Solution manual The Elements of Polymer Science and Engineering,3rd Edition Alfred Rudin, Phillip Ch

SOLUTION MANUAL COMPLETE

Rudin & Choi: The Elements of Polymer Science & Engineering, 3rd Edition Chapter 1 Solutions Manual

1-1

1-2

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

1-2

1-3 (a) (b) (c) (d)

100,000 / 113 = 885 100,000 / 192 = 521 100,000 / 97 = 1,031 100,000 / 162 = 617

1-4 (a) (i) 2 (ii) 2 (iii) 0 (b) same as in (a)

(iv) 4

1-5

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

1-3

1-6 (a) (b) (c) (d) (e) (f) (g) (h) (i) (j)

polyisobutene (polyisobutylene) poly(vinyl chloride) (PVC) poly(tetramethylene terephthalate) (poly(butylene terephthalate)) poly(ethyl methacrylate) poly(vinyl acetate) poly(methyl acrylate) polycaprolactam (nylon 6) poly(phenylene oxide) poly(para-bromostyrene) poly(ethylene-2,5-toluene-urethane) (poly(ethylene-2,5-tolyl-urethane))

1-7 (a) is too weak mechanically while (c) is too difficult to shape into pipe form. The correct answer is (b).

1-8 The functionality of glycerol is 3 in every case listed.

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

1-4

1-9

React 1 mole of HOCH2CH2OH with 2 moles as in reaction (1-12). Then react the product of this reaction with H-(OCH2CH2CH2CH2-O)m-H as in reaction (1-13).

1-10

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

1-5

1-11 The polyisobutene requires a fairly high molecular weight for use in inner tubes and cable coatings, in order to ensure good mechanical strength. Reasonably low viscosity is needed in adhesive and oil additive applications and this calls for lower molecular polymers.

1-12 (a) 4 (b) 2 (c) 1 1-13 (a) 2 (b) The functionality of (2) is 5 since there are 5 active hydrogens on the nitrogens. (c) Yes. 1-14

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

1-6

1-15 (a) and (d).

1-16

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

1-7

1-17 Polyethylene crystallizes very rapidly because its structure is regular and symmetrical and the chain carries no bulky substituents (see also Sect. 4.3). Polyisobutene is amorphous (Sect. 4.5.1) and can therefore exhibit recovery from high extensions.

1-18 (a)

(b) There are a few head-to-head linkages in the poly(vinyl acetate) which produce 1,2diol structures on alcoholysis to poly(vinyl alcohol). There are insufficient head-tohead placements to register in terms of HIO4 consumption but it only takes an average of one such linkage per macromolecule to result in a halving of the molecular weight, as shown. In this case the average number of vinyl alcohol repeating units in the initial macromolecule is 250,000 / 44 = 5700 (where the formula weight of the repeating unit is 44 g/mol). It only requires about 1 head-tohead linkage out of an average of 2300 such bonds to result in the observed reduction in molecular weight.

1-19 The glycol should be 1,3-propane diol. This molecule is symmetrical and would produce polyester structures like

which will crystallize. Polymers from 1,2-propane diol have the following structures:

and A mixture of such structures in the polymer chains would inhibit crystallization because of the different steric requirements of the two positional isomers.

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

1-8

1-20 (a) Formula weight for propylene = 42.

Degree of polymerization = 105 / 42 = 2376.

Number of C-C bonds = 2 x 2376 = 4752 (there are 2 bonds in the main chain per repeating unit). 〈

〉

〈

〉

√〈

( 〉

) (

)

(rms end-to-end distance)

(b) The most highly extended form involves the all-trans conformation of the polymer backbone:

The projection of the bond length on the chain length direction is (1.54 x 10-8) cos 35.25° = 1.26 x 10-8 cm.

For σ = 4753 the end-to-end distance of the fully extended chain = 4753(1.26 x 10-8) = 5989 x 10-8 cm.

Extensibility = (5989 – 150) / 150 = 39 times. (c) Because of the lack of rotational freedom of the real chain and because two segments of the chain cannot occupy the same space (excluded volume effect) the real macromolecule will have a longer rms end-to-end distance than that calculated in part (a) (cf. equation (1-31)). The ratio calculated in part (b) will therefore be lower.

Conclusion: the real chain is less extensible.

1-21 √〈 〉 √ This is a straightforward extension of the ideas behind equation (1-28).

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

1-9

1-22 For polyethylene, the torsional potential energy difference between the trans and gauche states (i.e., ∆e) is 2,100 J/mol (see example 1-2).

At 200°C (473K), the fraction of bonds in the trans state is given by: (

(

)

)

(

)

where R = 8.314 J/(mol-K).

The fraction of bonds in the gauche states is 1-0.46=0.54 (0.27 for each gauche state). 〈 〉 For polyethylene, (

(see Example 1-3).

)(

The Kuhn length, (

)

(

)

. Here l = 0.154 nm.

)

1-23 For polyethylene, when temperature increases, the number of bonds in gauche conformation increases and chain dimension decreases. For silicon polymers, when temperature increases, the number of bonds in trans conformation increases, and chain dimension increases.

1-24 ( ) ( )

(a)

In order to determine the r at which ( ) and

.

]

(

is maximum, let

( )

[ ( ) (

)

(

)

) Copyright © 2013 Elsevier Inc. All rights reserved.

(

) and

Solutions Manual

1-10

[ ( )

Let

]

or

. or

From

,

√

𝑃(𝑟) 𝜋𝑟

𝑟

(b) 〈 〉

∫

( )

∫

( )

∫

( )

∫

(

(

) ∫

(

)

(denominator = 1)

)

( ) (

)

(c) 〈 〉 〈 〉

(d)

√

.

Copyright © 2013 Elsevier Inc. All rights reserved.

Rudin & Choi: The Elements of Polymer Science & Engineering, 3rd Edition Chapter 2 Solutions Manual

2-1 .

For the mixture: ̅

(

∑ ̅​̅​̅ (

̅

)

)

(̅ )

∑

(

)

(

)

.

2-2 Molecular weight of calcium stearate = 607. ̅

∑ ̅​̅​̅ (

( ̅ (

)̅

̅

∑

(

̅

)

) )̅

( )

.

2-3

̅

∑ (

) (

)

(

)

(

)

2-4 ̅​̅​̅​̅

(̅​̅​̅​̅) ∑ ( ) and ̅​̅​̅​̅ ( ) ( ) ( ) ( ) ( ) ( ) ( ) The formula weight of the repeating unit = M0 = 104. ̅ ̅​̅​̅​̅ .

Similarly: ̅​̅​̅​̅ and ̅ . ∑ ̅ Since

∑

(

(

)

Variance of the number distribution ̅ ̅ ̅ (standard deviation = 1060).

