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Solutions Manual for Polymer Science and Technology Third Edition Joel R. Fried Upper Saddle River, NJ • Boston • Indianapolis • San Francisco New York • Toronto • Montreal • London • Munich • Paris • Madrid Capetown • Sydney • Tokyo • Singapore • Mexico City This text is associated with Fried/Polym...

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Solutions Manual for

Polymer Science and Technology Third Edition

Joel R. Fried

Upper Saddle River, NJ • Boston • Indianapolis • San Francisco New York • Toronto • Montreal • London • Munich • Paris • Madrid Capetown • Sydney • Tokyo • Singapore • Mexico City This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

The author and publisher have taken care in the preparation of this book, but make no expressed or implied warranty of any kind and assume no responsibility for errors or omissions No liability is assumed for incidental or consequential damages in connection with or arising out of the use of the information or programs contained herein. Visit us on the Web: InformIT.com/ph Copyright © 2015 Pearson Education, Inc. This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted. The work and materials from it should never be made available to students except by instructors using the accompanying text in their classes. All recipients of this work are expected to abide by these restrictions and to honor the intended pedagogical purposes and the needs of other instructors who rely on these materials. 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 5

M n = ∑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 =1

i

5

∑W

=

300,000 = 60,000 5

=

4 × 108 + 16 × 108 + 36 × 108 + 64 × 108 + 100 × 108 = 73,333 3 × 105

i

i =1

5

Mz =

∑W M i =1 5

i

2 i

∑W M i =1

i

i

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 1 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

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

6

M n = ∑Wi N = 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 =1

6

∑W

i

=

7,930,000 = 158,600 50.0

i

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 2 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

∫

∞

0

∞ a b +1 a b +1 1 ∞ b 1 b d t a t exp ( −t ) dt = − = Γ ( b + 1) = 1 t a exp 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 b −1 exp ( −t ) dt =

a b +1 1 Γ (b) = Γ ( b + 1) a b

a a Γ (b) = bΓ ( b ) b

Mn =

1 b = ab a

Mw =

∫

∞

0

∫

WMdM

∞

0

∞

= ∫ WMdM = 0

WdM

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 = M a = 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) = ∫

∞

0

( aM )

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).

3 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

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 =1 3

i

2 i

∑W M i =1

i

1(10,000 ) + 2 ( 50,000 ) + 2 (100,000 ) 2

=

2

1(10,000 ) + 2 ( 50,000 ) + 2 (100,000 )

2

= 80,968

i

(b) Calculate the z-average molecular weight, M z , of the continuous molecular weight distribution shown in Example 1.2. Solution 105

Mz

∫ = ∫

3

M 2 dM

10 105 103

MdM

( M 3) = ( M 2) 3

105

2

103 105

= 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.

4 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

Solution ∞

Xz =

∑ X 2W ( X )

∞

∑X

3

p x −1

∑ XW ( X ) ∑ X

2

p x −1

1 ∞ 1

=

1 ∞ 1

Let ∞

A = ∑ Xp x −1 = 1 + 2 p + 3 p 2 + " = 1

1 1− p

(geometric series)

∞

B = ∑ X 2 p x −1 = 1 + 22 p + 32 p 2 + " 1

∞

C = ∑ X 3 p x −1 = 1 + 23 p + 32 p 2 + " 1

Can show that B (1 − p ) = A (1 + p ) Therefore B =

1+ p

(1 − p )

3

∞

∞

∞

x =1

x =1

x =1

Write C (1 − p ) = ∑ 3 X 2 p x −1 − ∑ 3 Xp x −1 + ∑ p x −1 = 3B − 3 A2 +

Therefore C =

1 1 + 4 p + p2 = 3 1− p (1 − p )

1 + 4 p + p2

(1 − p ) ∞

and finally X z =

4

∑X

3

p x −1

∑X

2

p x −1

1 ∞

2 1 + 4 p + p2 1 + 4 p + p2 C 1 + 4 p + p (1 − p ) = = = = 4 1 − p2 B (1 − p )(1 + p ) (1 − p ) (1 + p ) 3

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

5 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

− kd t = ln (1 2 ) = −0.693

kd =

0.693 0.693 h = = 8.52 × 10−5 s -1 t 22.6 h 3600 s

2.2 Styrene is polymerized by free-radical mechanism in solution. The initial monomer and initiator concentrations are 1 M (molar) and 0.001 M, respectively. At the polymerization temperature of 60°C, the initiator efficiency is 0.30. The rate constants at the polymerization temperature are as follows: kd = 1.2 × 10-5 s-1 kp = 176 M-1 s-1 kt = 7.2 × 107 M-1 s-1 Given this information, determine the following: (a) Rate of initiation at 1 min and at 16.6 h Solution Ri = 2 fkd [ I ] = 2 ( 0.30 ) (1.2 × 10−5 ) [ I ] = 7.2 × 10−6 [ I ]

