228 5 Phase Equilibrium
EXAMPLE5.12
Find the additional pressure inside a water droplet of radius 1. 00 μmat25◦C.
Solution
At this temperature, the surface tension of water is equal to 0.07197 N m−^1.
P(l)−P(g)
(2)(0.07197 N m−^1 )
1. 00 × 10 −^6 m
1. 44 × 105 Nm−^2 1 .41 atm
If the pressure on a sample of liquid is increased, its vapor pressure increases, as
seen in Eq. (5.3-18). Using the relation of Eq. (5.5-14) for the additional pressure on
the liquid phase in Eq. (5.3-18) we obtain
PvapP
(p)
vapexp
[
Vm(l) 2 γ
rRT
]
(5.5-15)
or
ln
(
Pvap
Pvap(p)
)
2 Vm(l)γ
rRT
(5.5-16)
whereP
(p)
vapis the vapor pressure at a planar surface.
EXAMPLE5.13
The vapor pressure of a planar surface of water at 298.15 K is 23.756 torr. Find the radius of
a water droplet that has a vapor pressure that is 1.000 torr higher than this value at 298.15 K.
At this temperature, the surface tension of water is equal to 0.07197 N m−^1.
Solution
r
2(1. 806 × 10 −^5 m^3 mol−^1 )(0.07197 N m−^1 )
(8.3145 J K−^1 mol−^1 )(298.15 K) ln
(
24 .756 torr
23 .756 torr
)
2. 544 × 10 −^8 m 2. 544 × 10 −^5 mm 25 .44 nm
Since small droplets of a liquid have a larger vapor pressure than large droplets,
small droplets tend to evaporate while large droplets grow by condensation. This is
the mechanism by which raindrops grow large enough to fall. The initial formation of
a small droplet from the vapor is calledhomogeneous nucleationand often requires
a very large partial pressure (large degree of supersaturation). Most raindrops appar-
ently nucleate on specks of solid material (this is calledheterogeneous nucleation).
Seeding of supersaturated air with small crystals of silver iodide has been used in
efforts to produce rain by providing nucleation sites. Silver iodide is used because it is
possible to make 10^15 particles from a single gram of solid silver iodide by spraying
a solution of silver iodide in acetone into the atmosphere through a suitable nozzle.
The crystal structure of silver iodide is also favorable for the attachment of water
molecules.