Copyright © 2013 Elsevier Inc. All rights reserved.

)

(

)

Solutions Manual

2-2

2-5 (a) see Section 3.3.1 ̅ the polymer sample is monodisperse and ̅ (b) if ̅

̅

̅

̅ etc.

2-6 For all the homopolymers formed, ̅ ̅

∑

(

)

Mol. wt. of monomer a) 42 b) 62.5 c) 100 d) 44 e) 28 ̅ *DP = /(mol. wt. of monomer) Monomer

DP* 1,479 994 621 1,411 2,218

2-7 (

)

(

)

∑

(̅ )

(

(

∑( (

)

) )

)

PDI = Mw/Mn = 187,000/78,400 = 2.39 Assuming that the number distribution of the molecular weight can be described by the normal distribution, √ √

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

2-3

2-8 ni 1,000 1,800 1,700 1,500 Sum = 6,000

Di 2.5 3.0 3.2 3.5

niDi3 15,625 48,600 55,706 64,313 184,244

niDi 2,500 5,400 5,440 5,250 18,590

niDi4 39,063 145,800 178,258 225,094 588,215

∑

(a)

∑

(b) ∑ ∑

(c) Let ρ be the density of the particles

2-9 Number

Diameter (cm)

Weight of 1 sphere*

2 3 4 2

1 2 3 4

1ρ 8ρ 27ρ 64ρ

Weight of all spheres of this size 2ρ 24ρ 108ρ 128ρ

11 *ρ = density ̅

( )

̅

∑

( )

262ρ

( )

( )

0.0076 0.0916 0.4122 0.4886 1.0000

( )

( )

Weight fraction, wi

( )

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

2-4

2-10

̅

∑

̅

∑

Fraction

Weight (g)

1 2 3 4 5 6

1.5 5.5 22.0 12.0 4.5 1.5

Weight fraction, wi 0.03 0.12 0.47 0.255 0.095 0.03

47.0

1.00

(

̅ 2,000 50,000 100,000 200,000 500,000 1,000,000

)

(̅ )

(

)

(

)

(

)

Alternatively: Number of moles, ni* 1 1.5 2,000 0.75 x 10-3 2 5.5 50,000 1.1 x 10-4 3 22.0 100,000 22 x 10-5 4 12.0 200,000 6 x 10-5 5 4.5 500,000 0.9 x 10-5 6 1.5 1,000,000 1.5 x 10-6 *ni = ci/Mi where ci = weight in grams, in this case (eq. (2-9)).

Fraction

̅

∑

̅

∑

Weight (g)

̅

∑ ∑

Standard deviation of the number distribution,

√̅ ̅

̅

or: ̅

Copyright © 2013 Elsevier Inc. All rights reserved.

̅

Solutions Manual

2-5

2-11 By definition, ∑ ∑ ∑

(

)

(in this case)

. ̅

(a)

∑

(

(b)

̅

̅ ) (eq. (2-35))

√

(

∑ ̅

∑

(c) (d)

)

∑

̅ ̅ ̅

2-12 (a) PDI = Mw/Mn Poisson:

(

Exponential:

)

(

)

(

(b) Poisson: when Mw/Mn = 1 (i.e.

)

Copyright © 2013 Elsevier Inc. All rights reserved.

)

Solutions Manual

2-6

Exponential: when Mw/Mn = 2 (i.e.

)

(c) Poisson: Exponential: 2-13 (a)

k1

fN(M)

M

Cumulative mole fraction

(b) 1 Form:

𝑒 𝑘𝑀

0 M

Notes: ∫

∫

(c)

(d)

∫

( )

∫

( )

∫

( )

∫

( )

, No,

∫

is given.

cannot be determined since

is not known.

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

2-7 ∑

(e) ∑

(

) ∑

∑

Divide the equation by

Copyright © 2013 Elsevier Inc. All rights reserved.

Rudin & Choi: The Elements of Polymer Science & Engineering, 3rd Edition Chapter 3 Solutions Manual

3-1 ̅

(∑ ̅

)

(

[ ] alternatively: [ ]

) (

.

)

∑(

)

.

3-2 c (g/100ml)

h (cm HCCl3)

0.57 0.28 0.17 0.10

2.829 1.008 0.521 0.260

h/c (cm HCCl3)(100ml)/g 4.96 3.60 3.06 2.60

plot (h/c) against c. The intercept at c = 0 = (h/c)0 = 2.30 (cm HCCl3)(100 cm3)/g Pressure exerted by a column of HCCl3 2.30 cm long: -

-

Dimensions of Dimensions of ( )

-

If π/c is in cm2/s2, R should be in ergs/(mol K) to yield ̅ in g/mol. ( )

( ̅

) -

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

3-2

3-3 (a) 22.0 21.0 20.0 19.0 π/c 18.0 (J/kg) 17.0

y = 0.35x + 16.03

16.0 15.0 14.0 0

1

2

3

4

5

6 7 8 c (kg/m3)

9 10 11 12

(b) Range: 0 – 8 or 9 kg/m3 (c)

(d)

3-4 (a) Also, (

)

[

(

[ (

)

) (

] )

]

[

(b)

(

[ [

( (

) )

)

]

] ]

At low c,

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

3-3

(

)

(

)

(

)

(

)

[

(c)

(

When

)

]

,

(

)[

Since

] ,

is negative when

.

ηr 1.290 1.632 2.026

ηsp 0.270 0.632 1.026

3-5 c (dl/g) 0.20 0.40 0.60

ηinh 1.27 1.22 1.18

plot data as in Figure 3-8 to obtain [η] = 1.32 dl/g. ̅ ̅

.

3-6 See the discussion connected with equation (3-97) in the text.

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

3-4

3-7 ̅ ̅

∑ ∑

( (

)

(

)

)

Light scattering measures ̅ . The results of such determinations will not be affected to within experimental uncertainty. Membrane osmometry measures ̅ . These results will be significantly affected.

3-8 for ̅ : for ̅ :

sample A - vapor phase osmometry sample B - membrane osmometry. use light scattering in both cases.

Gel permeation chromatography could be used for ̅ and ̅ for both samples with a suitable calibration.

3-9 No. See Section 3.3.2.

3-10 (a) vapor phase osmometry (b) light scattering or gel permeation chromatography (with calibration) (c) membrane osmometry.

3-11 (a) No. Osmometry yields ̅ (eq. (3-29)) and light scattering gives ̅ (eq. (3-56)).

The two averages are the same within experimental error only for polymers with very narrow molecular weight distributions. (b) No. The slope of the light scattering zero concentration line will be ideally twice that in the osmometry plot (compare equations (3-24) and (3-53)).

