MS Word doc - Inside Mines

CH353 Homework #4
Answer the following two questions, based on the posted reading “Kinetics of Phase
Transformations” (from The Principles of Engineering Materials, Barrett, Nix and
1. Suppose that nuclei formed during a phase transformation (for instance in forming a
solid crystal from a liquid) are cubic in shape rather than spherical. If a is the
dimension of the cube, compute a value for the critical nucleus size, a*, and critical
free energy, G*, necessary to form a stable nuclei, in terms of the surface tension, ,
and latent heat of melting, Ev.
The reasoning for the formation of the cubic critical nucleus is the same as for the
spherical critical nucleus, except that the volume and surface area terms that
appear in the free energy cost of forming the nucleus, Gtot , are now a3 and 6a2 (the
volume and surface of a cube of side length a) respectively:
Gtot  a 3GV  6a 2
Note that the free energy of forming a nucleus still has the qualitative size3 versus
size2 terms, and thus has a maximum corresponding to the critical nucleus size:
d 3
Gtot 
(a GV )  (6a  )
 3a GV  12a
a* 
a  a*
As before, since SV and HV don’t vary much with T, we can use Eqn (5-23) in the
text that
EV T
GV 
a* 
EV T
G*  a* GV  6a*   a* GV  6a* 
32 3 Tmelt 2
EV 2 T 2
2. A small droplet (1 mg) of liquid copper is cooled until it solidifies. Calculate the
amount of undercooling (i.e., the temperature at which at least one stable nucleus
forms) if Tmelt = 1083°C,  = 200 ergs/cm2, density = 10 g/cm3, and Ev = -50 cal/g.
(Hint: You will need to numerically solve an implicit equation (by iteration or other
method) for T such that n*=1.) You can assume that the density of nuclei sites is 1020
per cm3. (This is the value given in the example in the reading. Actually, it is also
possible to estimate the density of potential nuclei sites as (r*) -3, where r* is the
critical nuclei radius. This expression for r* can then be used to solve for n*, but this
will lead to a negligible change in the final value for n*, since the temperature
dependence occurs primarily in G*, not s.)
The number of critical nuclei, n*, is given by the number of nucleation sites times
the probability of a stable critical nucleus forming per site:
 G* 
n*  s exp 
 kT 
where the number of nuclei sites, s, is the density of nuclei sites times the volume of
the sample:
s  (1020 cm 3 ) (.001g / 10 g cm -3 )  1016 nuclei sites
and the free energy for forming a spherical critical nucleus is (Eqn 5-28)
16 3 Tmel t2
G 
3EV 2 T 2
For n*=1, then we solve the equation
16 3Tmel t2
1  1016 exp 
 3EV kB T(Tmel t  T) 
for T. Plugging in the numbers above, and using, for instance, Mathematica, we
T 1028C
for the temperature of undercooling.
Answer the following, based on the posted reading “Surface Phenomena” (from Physical
Chemistry, 6th ed., Barrow):
3. Suppose that water in a certain container boils when the liquid is at a temperature at
which bubbles of 10-7 m diameter can be formed by the equilibrium vapor. Estimate
from approximate relations and data given in the reading (a) the vapor pressure that
must be reached by the liquid for these vapor bubbles to be formed, and (b) the
temperature at which boiling will occur at a pressure of 1 bar.
(a) From Eqn (7-20), the excess pressure in the bubble is given by
2 2(0.05885 Nm -1 )
 23.54 bar
5 x 10 8 m
(b) From the Clausius-Clayperon equation, which tells us how P varies with T along
the coexistence curve (on which P is the vapour pressure), we have
Hvap 
 1  1 
ln( P2 / P1 )  
R T2 T1 
where P1 = 1 bar, and T1 = 100°C = 373K. Under these conditions, Hvap = 40.7
kJ/mol. Thus P2 = 24.54 bar, and we can numerically solve for T2 = 493K = 223°C.
Thus for bubbles of this size, the excess pressure required allows superheating to
occur to quite high temperatures before boiling begins.
P 
4. What is the minimum size of the droplets that could grow spontaneously from a vapor
with twice the equilibrium vapor pressure at 20°C if the vapor is (a) benzene, and
(b) water. You can use a value of .9 g/cm3 for benzene. (At 20°C the vapor pressure
of benzene is about 0.1 bar, and for water it is about 0.023 bar. Do we need this
information to solve the problem?)
(a) Benzene: Since the vapour pressure of small liquid drops, P = 2 P0, is higher
than that of the bulk fluid,P0, as given by (Eqn. 7-24)
ln( P / P0 )  ln 2 
where M = 78.1 g/mol is the molar mass of benzene, and its surface tension  =
0.0289 N/m (from Table 7-1), we can solve for the droplet size, r, from this equation
to give r = 3.0 x 10-9 m = 3 nm.
(b) Water: We use the same approach, but using the values for water of M = 18.02
g/mol and  = 0.07275 N/m, we find r = 1.56 x 10-10 m = 1.56 Å. Of course these
quantities are on the scale of an atom, and so we would expect that at these
pressures,any droplet would grow spontaneously, being composed of many