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.
The original form of the Kelvin equation, published in 1871, is: [1] where:
p(r)
r
P
r=infty
peq
\gamma
\rho\rm
\rho\rm
r1
r2
This may be written in the following form, known as the Ostwald–Freundlich equation:where
p
p\rm
\gamma
Vm
R
r
T
Equilibrium vapor pressure depends on droplet size.
r
p>p\rm
r
p<p\rm
As
r
p
psat
If the vapour is cooled, then
T
p\rm
p/p\rm
\gamma
Vm
r
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.
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]
The formal definition of the Gibbs free energy for a parcel of volume
V
P
T
G=U+pV-TS,
where
U
S
dG=-SdT+VdP+
k | |
\sum | |
i=1 |
\muidni,
where
\mu
n
x
x
r
nx
\DeltaG=Gd-Gv,
where
Gd
Gv
Ni
Nf
Nf=Ni-nx.
Let
gv
gl
\DeltaG=Nfgv+nxgl+4\pir2\sigma-Nigv,
where
4\pir2\sigma
r
\sigma
\DeltaG=(Ni-nx)gv+nxgl+4\pir2\sigma-Nigv=nx(gl-gv)+4\pir2\sigma.
Let
vl
vv
nxvl=
4 | |
3 |
\pir3.
The number of molecules in the drop is then given by
nx=
4\pir3 | |
3vl |
.
The change in Gibbs energy is then
\DeltaG=
4\pir3 | |
3vl |
(gl-gv)+4\pir2\sigma.
The differential form of the Gibbs free energy of one molecule at constant temperature and constant number of molecules can be given by:
dg=(vl-vv)dP.
If we assume that
vv\ggvl
dg\simeq-vvdP.
The vapor phase is also assumed to behave like an ideal gas, so
vv=
kT | |
P |
,
where
k
\Deltag=-kT
P | |
\int\limits | |
Psat |
dP | |
P |
,
where
Psat
x
P
\Deltag=gl-gv=-kTlnl(
P | |
Psat |
r).
The change in the Gibbs free energy following the formation of the drop is then
\DeltaG=-
4 | |
3 |
\pir3
kT | |
vl |
lnl(
P | |
Psat |
r)+4\pir2\sigma.
The derivative of this equation with respect to
r
\partiall(\DeltaGr) | |
\partialr |
=-4\pir2
kT | |
vl |
lnl(
P | |
Psat |
r)+8\pir\sigma.
The maximum value occurs when the derivative equals zero. The radius corresponding to this value is:
r=
2vl\sigma | |||||
|
.
Rearranging this equation gives the Ostwald–Freundlich form of the Kelvin equation:
lnl(
P | |
Psat |
r)=
2vl\sigma | |
rkT |
.
p
c
r
p\rm
c\rm
ln
c | |
c\rm |
=
2\gammaVm | |
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 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−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]
ln
c | |
c\rm |
=
Vm | \left( | |
RT |
2\gamma | |
r |
+\Pi\right).
where
\Pi
\left(2\gamma/r\right)
\Pi
c
c\rm