3-12 (a) If each polymer molecule contained 2 terminal –COOH groups there are moles in the sample. ̅ is then given by (b) It is assumed that there are 2 carboxylic acid groups per molecule.

Copyright © 2013 Elsevier Inc. All rights reserved.

.

Solutions Manual

3-5

3-13

(

)

R is defined per unit volume (see text in connection with eq. (3-47)). (

)

3-14 (

[ ]

)

[ ]

[ ] [ ] Huggins’ constant is dimensionless.

3-15 Standard deviation of the weight distribution, ̅ √̅ ̅ By extension from equation (2-39) measure of skewness of the weight distribution is: ̅

̅

̅

̅

(̅

̅

̅ )

̅

̅

3-16

(

)

(c = concentration in g/dl = g/100cm3).

c grams solute has volume = c/ρ cm3 (ρ = apparent density of solute). ∴ volume fraction solute at concentration of c g/100 ml solution (

) (

. (

)

(v = specific volume)

)

Copyright © 2013 Elsevier Inc. All rights reserved.

Rudin & Choi: The Elements of Polymer Science & Engineering, 3rd Edition Chapter 4 Solutions Manual

4-1 Nylon 6,6 is a crystallizable polymer. It can be oriented at temperatures between Tg and Tm to enhance its tensile strength. Release of the orienting stress does not result in loss of orientation because the fibre structure is anchored by crystallites formed during the drawing process. Polyisobutene is elastomeric and normally not crystalline. Orientation at any temperature above the low Tg of this polymer has no permanent effect. The structure retracts when the drawing load is removed.

4-2 Cross-link or copolymerize with comonomers whose homopolymers have higher Tg’s than polystyrene. Alpha-methylstyrene, metha-crylonitrile, maleic anhydride or acrylonitrile could be used as comonomers.

4-3 Densities of crystalline and amorphous polyethylene:

When sample

,

When sample

,

When sample

,

Assumed that the mass and volume of the crystalline and amorphous parts are additive.

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

4-2

4-4 n 8 10 12 14 16 24 32 ∞

Tm (°C) observed -56.8 -29.7 -9.7 2.5 14.7 47.6 67.0 -

Tm (°C) calculated -52.2 -29.2 -11.1 3.5 15.7 48.6 68.0 143.5

160.0 140.0 120.0 100.0 80.0 T (°C)

60.0 40.0 20.0 0.0 -20.0

0

50

100

150

200

-40.0 -60.0

When

n

,

( ) (

)

(

) .

Copyright © 2013 Elsevier Inc. All rights reserved.

250

Solutions Manual

4-3

4-5

∫

∫

However, (

(vol. change of melting) )

where M = 28 g/mol. ρ before melting = 955 kg/m3 = 0.955 g/cm3 ρ after melting = 855 kg/m3 = 0.855 g/cm3 (

)

4-6 (b) has a higher Tm and higher Tg because the phenyl group stiffens the macromolecular chain.

4-7 (

(

))

(

(

))

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

4-4

4-8 (

(a)

) (

)

( WLF: (

) (

)

(

)

)

(

)

(

(

)

(

)

(

)

(

)

(

)

(

)(

)

)

T = Tg,

(b) C1 = 17.4, C2 = 51.6, Tg = 200 K, R = 8.314,

(c) ( )

(

)

(

(

)

)

( ) (

)

( )

4-9

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

4-5

For small amount of solvent Also,

.

.

Let M be the mass of polymer + solvent per mole of mixture and Me be the mass of the polymer.

Let

4-10 (

)

(

)(

)

. (

(

)(

)

)

Original length of specimen = 10 cm.

Therefore, the specimen will elongate 0.76(10) cm = 7.6 cm.

4-11 This is because work done on stretching the rubber must result in a release of heat under isothermal condition. This leads to a positive ( ) ; therefore ( ) is negative at small extensions.

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

4-6

4-12

[

(

)

(

)

(

)

(

)

(

)(

)

(

)(

)]

(no volume change)

At 27 °C, 1% elongation, i.e. (

)(

)

(

. )

4-13 There are 150 C atoms between cross-links on the average. This corresponds to 150/4 = 37 monomer units between cross-links (see structure 1-19).

Formula weight of an isoprene monomer = 68 g/mol. molecular weight between cross-links, (

)(

)

)(

4-14 (a) Ideal rubber 〈

〉

〈

〉

(

)

. (

)

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

4-7

(b) (c) Upon stretching, the entropy of the rubber sample decreases.

4-15 Refer to Section 4.8.

Units of force = ml/t2 Units of elongation = l Units of area under force-elongation curve

()

units of work.

4-16 From Table 4.2 for polyisobutene,

We have to calculate a temperature (T) change corresponding to a shift factor of 10, i.e., (

) (

)

(

)

4-17 (a)

( ) ( )

( ) ( )

(b)

: ( ) : ( )

(

)

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

4-8

( )

(c)

( ) Percentage of original stress

(

)

(

)

(d) It is because the Maxwell model would yield zero stress when residual stress in this model.

4-18 (

(

)

)

(

)

Energy dissipated (

)

4-19 (a)

( )

( )

(b)

( )

( )

Copyright © 2013 Elsevier Inc. All rights reserved.

while there is a

Solutions Manual

4-9 (

( )

(c)

) (

(

(d)

) )

(e)

0

γ0

π/2

0

π

-γ0

3π/2

0

2π

γ0

(

Note:

(

)

(

)

)

(f) At low ω, the material has sufficient time to relax, and (i.e., storage compliance > loss compliance). On the other hand, at high ω the opposite is true: .

4-20 ( )

Boundary conditions: ( ) ∫

( )

∫ )| ( )

( ( )

(

but, ( ) ( )

| (

))

in this case and so: (

(

( )

))

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

4-10

4-21 Cut tensile specimens at right angles to each other from the film. The more highly oriented samples will have higher moduli and lower elongations to break. These will be machine direction specimens in films with unbalanced properties. (As a check, the shrinkage of the samples can be measured, e.g., after immersion in boiling water for a given time. The more highly oriented direction will exhibit greater shrinkage.)

4-22

(

)

For brittle fracture, √

√

Since is significantly above the yield stress ( yielding.

), the pipe will fail by

When it fails,

Copyright © 2013 Elsevier Inc. All rights reserved.

Rudin & Choi: The Elements of Polymer Science & Engineering, 3rd Edition Chapter 5 Solutions Manual

5-1

(

)

( (

(

)

) )(

(

)

)

5-2 For a statistical copolymer, ∑ where δi is the solubility parameter of the polymer of monomer i, and wi is the weight fraction of monomer i in the copolymer.

M(butadiene) = 54 M(methyl methacrylate) = 100 There are 3 butadiene units in copolymer for every methyl methacrylate residue.

