Solutions Manual for
Polymer Science and
Technology
Third Edition
Joel R. Fried
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ISBN-10: 0-13-384559-1
ISBN-13: 978-0-13-384559-4
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
SOLUTIONS TO PROBLEMS IN POLYMER SCIENCE AND TECHNOLOGY,
3RD EDITION
TABLE OF CONTENTS
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 7
Chapter 11
Chapter 12
Chapter 13
1
5
14
24
28
36
40
51
52
CHAPTER 1
1-1 A polymer sample combines five different molecular-weight fractions, each of equal weight. The
molecular weights of these fractions increase from 20,000 to 100,000 in increments of 20,000.
Calculate M n , M w , and M z . Based upon these results, comment on whether this sample has a
broad or narrow molecular-weight distribution compared to typical commercial polymer samples.
Solution
Fraction #
1
2
3
4
5
Σ
=
Mn
5
Wi N
∑=
i =1
Mi (×10-3)
20
40
60
80
100
300
Wi
1
1
1
1
1
5
Ni = Wi/Mi (×105)
5.0
2.5
1.67
1.25
1.0
11.42
5
=
43,783
1.142 × 10−4
5
=
Mw
∑W M
i
i
=
5
∑Wi
i =1
300,000
= 60,000
5
i =1
5
Mz
=
∑W M
i
2
i
=
∑Wi M i
i =1
5
4 × 108 + 16 × 108 + 36 × 108 + 64 × 108 + 100 × 108
= 73,333
3 × 105
i =1
M z 60,000
=
= 1.37 (narrow distribution)
M n 43,783
1-2 A 50-gm polymer sample was fractionated into six samples of different weights given in the table
below. The viscosity-average molecular weight, M v , of each was determined and is included in the table.
Estimate the number-average and weight-average molecular weights of the original sample. For these
calculations, assume that the molecular-weight distribution of each fraction is extremely narrow and can
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
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1
be considered to be monodisperse. Would you classify the molecular weight distribution of the original
sample as narrow or broad?
Fraction
1
2
3
4
5
6
Weight
(gm)
1.0
5.0
21.0
15.0
6.5
1.5
Mv
1,500
35,000
75,000
150,000
400,000
850,000
Solution
Let M i ≈ M v
=
Mn
6
W N
∑=
i
i =1
Fraction
Wi
Mi
1
2
3
4
5
6
Σ
1.0
5.0
21.0
15.0
6.5
1.5
50.0
1,500
35,000
75,000
150,000
400,000
850,000
Ni = Wi/Mi
(×106)
667
143
280
100.
16.3
1.76
1208
WiMi
1500
175.000
627,500
2,250,000
2,600,000
1,275,000
7,929,000
50.0
= 41,322
1.21 × 10−3
6
=
Mw
∑W M
i
i
=
6
∑Wi
i =1
7,930,000
= 158,600
50.0
i =1
M w 158, 600
=
= 3.84 (broad distribution)
Mn
41,322
1-3 The Schultz–Zimm [11] molecular-weight-distribution function can be written as
=
W (M )
a b +1
M b exp ( − aM )
Γ ( b + 1)
where a and b are adjustable parameters (b is a positive real number) and Γ is the gamma function (see
Appendix E) which is used to normalize the weight fraction.
(a) Using this relationship, obtain expressions for M n and M w in terms of a and b and an expression for
M max , the molecular weight at the peak of the W(M) curve, in terms of M n .
Solution
Mn =
∫
∞
0
∞
WdM
∫ (W
0
M ) dM
let t = aM
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
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∫
∞
0
∞
1 ∞ b
1
a b +1
a b +1
b
d
t
a
t exp ( −=
t ) dt
−
=
Γ ( b=
+ 1) 1
exp
t
a
t
(
)
(
)
(
)
∫
Γ ( b + 1) 0
Γ ( b + 1) a b +1 ∫0
Γ ( b + 1)
=
WdM
∞
∫0 (W
M=
) dM
∞
a b +1
a b +1 1
b −1
−
=
d
t
a
t
a
t
exp
(
)
(
)
(
)
Γ ( b + 1) ∫0
Γ ( b + 1) a b
∫
∞
0
−t ) dt
t b −1 exp (=
a b +1 1
=
Γ (b)
Γ ( b + 1) a b
a
a
Γ (b) =
bΓ ( b )
b
1
b
=
ab a
M
=
n
∫
M w=
∞
0
∫
WMdM
∞
0
=
WdM
∫
∞
0
WMdM =
b +1
∞
a b +1
a b +1 Γ ( b + 2 )
t
exp
−
=
=
t
d
t
a
a
(
)
(
)
(
)
Γ ( b + 1) ∫0
Γ ( b + 1) a b + 2
( b + 1) Γ ( b + 1) = b + 1
aΓ ( b + 1)
a
(b) Derive an expression for Mmax, the molecular weight at the peak of the W(M) curve, in terms of M n .
