi \right).}$

where ${\displaystyle \Pi }$ is the disjoining pressure that accounts for the intermolecular interactions between the sessile drop and the solid and ${\displaystyle \left(2\gamma /r\right)}$ is the Laplace pressure, accounting for the curvature-induced pressure inside the liquid drop. When the interactions are attractive in nature, the disjoining pressure, ${\displaystyle \Pi }$ is negative. Near the contact line, the disjoining pressure dominates over the Laplace pressure, implying that the solubility, ${\displaystyle c}$ is less than ${\displaystyle c_{\rm {sat}}}$. This implies that a new phase can spontaneously grow on a solid surface, even under saturation conditions.^[6]

See also

• Condensation
• Gibbs–Thomson equation
• Ostwald–Freundlich equation

References

Further reading

• W. J. Moore, Physical Chemistry, 4th ed., Prentice Hall, Englewood Cliffs, N. J., (1962) p. 734–736.
• S. J. Gregg and K. S. W. Sing, Adsorption, Surface Area and Porosity, 2nd edition, Academic Press, New York, (1982) p. 121.
• Arthur W. Adamson and Alice P. Gast, Physical Chemistry of Surfaces, 6th edition, Wiley-Blackwell (1997) p. 54.
• Butt, Hans-Jürgen, Karlheinz Graf, and Michael Kappl. "The Kelvin Equation". Physics and Chemistry of Interfaces. Weinheim: Wiley-VCH, 2006. 16–19. Print.
• Anton A. Valeev,"Simple Kelvin Equation Applicable in the Critical Point Vicinity",European Journal of Natural History, (2014), Issue 5, p. 13-14.