Therefore, for every 100 g (1 mol) methyl methacrylate the copolymer contains 3(54) = 162 g butadiene. Weight fraction of methyl methacrylate = 100/(100+162) = 0.38.

From Table 5.3, δ(PMMA) = 9.3 δ(PBD) = 8.4 then (

)

(

)

(

)(

)

(

Copyright © 2013 Elsevier Inc. All rights reserved.

)

Solutions Manual

5-2

5-3

From Table 5.2: Group -CH2 >CH-O(ether) -CH3

Fi 131.5 85.99 114.98 148.3

ΣFi 526.0 85.99 114.98 148.3 875.27

No. Groups 4 1 1 1

Density = 1 g/cm3 Mr = 108 g/mol (

)( )

(eq. (5-13))

5-4 (a) The copolymer is soluble in poorly hydrogen-bonded solvent mixtures with ( ) . Take as a target value. From the brief list of solubility parameters in Table 5.3 methylene chloride ( ) and nitromethane ( ) are possible cosolvents.

Let Varsol be solvent a, and the other liquid be solvent b. where φ are the volume fractions.

With solvent b = methylene chloride, ( ) ( )

and:

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

5-3

φa = 0.33 = volume fraction of Varsol 2 in a mixture with methylene chloride.

Similarly, the volume fraction of Varsol required in a mixture with nitromethane = 0.73.

(The specific gravities of these solvents are approximately as follows: Varsol 0.7, methylene chloride 1.33, nitromethane 1.14. As a result, the first mixture would be only about 21% by weight of Varsol while the second blend would contain 62%.) (b) The copolymer would probably form stable mixtures with polyethylene, depending on its vinyl acetate content. If there were any appreciable lengths of ethylene segments in the copolymer these would co-crystallize with polyethylene (2(b), Table 5.5).

5-5 (a) Solvent

δ (Table 5.3)

H-bonding

tetrahydrofuran n-hexane 1-butanol dioctyl phthalate

9.1 7.2 11.4 7.9

mod. poor strong mod.

Three-Dimensional Sol. Parameters (Table 5.4) δd δp δH 8.2 2.8 3.9 7.2 0 0 7.8 2.8 7.7 8.1 8.1 1.5

The δd values of the different solvents do not differ enough to warrant concern. It is then necessary only to match the δp and δH values of the solvent mixture and tetrahydrofuran.

Let: φ1 = volume fraction of n-hexane in the mixture φ2 = volume fraction of 1-butanol in the mixture φ3 = volume fraction of dioctyl phthalate in the mixture φ3 = 1 - φ1 – φ2 (by definition of volume fractions) (

)

( )

(

)

(

)(

)

( )

(

)

( )

(

)

(

)(

)

( )

( ) ( )

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

5-4

Simultaneous solution of these two equations yields:

Check on δd of the mixture: ( )

(

)

(

)

(

)

(b) An alternative route can be tried using equation (5-15) to calculate δV and matching the δV and δH of the blend to the corresponding values of tetrahydrofuran. In this instance negative volume fractions are calculated indicating that this blend cannot match the solvency of tetrahydrofuran. The contradiction between the calculations of solutions (a) and (b) to this problem emphasizes the empirical nature of subparameter solubility parameter models. An empirical model can be very useful but it will eventually fail to match some natural phenomenon.

5-6 Block copolymers are prime choices as viscosity reducing additives. Examples are ethylene-propylene block copolymers admixed with polyethylene (A. Rudin, Polym. Eng. Sci., 10, 94 (1970)) and styrene-butadiene block copolymers in SBR (C. K. Shih, in “Science and Technology of Polymer Processing,” N. P. Suh and N-H. Sung. ed., MIT Press, Cambridge, Mass., 1979). Graft copolymers could also be used in some instances but the structure of block copolymers is more accurately known and controlled. The blocks which are similar in structure to the host polymer ensure good interphase adhesion while the second blocks provide the required controlled immiscibility in the melt phase.

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

5-5

5-7 From Table 5.3: Solvent

δ

Volume fraction

toluene 1-butanol

8.9 11.4

methyl ethyl ketone

8.4

methyl isobutyl ketone diethylene glycol monomethyl ether

8.4

0.50 0.13 0.12 (= 0.32 × 0.37) 0.20

10.2

0.05

(

)

(

)

(

)

(

)

(

)

The bulk of the solvents are medium or poorly hydrogen-bonded according to the table.

By inspection, tetrahydrofuran should be as good a solvent for nitrocellulose as the mixture listed.

5-8 (

(a) (

√

√

√

)

√

(

(b)

)

)

At 200 K, ( (

) (

) (

)

)

( (

)

)

At 600 K, ( (

) )

(

)

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

5-6 (

)

( (

)

)

At 200 K while at 600 K, exhibit a UCST phase behavior.

. Therefore the blend should

5-9 (a) -

(b) Both LCST & UCST behaviour is expected. The first term of the empirical equation implies

, while the second term implies

:

χ χ 2nd term 1st term

T

(c)

√

√ For A,

For B,

(

√

√

)

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

5-7

(d) Set

and solving for T:

0.000201

χ

231

T(K)

357

5-10 (a)

and (<0.5) Therefore, the solvent and polymer are miscible under the given conditions.

(b) Based on the equation shown in part (a) of Example 5-2. (

)

Rearranging the above equation yields: ( )

5-11 (a) Molar volume of PS Molar volume of cyclohexane Using the molar volume of cyclohexane as the reference volume, the chain length of PS is

.

Volume of 1 g PS with Total volume of the solution = 0.95 + 108 = 108.95 cm3

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

5-8

T = 400 K (

)

(

) (

(

(

(b)

√

√

)

.

(c) It has a UCST. (d) At θ temperature,

. (

)

(

)

T = 343.3 K. 5-12 (a) The failure strains of fibers and matrix are given

Therefore, the fibers fail before the matrix does. (b) When the fibers fail, the failure stress,

Copyright © 2013 Elsevier Inc. All rights reserved.

)(

))

Rudin & Choi: The Elements of Polymer Science & Engineering, 3rd Edition Chapter 6 Solutions Manual

6-1 (a)

( ( ))

(

( (

)) ( (

( )

( (

)

)

(

)

(

)

(

( (

))

( )

)

( )

))

(

)

(

)

( (

))

(

)

(

)

(

(

( )

(

( ) ( )

)

)

(

)

Finally: ( ( ))

)

( (

From the Taylor series expansion of the (

)

(

) … (1)

) for very small x we have:

)

So as the n/N is very small we can say: (

)

( )

(

)

( )

( )

So applying this in Eq. (1) we have: ( ( ))

(

( )

)

[

( )

( ) ]

[

( ) ] ( )

(

)

( ) (

)

Now if assuming that N is large enough that ( ( ))

(

then we have:

)

Copyright © 2013 Elsevier Inc. All rights reserved.

