The vapor pressure of small liquid drops depends on the drop size.
Although the surface properties of a liquid are different from those of the bulk liquid, the special surface properties can be ignored except in a few situations. One is the case in which a liquid is dispersed into fine droplets and the surface then constitutes a large fraction of the total material. A similar situation occurs with finely divided material.
Consider the transfer of dn mol of liquid from bulk liquid to a droplet of radius r. if the normal vapor pressure of the liquid is P0 and of the droplet is P, the free energy change for this can be written, according as
dG = dn RT In P/P0
the free energy change can also be calculated from the surface energy change of the droplet that results from the surface area increase due to the addition of dn mol of the substance with molar mass M. this addition produces a volume increase of M dn/p.
The volume adds a spherical shell, whose area is 4∏r2. The increase in the radius of the droplet dr is given by the relation
M dn/p = 4∏r2 dr
Or
dr = M/4∏r2p dn
The increase in surface energy is γ times the increase in the surface area that results from the increase dr in the droplets radius; i.e.
dG = γdA = γ [4∏ (r + dr)2 - 4∏r2] = 8γ∏r dr
substitution of equation gives
dn RT In P/P_0 = 2γM/pr dn
And In P/P_0 = 2γM/prRT
if as is assumed here, SI units are used, care must be taken to state the density in kilograms per cubic meter instead of the often used grams per millimeter. The conversion is p(kg m-3) = 103 p(g mL-1).
Vapor pressure of water as a function of radius of curvature of surface at 25°C (P0 = 0.03167 bar and γ = 0.07197 Nm-1)
|
m
|
nm
|
P/P0
|
|
10-6
|
103
|
1.001
|
|
10-7
|
102
|
1.011
|
|
10-8
|
101
|
1.111
|
|
10-9
|
100
|
2.88
|
Equation relates the vapor pressure P of a droplet with a highly curved surface to the vapor pressure P0 of the bulk liquid. The appearance of r in the denominator implies the dependence of vapor pressure on droplet size that is illustrated in the table.
These data produce something of a dilemma when condensation of a vapor to a liquid is measured. The creation of an initial small droplet of liquid would lead to a particle with such a high vapor pressure, according to, that it would evaporate even if the pressure of the vapor were greater than the vapor pressure of the bulk liquid. Condensation can take place on dust particles or other irregularities so that the equilibrium thermodynamic result can be circumvented by some mechanism that avoids an initial slow equilibrium growth of droplets.
Similar condensations are necessary when the reverse process, the boiling of a liquid, which requires the formation of small vapor nuclei, is treated. Chemically, one also encounters this phenomenon in the difficulty with which some precipitates form and in the tendency for liquids to supercollider. Likewise, the digestion of a precipitate makes use of the high free energy of the smaller crystals for their conversion to larger particles.