[ I] = [ Io ] exp ( −kd t ) at 1 min: [ I] = 0.001( 0.9993) = 0.0009993 M

Ri = ( 7.2 × 10−6 ) ( 0.0009993) = 7.19 × 10−9 M s -1

at 16.6 h: [ I] = 0.001( 0.488) = 0.000488 M

Ri = ( 7.2 × 10−6 ) ( 0.000448 ) = 3.51 × 10−9 M s -1

(b) Steady-state free-radical concentration at 1 min Solution 12

fk [ IM x ⋅] = d kt at 1 min:

[ I]

12

( 0.30 ) (1.2 × 10−5 )

[ IM x ] =

7.2 × 107

12

( 0.0009993)

12

= 7.08 × 10−9 M

(c) Rate of polymerization at 1 min Solution Ro = kp [ IM x ⋅][ M ]

[ M ] = [ M ]o exp ( −kp [ IM x ⋅] t ) = (1) exp −176 ( 7.08 × 10−9 ) 60 Ro = 176 ( 7.08 × 10−9 ) ( 0.9999 ) = 1.24 × 10−6 M s −1

= 0.9999 M

(d) Average free-radical lifetime, τ, at 1 min, where τ is defined as the radical concentration divided by the rate of termination Solution

τ=

[ IM x ⋅] 2 2k t [ IM x ⋅]

=

1 1 = = 0.981 s 7 2k t [ IM x ⋅] 2 ( 7.2 × 10 )( 7.08 × 10−9 )

6 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

(e) Number-average degree of polymerization at 1 min Solution Rp 1.24 × 10−6 1.24 × 10−6 1.24 × 10−6 Xn = = = == = 172 Rt 2k t [ IM x ⋅]2 2 ( 7.2 × 107 )( 7.08 × 10−9 )2 7.22 × 10−9 2.3 It has been reported that the rate of a batch photopolymerization of an aqueous acrylamide solution using a light-sensitive dye is proportional to the square of the monomer concentration, [M]2, and the square root of the absorbed light-intensity, I1/2. Note that, although this polymerization is free radical, the apparent kinetics appear not to be typical of usual free-radical polymerization for which the rate of polymerization is proportional to the first power of monomer concentration and to the square root of the initiator concentration (eq. (2.25)). The following polymerization mechanism has been proposed to explain the observed kinetics: Initiation M+D

k1 ,hν

R r2 = 22 , there is a tendency for consecutive homopolymerization since M1 (styrene) will k12 k21 polymerize until it is completely consumed and then M2 (vinyl acetate) will polymerize. r1 =

2.11 Explain why high pressure favors the propagation step in a free-radical polymerization. How would the rate of termination be affected by pressure? Answer It would be expected that pressure would increase the rate of propagation but decrease the rate of termination. This is supported by studies of styrene polymerization (Ogo, Macromol. Sci.-Rev. Macromol. Chem. Phys. C24, 1 (1984)). One argument that can be made is that pressure increases viscosity and, therefore, the diffusion of long-chain radicals is reduced (i.e., the rate of termination decreases). Kiran & Saraf (J. Supercritical Fluids 3, 198 (1990) discuss volume production arguments. Net volume decreases during propagation and is, therefore, favored at high pressure. On the other hand, net volume increase during termination and unfavored at high pressure. 2.12 From data available in Section 2.2.1, calculate the activation energy for propagation for the freeradical polymerization of styrene. Do you expect the activation energy to be dependent upon solvent in a solution polymerization? Solution Using data for styrene bulk polymerization in Table 2-3 and kp = Ap exp ( − Ep RT ) R = 8.3144 J mol-1 K-1 Plot gives slope = –3.925×103; using this value gives Ep = 32.6 kcal mol-1 Polymer Handbook, 4th ed., cites a value of 31.5 kcal mol-1

11 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

6 5 4 3 2 1 0 0.003

0.0031

0.0032

0.0033

0.0034

-1

1/T (K )

2.13 Draw the chemical structures of the two ends of a terminated polystyrene chain obtained by the atom transfer radical polymerization of styrene using 1-phenylethyl chloride (1-PECl) as the initiator, CuCl as the catalyst, and 2,2’-bipyridine as the complexing agent. Solution 1-phenyl chloride initiator Cl CH3

CH2

CH

See Wang & Matyjaszewski, Macromolecules 28, 7901 (1995) or Coessens et al., Progr. Polym. Sci. 26, 337 (2001). 2.14 Show that the rate of polymerization in atom transfer radical polymerization is proportional to the equilibrium constant defined in eq. (2.50). Solution Rp = kp [ Pn ⋅][ M ] = K e

[ Pn − X ][Cu(I)X ] M [ ] [Cu(II)X 2 ]

2.15 Show that azeotropic copolymerization occurs when the feed composition is given as f1 =

Solution At azeotrope:

d [ M1 ]

1 − r1 . 2 − r1 − r2

[ M1 ] d [M2 ] [M2 ] =

12 This text is associated with Fried/Polymer Science and Technology, Third Edition (9780137039555) Copyright 2014, Pearson Education, Inc. Do not redistribute.