( )

)

Solutions Manual

6-2 (

)

And finally: ( ) (b) ( ) ∫

( )

And as ( )

√

∫

√

√

∫

( )

the final form of probability will be:

√

6-2 〈(

) 〉

( )

∫

∫

√

∫

√

Now with change of variable: √

√

We have: 〈(

) 〉 √

√

√

∫

∫

√

( )

We know that the Gamma function of ½ is √ so finally: 〈( ) 〉 Now combining this result with the time dependency of number of steps we will end up with amazing results (see following example). Doing this we have: 〈(

) 〉

( )

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

6-3

6-3 If we take the Fourier transform of the Eq. (6-22) we have: (

)

(

)

The solution to this ordinary differential equation is: (

)

The initial condition for the point source is the positive concentration at x=0 and zero concentration elsewhere in the starting point of the process. So it is: ( ) ( ) where ( ) is Dirac delta function.

The Fourier transform of the initial condition is: (

)

√

Now if we take the inverse Fourier transform of the result we have: (

)

( )

∫

Now if we change the variables as follow:

√ √

We have: (

)

√

∫

( )

And with solving integral we will end up with: (

)

√

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

6-4

6-4 First consider the change of variable as: 〈 ̇〉 So we have: (

)

Now consider another change of variable as follow:

Now we have:

Now putting everything back in the Original equation and applying the initial condition we have: 〈 ( ) ̇ ( )〉 [ ( )] As a next step, we note that 〈 ( ) ̇ ( )〉

〈 ( )〉, so that the mean squared

displacement of the particle can be found by an additional integration over the above equation which is: 〈 ( )〉

[

(

)]

6-5 In this system the potential energy is: ( ) . The behavior of ( ) is governed by:

where ()

()

()

Solving this initial value problem using Laplace transformation we have: ( )

( )

∫

(

́)

( ́) ́

The x after infinite time would reach to its equilibrium value which is: ∫

(

́)

( ́) ́

∫

(

́)

( ́) ́

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

6-5

So if we assume that the process has been started at , we don’t need initial value, ( ), as it will be dampened after such long time and x(t) can be written as: ( )

(

∫

́)

( ́) ́

Now we can calculate the time correlation function of x as follow: 〈 ( ) ( )〉 ∫

〈∫

(

∫ (

∫

)

)

( ) ( )〉

〈 ( ) ( )〉

But from Langevin dynamics we know that: 〈 ( ) ( )〉 where

(

)

is the Dirac delta function.

So we have: 〈 ( ) ( )〉

(

∫

)

Accordingly: 〈 ( ) ( )〉

〈

〉

And finally for mean square displacement we have: 〈( ( )

( )) 〉

(

〈 ( )

( ) ( )〉

〈

〉

)

Now in the limit of 〈( ( )

( )

the Taylor series expansion results in:

( )) 〉

which exactly matches to our expectation.

Copyright © 2013 Elsevier Inc. All rights reserved.

〈 ( ) ( )〉

Rudin & Choi: The Elements of Polymer Science & Engineering, 3rd Edition Chapter 7 Solutions Manual

7-1 Two NH2 groups react with one COOφ group and the number of NH2 groups is twice that ̅ of COOφ, so f of each monomer = fav = 2. Since here, ̅ in this case. ̅

7-2 The number of molecules cannot change and the number of repeating units cannot change in this closed system. Therefore ̅ does not change. ̅

(∑

)

(

)

The molecular weight distribution of the mixture of polyamides becomes randomized however. Then ̅ ( ) ̅ (eq. (7-36). Since these are presumably high conversion commercial polymers, p is close to 1 and ̅ is close to 144,000.

7-3 pentaerythritol phthalic anhydride tricarballylic acid

moles 1.21 0.50 0.49 2.20

f 4 2 3

equivalents 4.84 1.00 1.47 7.31

Total OH equivalents = 4.84 Total acid equivalents = (1.00 + 1.47) = 2.47.

Since the acid equivalents are in deficient supply: (

)

̅ When the mixture gels ̅ is infinite and ∴ p at gel point = 0.89.

.

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

7-2

7-4 p = 0.99 (

) (

) (

)

7-5 Total acid equivalents = 1.0 + 2(1.31) = 3.62 Total OH equivalents = 3(1.30) + 2(0.50) = 4.90 (

(a)

)

̅

(b) f 1 2 3

moles 1.0 1.31 1.30

oleic acid phthalic acid glycerol

equivalents 1.0 2.62 3.90

total acid equivalents = 3.62 total OH equivalents = 3.90 (

)

̅

(

)

when the mixture gels.

The mixture will gel at 99.7% conversion 7-6 (a)

( ) with i = 1 and p = 0.9, ( ) ( )

(1 percent by weight)

(b) ̅ ̅ ̅

̅

∑ but

(

∴

∑ (

) )

(

) ∑

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

7-3

using the summation in eq. (7-31c):

∴ ̅ ̅ ̅ ̅

(

)

(

)(

( (

)

(

)

) ) (

)

.

7-7 (a)

̅ (for later use: ̅

)

(b) = weight fraction of polymer in reaction mixture. (c) Mixture consists of monomer w1 = 0.16, and polymer wp = 0.84. ̅ ̅ ∑ ( ) ̅ ̅ = weight average degree of polymerization of the polymer.

7-8 Formula weight of ̅

= 164.

Need to make sure ̅ does not exceed 305 at ( ∴ add a monofunctional impurity like

).

or How much to add? ̅ ̅

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

7-4

moles

f

equivalents

1

2

2

1 ∑ ∑

. Add 0.0066 moles monofunctional ingredient for every mole of monomer.

7-9 (a)

( (

) )

(b)

( (

) )

(

)

.

7-10 HOCH2CH2OH

Moles 1

Functionality 2

Equivalents 2

0.99

2

1.98

0.01

1

0.01

Total acid equivalents = 1.99 Total OH equivalents = 2 ∴

(

)

At p = 1, ̅ In the absence of the benzoic acid impurity 19)).

and ̅ at

Copyright © 2013 Elsevier Inc. All rights reserved.

is very large (eq. (7-

Solutions Manual

7-5

7-11 moles 1.21 0.54 0.49 2.20

pentaerythritol phthalic anhydride tricarballylic acid

f 4 2 3

equivalents 4.84 1.08 1.47

Total OH equivalents = 4.84 Total acid equivalents = (1.08 + 1.47) = 2.55.

Since the acid equivalents are in deficient supply: (

)

̅ When the mixture gels ̅ is infinite and ∴ p at gel point = 0.86.

.

( ) ( ) Both equations (i.e., and ) are derived based on bifunctional monomers. Since the monomers used contain multiple functional groups, the equations cannot be used to calculate and .

