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Polyethylene
Spring 2005
Homework #3
(Solution)
Problem 1
Diffusion is a thermally activated process.
Parameters of a thermally activated process often
obey an exponential equation.
For diffusion coefficients it has the form:
D
=
D
0
e
−
Q
D
RT
D is a function of temperature.
More, precisely, log(D) is a linear function of 1/T, as shown in
the figure in Homework #3.
In this equation you have two unknown values, D
o
and Q.
You
just need to read out two points (D
1
,T
1
) and (D
2
,T
2
) on the curve and then you can determine
the values of D
o
and Q.
Pay attention to the axes of the figure: logD and 1/T.
Taking natural logarithmic of this equation yields:
ln
D
=
ln
D
0
−
Q
D
R
1
T
From the graph
(i)
when D = 10
13
m
2
/s; T=1183˚K
(ii)
when D=10
15
m
2
/s, T=968˚K
Substitute the values of D and T above into the equation gives
QD = 204 kJ/mol
D0 = 1.011 10
4
m
2
/s
Problem 2
It is important to recognize that this is a situation conforming to:
D
1
t
1
= D
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 Summer '07
 MarwanDarwiche

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