From eq. 2.42: r1 [ M1 ] + [ M 2 ] = [ M1 ] + r2 [ M 2 ] Dividing both side by [M2] gives: 1− r [ M1 ] = 2 [ M 2 ] 1 − r1 Dividing by [M1]+[M2] gives 1 − r2 f1 = (1 − f1 ) 1 − r1 Rearrangement gives 1 − r2 2 − r1 − r2 Substituting values of r1 and r2 from problem 2.9 gives the same result, f1 = 0.580 at the azeotrope. f1 =

2.16 Methyl methacrylate is copolymerized with 2-methylbenzyl methacrylate (M1) in 1,4-dioxane at 60°C using AIBN as the free-radical initiator. (a) Draw the repeating unit of poly(2-methylbenzyl methacrylate). Solution CH3 CH2

C C

O

O CH2 H3 C

(b) From the data given in the table below, estimate the reactivity ratios of both monomers. f1

F1*

0.10 0.25 0.50 0.75 0.90

0.14 0.33 0.52 0.70 0.87

* From 1H-NMR measurements

Solution Data can be fitted to Eq. 2.45 using nonlinear regression analysis. Alternately (and less preferred) is the traditional linearization of the instantaneous copolymerization equation in the form (see Flory, Principle of Polymer Chemistry, pp. 185–189) G = r1 H − r2 where f F G = 1 1− 2 f2 F1 and 2

f F2 H= 1 f 2 F1 A plot of G versus H gives r1 from the slope and r2 from the intercept.

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Solutions reported in I. Erol, C. Soykan, J. Macromol. Sci.: Part A – Pure & Applied Chemistry A39, 953 (2002) give average values of r1 = 1.03 and r2 = 0.77. A nonlinear regression (Matlab) gives r1 = 0.6311 and r2 = 0.5328. CHAPTER 3 3.1 Polyisobutylene (PIB) is equilibrated in propane vapor at 35°C. At this temperature, the saturated vapor pressure (p1o) of propane is 9050 mm Hg and its density is 0.490 g cm-3. Polyisobutylene has a molecular weight of approximately one million and a density of 0.915 g cm-3. The concentration of propane, c, sorbed by PIB at different partial pressures of propane (p1) is given in the following table. Using this information, determine an average value of the Flory interaction-parameter, χ12, for the PIB– propane system. p1 (mm Hg) 496 941 1446 1452

c (g propane/g PIB) 0.0061 0.0116 0.0185 0.0183

Solution CH3 CH2

C

n

CH3

M o = 56.11 ; r =

1 106 = 1.78 × 104 and 1 − = 0.9999 ≈ 1 r 56.1

w1 ; w1 = ρ1V1 ; w2 = ρ 2V2 w2 ρV V ρ c = 1 1 or 2 = 1 ρ 2V2 V1 ρ 2 c ρ V1 V 1 φ1 = =1+ 2 =1+ 1 or φ1 ρ2c V1 + V2 V1 p p a1 = 1o = 1 p1 9050 c 0.0061 0.0116 0.0183 0.0185

φ1 0.01126 0.02120 0.03304 0.03339

p1

496 941 1452 1446

lna1

-2.9039 -2.2636 -1.8298 -1.8340 ave.

χ

c=

12

0.6075 0.6381 0.6559 0.6410 0.64

See S. Prager, E. Bagley, and F. A. Long, JACS 75, 2742 (1953).

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3.2 The following osmotic pressure data are available for a polymer in solution: c (g dL-1) 0.32 0.66 1.00 1.40 1.90

h (cm of solvent) 0.70 1.82 3.10 5.44 9.30

Given this information and assuming that the temperature is 25°C and that the solvent density is 0.85 g cm-3, provide the following: (a) A plot of Π/RTc versus concentration, c Solution Π = ρ gh = 0.85 ( 980.665 ) h h (cm)

( RTc ) ×106

0.32 0.66 1.00 1.40 1.90

0.70 1.82 3.10 5.44 9.30

7.310 9.216 10.360 12.985 16.358

Π

c (g dL-1)

18 16 14 12 10 8 6 4 2 0 0

0.2

0.4

0.6

0.8 c (g

1

1.2

1.4

1.6

1.8

2

dL-1 )

(b) The molecular weight of the polymer and the second virial coefficient, A2, for the polymer s...