7-12 (a) where N is the number of bifunctional monomers. (b) It won’t as ̅

when p = 1.

( ) (c) According to equation (7-22), unreacted.

Probability of finding unreactive monomers Probability of finding sensitive monomers (d)

(

)(

. When

, co-reactive sites are . (

)

.

)

(e) According to Figure 7.5(a), the monomer is the most prevalent species on a mole basis at p = 0.99 and x1 = (1 - 0.99) = 0.01.

According to equation (7-27), . (

) (

)

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

7-6

(f) According to equation (7-36), polydispersity =

According to equation (7-38): (

√

)

.

7-13 (a) Polydispersity, (

∴ (

(eq. (7-36)) )

) √

(b) See (c). (c) Monomer A AA

Functionalities

Moles

Equivalents

3

875/350=2.5

3×2.5=7.5

BB

2

2000/1000=2.0

2×2.0=4.0

CC

2

15/150=0.1

2×0.1=0.2

C CC

3

225/225=1.0

3×1.0=3.0

(d) Total equivalents of A = 7.5 Total equivalents of B & C = 4.0 + 0.2 + 3.0 = 7.2

(e) At gelation, ∴

(f) Polydispersity (g) √

(

) Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

7-7

7-14 (a) Component NCO% or OH no. polyisocyanate 11.7 polyol 1 35 polyol 2 107.5 polyol 3 290

f equivalent weight mol. wt.

3 359 1077 2 1603 3206 2 522 1044 3 193 579

(b) 359 g polyisocyanate  1 mol NCO 1000 g polyisocyanate  2.786 mol NCO

∴mol OH needed = 2.321.

Let x be the mass of the polyol mixture:

(c) Component amount (g) polyisocyanate 1000 polyol 1 1862.5 polyol 2 241.8 polyol 3 134.3

ni nifi 0.9285 2.7855 0.5809 1.1618 0.2316 0.4632 0.2320 0.696 1.973

OH groups deficient (

)

(d) When gelation occurs ̅

It would gel. (e) The material is expected to have higher density of crosslinking, therefore more rigid and showing higher modulus.

Copyright © 2013 Elsevier Inc. All rights reserved.

Rudin & Choi: The Elements of Polymer Science & Engineering, 3rd Edition Chapter 8 Solutions Manual

8-1 ( )

[ ]

(a) ̅​̅​̅​̅

(

[ ]

(b)

)(

)

( ) (

)(

.

)

8-2 ̅​̅​̅​̅

(̅​̅​̅​̅ )

[ ] [ ] [

[

]

]

.

8-3 ̅​̅​̅​̅

[ ]

[(

(̅​̅​̅​̅

)

)(

)] .

For polymerization by thermal decomposition of an initiator: [ ]

(

(

[ ]) )[(

)( -

-

)(

)]

.

Total rate of polymerization

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

8-2

-

-

-

8-4 [ - ̇

[

]

][

rate of initiation

[ - ̇

[ ][ rate of termination

] [

][

steady state assumption:

[

[

i.e. [

[

]

] ] ] ]

][

]

[

][

[ ][

]

]

]

[ ]

rate of polymerization

[ ][

]

8-5 (a) [ ]

(M0 = 71 for acrylamide)

[ ]

(

(

) (

[ ]) )[(

)(

)( -

( [ ]

([ ] )

(b)

(

) [

-

) (

(

)]

)

[]

] [

) (

-

-

)(

)(

) (

)

]

[ ]

([ ] ) [ ] [ ]

[ ] [ ] [ ] [ ] [ ] 0.886(100) = 88.6 g/l = amount of polymer made in 10 minutes.

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

8-3

8-6 Density of MMA = 0.94 g/cm3 [ ] []

-

(a)

[ ]

Kinetic chain length, ( [

(

[]

[ ])

)

)(

)(

[]

[ ][

(b)

( )]

[ ]

]

[]

( )

( ) [ ]

Solve as for problem 8-5 to obtain [ ]

.

Amount of monomer consumed = [M]0 – [M] = (9.40 – 9.31)(100) = 9 moles and amount of monomer made = 9(100) = 900 grams.

8-7 [

][

]

[ ] At steady state: [ [

][

]

] [

][

[ ][

]

{

[

]

]

} [ ]

(

[

][

]

)

8-8 (a) (i)

is increased 10 fold (ii) ̅​̅​̅​̅ is increased 10 fold is decreased by a factor of (√ ) (ii) ̅​̅​̅​̅ is increased by a factor of √ .

(b) (i)

8-9 See Section 8.8.5.

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

8-4

8-10 [

]

[

[ ] [

][

[ ][

]

[

[ ]

] ]

]

[ ]

[

]

[ ]

[ ]

[

]

[ ]

[ ]

Integrate between [ ] [ ] [ ] [ ] and [TH], [M] at t: [

([

] ]

[ ]

)

([ ] )

(

)

(

)

.

8-11 The parameters which pertain to the initial polymerization (which is to be modified) as subscripted zero. For initiation by thermal decomposition of an initiator: [ ](

[ ])

[] ̅​̅​̅​̅

and is to be unchanged when the reaction conditions are modified. If must also be doubled to keep ̅​̅​̅​̅ constant, i.e.

doubled then in final polymerization mixture [ ].

Then, double the initiator concentration. At this stage √

[ ]√

To increase √ √

√

to [√

. multiply [M] by √ then

]√[ ]

.

Conclusion: double [I] and multiply the initial [M] by √ .

Copyright © 2013 Elsevier Inc. All rights reserved.

is

Rudin & Choi: The Elements of Polymer Science & Engineering, 3rd Edition Chapter 9 Solutions Manual

9-1 (a)

[

]

[

]

( ) ( )

The copolymer contains 2.64 moles methyl methacrylate for every mole of 5-ethyl-2vinyl pyridine. [

]

(b) [[

]

]

.

9-2 Using equations (9-44) and (9-45):

if dM3 = 1 for every mole of M3 in the copolymer, dM1 = 2.867 and dM2 = 2.608.

Therefore,

.

9-3 Use the reactivity ratio from problem 9-2 [

]

[

]

(equations (9-49) and (9-57))

When f1 = mole fraction of methacrylonitrile, [ [

]

] [

]

f2 = mole fraction styrene in copolymer [ [

[ ]

[

]

]

] [

]

̅ = average sequence length of methacrylonitrile = 0.44(0.43) + 1 = 1.19.

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

9-2

̅ = average sequence length of styrene units [ [

] ] [ [

(equations (9-58) and (9-50))

] ]

.

