Page 51 - Basic physical chemistry for the atmospheric sciences

Chemical thermodynamics 37

chance collisions to a radius that just exceeds r will continue to grow
by condensation from the vapor phase, because, by so doing, they
s
cause a decrease in the total energy of the y stem.
In the region just around R = r, a droplet can grow or evaporate
l
s
infinitesimal y without any change in the energy of the y stem. We can
obtain an expression for r in terms of e by setting a(tl.E)/aR = 0 at
R = r. Hence, from Eq. (2.54)
2 a-
r = ---- (2. 5 5)
nkT 1n(e/e5)

Equation (2. 5 5) is referred to as Kelvin 's f o rmula. It is important, for
example, in understanding the condensation and evaporation of small
d
water drops in air, such as those in clou s .
Exercise 2 . 9 . What is the relative humidity t l 0 °C u s t above the
a
j
surface of a water droplet with a radius of 0.01 0 µ.,m? If the droplet
were placed in an enclosure having this same relative humidity, would
the droplet be in stable or unstable equilibrium? (The surface energy
of water at 1 0 °C is . 0 76 J m - 2 , and the number density of molecules
0
2
3
in water is . 3 x 1 0 8 m- -)
3
Solution. Kelvin' s formula [Eq. (2.55)) allows us to determine the
radius of a droplet that is in equilibrium (i. e . , neither evaporating nor
condensing) with a water vapor pressure e. However, if a droplet
neither evaporates nor grows when the air surrounding it has vapor
pressure e, it is because the vapor pressure just above the surface of
the droplet is equal to e. Th s :we can interpret Eq. (2.55) as giving
u
the vapor pressure e (or relative humidity RH = 1 0 0 e/e5) just above
the surface of a droplet of radius r. Hence, from Eq. (2.55)

e ( 2 a- )
RH = 00 1 es = l OO exp (2.56)
nkTr
2
2
Substituting, a- = 0.076 J m- , n = 3 . 3 x 1 0 8 m - 3 , k = l .381 x l o - n
1
J deg- molecule - 1 , T = 283 K , and r = 0.01 0 µ.,m = 1 . 0 x 1 0 - s m into
Eq. (2.56) gives RH = 1 1 0%. (Not : Substitution of the numerical
e
values given in this problem into Eq. (2.56) gives RH = 1 1 3 . How­
%
ever, the numerical values are given to an accuracy of just two signifi­
cant figures. Therefore, the derived value of RH is also accurate to
just two significant figures; hence 1 3% must be rounded to 1 0% )
.
1
1
I f the droplet is placed in n enclosure with a relative humidity of
a
1 1 0 % , it will be in equilibrium. However, if the droplet should evapo-