Solution
dW
a b +1
bM b −1 exp ( − aM ) + M b ( − a=
=
) exp ( −aM ) 0
dM Γ ( b + 1) 
bM b − a = aM b
b
a
= M
=
M n (i.e., the maximum occurs at M n )
a
(c) Show how the value of b affects the molecular weight distribution by graphing W(M) versus M on the
same plot for b = 0.1, 1, and 10 given that M n = 10,000 for the three distributions.
Solution
b
a=
10,000
b
a
=
W
0.1
1×10-5
1
1×10-4
10
1×10-3
a b +1
M b exp ( − aM ) dM
Γ ( b + 1)
where
=
Γ ( b + 1)
∞
∫ ( aM )
0
b
exp ( − aM ) dM .
Plot W(M) versus M
Hint:
∫
∞
0
x n exp ( − ax ) dx =
Γ ( n + 1) a n +1 =
n ! a n +1 (if n is a positive interger).
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
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1-4 (a) Calculate the z-average molecular weight, M z , of the discrete molecular weight distribution
described in Example Problem 1.1.
Solution
3
=
Mz
∑W M
i
2
i
=
∑Wi M i
i =1
3
1(10,000 ) + 2 ( 50,000 ) + 2 (100,000 )
= 80,968
1(10,000 ) + 2 ( 50,000 ) + 2 (100,000 )
2
2
2
i =1
(b) Calculate the z-average molecular weight, M z , of the continuous molecular weight distribution
shown in Example 1.2.
Solution
M dM ( M 3)
∫=
=
MdM
( M 2)
∫
105
Mz
=
2
3
105
2
103
105
3
10
105
103
66,673
103
(c) Obtain an expression for the z-average degree of polymerization, X z , for the Flory distribution
described in Example 1.3.
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Solution
∞
∑1 X 2W ( X )
Xz =
=
∞
∑ XW ( X )
1
∞
∑X
3
p x −1
∑X
2
p x −1
1
∞
1
Let
∞
1
A =∑ Xp x −1 =1 + 2 p + 3 p 2 +  =
1− p
1
(geometric series)
∞
1 + 22 p + 32 p 2 + 
B=
∑ X 2 p x −1 =
1
∞
C=
1 + 23 p + 32 p 2 + 
∑ X 3 p x −1 =
1
Can show that B (1 − p ) = A (1 + p )
Therefore B =
1+ p
(1 − p )
Write C (1 − p ) =
3
∞
=
x 1
Therefore C =
∞
∞
∑ 3 X 2 p x −1 − ∑ 3 Xp x −1 + ∑ p x −1 = 3B − 3 A2 +
=
x 1=
x 1
1
1 + 4 p + p2
=
3
1− p
(1 − p )
1 + 4 p + p2
(1 − p )
4
∞
∑X
p x −1
3
2
C 1 + 4 p + p (1 − p )
1
and finally X=
= =
=
z
∞
4
B
2 x −1
p
1
−
1
+
p
(
)
(
)
∑X p
3
1 + 4 p + p2
1 + 4 p + p2
=
1 − p2
(1 − p )(1 + p )
1
Mz = Mo X z
CHAPTER 2
2.1 If the half-life time, t1/2, of the initiator AIBN in an unknown solvent is 22.6 h at 60°C, calculate its
dissociation rate constant, kd, in units of reciprocal seconds.
Solution
=
[ I] [ I]o exp ( −kd t )
[ I]=
[ I]o
1
= exp ( −kd t )
2
This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555)
Copyright 2014, Pearson Education, Inc. Do not redistribute.
5