9-4 e

Q

Butadiene

-1.05

2.39

Vinyl acetate

-0.22

0.03

Styrene

-0.80

1.00

Acrylonitrile

1.20

0.60

Methyl methacrylate

0.40

0.74

Remarks negative e, high Q: i.e. electronrich double bond (relatively nucleophilic) and resonance stabilized. low Q – no resonance stabilization, probably won’t copolymerize with butadiene by free radical means similar to butadiene positive e: i.e. relatively electrophilic double bond like acrylonitrile has positive e and is an electron acceptor

Use r1r2 as a measure of alternation. Alternation increases as and r2 are < 1. [ ( ) ] (a) butadiene/vinyl acetate: [ ( Check:

) ]

[

(

)]

(butadiene = M1) (eq. (9-70))

[

(

)]

(vinyl acetate) (eq. (9-71))

Conclusion: This is probably not a practical copolymerization. (b) butadiene/styrene: [ (

) ]

(c) butadiene/acrylonitrile: [ (

) ]

(d) butadiene/methyl methacrylate:

Copyright © 2013 Elsevier Inc. All rights reserved.

provided both r1

Solutions Manual

9-3 [ (

) ]

Therefore, order of increasing alternation = styrene < methyl methacrylate < acrylonitrile; vinyl acetate does not copolymerize.

9-5 (a)

(

)

( (

(

)

) )

( ) ( ) ( ) = fraction of acrylonitrile sequences with one acrylonitrile unit. ( ) ( ) ( ) = fraction of acrylonitrile sequences containing 2 acrylonitrile units.

Fraction of acrylonitrile sequences with 3 or more acrylonitrile units ( ) ( ) . (b) ( ) = mole fraction of acrylonitrile.

9-6 ̅

p = extent of reaction at which gelation occurs, ρ0 = fraction of reactive groups which are part of a multifunctional cross-linking agent and ̅ is the weight average degree of polymerization which the polymer would have had if all monomers were bifunctional.

Increase of initiator concentration results in a decrease in ̅ (Chapter 8).

9-7 To avoid having the cross-linkable units bunched together the reactivity ratio of the acidcontaining vinyl monomer should be close to zero in its reaction with vinyl toluene. Also, the reactivity ratio for vinyl toluene should be < 1 to ensure that the polymerization will not just yield poly(vinyl toluene) without incorporating the acid vinyl comonomer.

Conclusions: r (acid vinyl monomer) = 0 r (vinyl toluene) < 1.

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

9-4

9-8 Take allyl acetate as monomer 1: [

(

)]

[

[

(

)]

[

( (

)

(

)] (

)]

) (

)

(

)

This is not a good idea. The reaction will not yield copolymers with appreciable concentrations of both monomers.

9-9 For a random copolymerization r1r2 = 1. Here r1 = 5 (ethylene = M1). Also, in a random copolymerization the simple copolymer equation reduces to: [

]

[

]

[ [

] ]

or: (

) (

)

Solving: f1 = 0.32 = mole fraction of ethylene in the gas phase. Gas phase composition = 32 mole % ethylene and 68 mole % propylene.

9-10

(

(

)

)

(

)

(positive root) Conclusion: need 0.931 mole fraction vinyl chloride (0.069 mole fraction vinyl acetate) in the feed.

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

9-5

9-11 The best product contains isolated acrylamide residues so that the eventual cross-links formed at these sites are spaced apart and not bunched together. That is to say, the average sequence length ( ̅ ) of acrylamide (monomer 2) is ideally = 1 in these polymers. ̅ Here f1 = 0.85, f2 = 0.15.

For the methyl methacrylate case: (

̅

)

.

There will be little difference in the styrene case (r2 = 0.6).

Conclusion: There is nothing to choose between the two cases in terms of isolated, useful cross-linking sites. (There will be other differences in costs and properties which may be important.)

9-12 (a) ethylene glycol adipic acid tricarballylic acid

moles 2 1.90 0.051 3.951

functionality, f 2 2 3

equivalents 4 3.80 0.153 7.953

There are a total of 3.953 acid equivalents. Therefore, only 3.953 alcohol equivalents are available for reaction (barring side reactions).

̅

p = extent of conversion at gel point. ρ0 = fraction of reactive groups which are part of a multifunctional cross-linking agent. ̅ = weight average degree of polymerization which the polymer would have if all monomers were bifunctional At high conversions the weight average degree of polymerization of the reaction mixture will be close to that of the polymer. Therefore, as a good approximation we will set ̅

(from eq. (7-35)) (

)

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

9-6

Note: This calculation of ρ0 uses only the useful equivalents. Calculation with the total number of equivalents (7.953) yields ρ0 = 0.01924. This makes no difference in this case.

Substituting in eq. (9-68): (

)

(b) ̅ (you could put ̅ (

̅ for the sake of consistency of symbols in this case).

)

at gel point ̅

and

(

)

,

.

The difference in the calculations reflects the fact that one estimate assumes that gelation occurs when ̅ is infinite whereas the other assumes that ̅ is infinite when the reaction gels. We know from Chapter 2 that ̅ ̅ and so the former calculation indicates the occurrence of massive cross-linking at an earlier stage of the polymerization. (Note that these calculations require the carrying of relatively many decimal places).

Copyright © 2013 Elsevier Inc. All rights reserved.

Rudin & Choi: The Elements of Polymer Science & Engineering, 3rd Edition Chapter 10 Solutions Manual

10-1 (a) no change (b) no change.

10-2 Add more initiator in interval II so [I] is doubled. The rate of polymerization and the number of particles per unit volume of aqueous phase will not be altered because the [I] value which is operative in these relations is that in interval I (see text for details). An alternative procedure would involve addition of chain transfer agents but the details of such procedures cannot be quantified without knowledge of the nature of polymerizing monomer and reaction temperature. Since this information was not provided the only way the question can be answered in general is as outlined.

10-3 The enhanced polymerization rate probably reflects the existence of more than one radical per particle because of the large particle size. The rate of mutual termination of such macro-radicals is also decreased by the autoacceleration effect (Section 8.13.2).

Copyright © 2013 Elsevier Inc. All rights reserved.

