Page 49 - Basic physical chemistry for the atmospheric sciences

Chemical thermodynamics .l"i

(2.49)
/Lvsat - � = kT lne 5
2
Also from Exercise . 8 ,
1-Lvsat /L e (2 .50)
=
where /Le is the chemical potential of a molecule in the liquid phase
that is at equilibrium with the vapor phase at temperature T. From
Eqs. (2.48) and (2.49)
e
1-Lv - /Lvsat = kT ln­
e,
or using Eq. (2. 5 0),
e
/J-v - /Le = kT ln­
e
s (2.5 1 )
u
Let s consider now the formation of a pure droplet by condensation
from its supersaturated vapor. In this process, which is referred to as
homogeneous nucleation, the first stage in the growth process is the
chance collisions of a number of molecules in the vapor phase to form
a small embryonic droplet large enough to remain intact. Let V be the
volume and A the surface area of such an embryonic droplet that has
formed at constant temperature and pressure. If /Le and /Lv are the
chemical potentials in the liquid and vapor phases, and n is the number
of molecules per unit volume of liquid, the decrease in the Gibbs free
energy of the system due to the condensation is n V(µ,v - /Le) . Now,
quite apart from any work associated with the change in volume of the
system, work is done in creating the surface area of the droplet. This
work may be written as Aa-, where a- is the work required to create a
unit area of vapor-liquid interface (called the interfacial energy be­
tween the vapor and the liquid, or the sur a ce energy of the liquid). If
f
this were an equilibrium transformation (which it s not) Aa- would be
i
equal to n V(µ,v - /Le) (see Exercise 2 . 1 8 ). Instead, the change in the
Gibbs free energy will, in general, differ from the work term Aa-. Let
us write
(2.52)

then, t:..E is the net increase in the energy of the system due to the
I
formation of the drop. Combining Eqs. (2.5 ) and (2.52) we obtain
t:.. = A a- - n V kT In (;) (2.53)
E