Solution manual for polymer science and technology 3rd edition by fried

Solution Manual for Polymer Science and Technology 3rd Edition by Frie
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Solutions Manual for

Polymer Science and
Technology
Third Edition

Joel R. Fried

Upper Saddle River, NJ • Boston • Indianapolis • San Francisco
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Solution Manual for Polymer Science and Technology 3rd Edition by Frie
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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)
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Solution Manual for Polymer Science and Technology 3rd Edition by Frie
Full file at />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

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1


Solution Manual for Polymer Science and Technology 3rd Edition by Frie
Full file at />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

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Solution Manual for Polymer Science and Technology 3rd Edition by Frie
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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).

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Solution Manual for Polymer Science and Technology 3rd Edition by Frie
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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 Manual for Polymer Science and Technology 3rd Edition by Frie
Full file at />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

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5