ChemE 178:
Homework #5 SOLUTIONS
Due Friday, October 17
1.) Consider a freelyjointed chain with a distance between joints of 0.6 nm.
a.) Compute the root mean square endtoend distance of such a chain that has 10000 links.
For a FREELY JOINTED CHAIN:
where n, the number of links is 10,000 and l, the distance between joints, is
0.6nm:
r
2
1
2
=
n
1
2
l
=
(10
,000)
1
2
*0
.6
nm
=
60
nm
b.) Compute the radius of gyration of the chain in part a.)
For a linear Gaussian chain, radius of gyration and mean squared endtoend
distance are related by:
R
g
2
=
1
6
r
2
R
g
2
1
2
=
1
6
r
2
1
2
=
1
6
*60
nm
=
24.5
nm
c.) Suppose that the links in the chain can be completed aligned in a straight line, by
applying a force to the end of the chain, for example.
What is the endtoend distance of
this (stretched or aligned) chain?
If we align the chain in a straight line, the endtoend distance is the same as the
countour length of the chain.
Calculate the contour length:
nm
85
.
899
,
4
2
5
.
109
sin
*
nm
6
.
0
*
000
,
10
2
sin
*
l
*
n
L
=
=
β
=
Note: It is not possible to align CCC bonds at a 180 degree angle, so the sin
(beta/2) term takes care of the tetrahedral symmetry of carbon.
2.)(similar to Y&L 3.5):
For the linear molecule of polyethylene with molecular weight 119,980
g/mol calculate
a.) The contour length
From Y&L 3.5 we know l=0.154nm.
We must find n, the number of links.
For
polyethylene, the molecular weight of the repeat unit (C
2
H
4
) is
M
o
=12*2+1*4=28 g/mol.
Neglecting the end groups, the degree of
polymerization is:
x
n
=
MW
M
o
=
119980
g
/
mol
28
g
/
mol
=
4285
2
2
nl
r
=
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 Fall '08
 SEGALMAN
 Polymer, gyration, endtoend distance, freely jointed chain

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