Rudin & Choi: The Elements of Polymer Science & Engineering, 3rd Edition Chapter 11 Solutions Manual

11-1 Isobutene is cationically active because of the inductive effects of the two electronreleasing methyl groups attached to one of the doubly bonded carbons (Figure 11.2). For this reason it does not respond to anionic initiators. Free radicals do not produce high polymers because chain transfer reactions on the methyls produce relatively stable radical species which do not propagate readily (cf. reaction 8-83). The electronegative –CN group in acrylonitrile decreases the electron density on the double bond. The facilitates attack by anionic initiators and discourages reactivity towards cations. In additions, anionic active centres can be stabilized to some extent by delocalization of the charge across the cyanide group:

11-2 (a) NaOCH3 in tetrahydrofuran or an alkane or a mixture of these solvents is an example. Almost any anionic initiator is strong enough to effect the polymerization of this monomer. (b) Lithium metal or lithium alkyls in hydrocarbon solvents. Alkali metal complexes of alpha-methylstyrene (Section 11.2.2.5) in tetrahydrofuran would also probably initiate this polymerization but the product polymers would not be a stereoregular and this would therefore not be a practical process. (c) A Lewis acid and Lewis acid salt is a suitable initiator for tetrahydrofuran. See reactions 11-60 and 11-61, for example. Polymerization can be carried out in bulk, liquid monomer or in an inert diluent like methylene chloride. Incidentally, the polymerization of this monomer is usually performed at room temperature. The ceiling temperature (Section 8.16.2) of polytetrahydrofuran is 85°C. Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

11-2

(d) Isotactic polypropylene: (C2H5)3Al/TiCl3 catalyst suspended in hydrocarbon liquid, 50°C/20atm.

Syndiotactic polypropylene: (C2H5)3Al/VCl4/anisole catalyst soluble in toluene, -70 °C. (e) Protonic acids (e.g. AlCl3) are all suitable (Section 11.4.2.1). Polymerizations can be carried out in concentrated solutions in aromatic hydrocarbons.

11-3 (a) See Section 11.2.6.3. Initiate styrene polymerization with butyl lithium in tetrahydrofuran. Add methyl methacrylate monomer and use a coupling agent like I2 to complete the process. (b) See Section 11.2.2.2 and 11.2.6.3. Initiate alpha-methylstyrene polymerization with butyl lithium in benzene solution. Form a polyisoprene block by addition of this monomer to the living alpha-methylstyrene ends and then add styrene to the isoprenyl ends. A little tetrahydrofuran will accelerate the attack of dienyl ends on styrene monomer. (c) See Table 11.1. Any process that forms living polystyryl ends can be used to initiate the methacrylonitrile blocks.

11-4 Polytetrahydrofuran, H-(OCH2CH2CH2CH2)m-OH can be made by methods sketched in the answer to the problem 11-2 (c): This hydroxyl-ended polymer is then reacted with 2 moles of 2,4-toluene diisocyanate to produce an isocyanate-ended prepolymer (cf. reaction 1-12). Polybutadiene is polymerized in hydrocarbon solvent with a lithium alkyl initiator (Figure 11.1) and then capped with ethylene oxide (reaction 11-22) to yield a hydroxyl-ended polybutadiene which can be made to react with 0.5 moles of isocyanateended prepolymer described above.

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

11-3

11-5 (a)

(b)

(c) See reaction 11-58. (d) See reactions 11-10 and 11-11. (e) See reactions 11-59.

11-6 See Section 11.2.3.2. ̅​̅​̅​̅ for initiations by electron transfer (eq. (11-11)) followed by dimerization of the monomeric radical anions (eq. (11-12)). Since the reaction was complete in the case d[M] = [M] and d[I] = [I].

Molecular weight of styrene, M0 = 104. (

)(

)(

)

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

11-4 . .

Since naphthalene is in excess the total no. moles initiator in 50 ml tetrahydrofuran is 0.043 moles. The no. moles initiator transferred in 1 ml of dark green solution moles.

Total volume of polymerization mixture = 50 ml + 1 ml = 51 ml. . ̅​̅​̅​̅ ̅

̅​̅​̅​̅

̅​̅​̅​̅ ̅​̅​̅​̅

̅​̅​̅​̅

̅

.

Copyright © 2013 Elsevier Inc. All rights reserved.

Rudin & Choi: The Elements of Polymer Science & Engineering, 3rd Edition Chapter 12 Solutions Manual

12-1 (

)

(

)

Integrating between temperatures T1 and T2: ( )

( )

(

(

(

)

-

-

( (

)

)

)

)

(a) equilibrium constant for polymerization: (

(

)

(

)

(

)

)

(

)

)

(

(

)

.

(b) equilibrium H2O concentration: ( ) ( ) -

(

(

)

(

)

)

(

)

( ) ( ) ( ).

The volume of head space in the polymerization reactor above the polymer melt is independent of temperature. The vapor pressure (1 atm steam) also does not vary with temperature. Let [ [

]

[

]

]

= concentration of water in head space at 250 °C.

assuming that the H2O vapor behaves as an ideal gas.

Let [H2Onylon] = equilibrium concentration of water in the polymer melt: [ [

(

)]( (

[

)

](

](

[

(

)]( (

)

)

) )

)

[

(

)]( (

)

)

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

(c) ̅

12-2

([

]

)

Compare ̅ ( ̅ (

)

̅ (

)

[

) and ̅ (

):

(

)

(

)[

](

)

(

)

(

)[

](

)

]

C0 is the same in both cases. Substituting the calculated values: ̅ (

)

̅ (

)

[

]

The molecular weight decreases as the final reaction temperature is lowered at fixed water vapor pressure.

12-2 The suspension polymerization process could be modified by any or all of the following techniques: (a) add electrolyte (e.g. NaCl) to aqueous phase to salt out the acrylonitrile; (b) add small amounts of water-soluble inhibitors (e.g. thiosulfates or thiocyanates) to the aqueous phase to prevent polymerization in this phase; (c) use an initiator with negligible water solubility; (d) add a water-insoluble solvent that acts as an extraction agent (an alkyl aryl ether could be used if its costs were acceptable and it could be left in the final copolymer product).

12-3 The three temperature/time stages are: (a) 80°C/6 hours (b) 90°C/5 hours (c) 110°C/12 hours.

The required kd values are then: (

(a)

)

(b)

(

(c)

(

(

)

(

)

) ).

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual (

12-3 (

) (

)

) (

(

) -

-

)

-

The following table gives the calculated kd values for some of the potential initiators listed in Table 8.1.

Temp. (°C)

kd

dibenzoyl peroxide

73

1.93 × 10-5

dibenzoyl peroxide

80

4.60 × 10-5

dibenzoyl peroxide

90

15.0 × 10-5

t-butyl peracetate

102

1.93 × 10-5

t-butyl peracetate

110

4.49 × 10-5

dicumyl peroxide

115

1.93 × 10-5

dicumyl peroxide

110

1.16 × 10-5

From these calculations benzoyl peroxide is reasonably close to the requirements for stage (b) and dicumyl peroxide is the best of the list for stage (c).

12-4 Since the effective monomer concentration in the polymerizing particles is reduced in (b) and (c) compared to (a) the molecular weights of the polymers made by the M ADD and E ADD processes will be lower than in the batch (B) process. The particle size distribution in the E ADD process will be very board because soap is being added and particles are being formed through the course of the whole polymerization. Some particles which are formed near the end of the polymerization will not have had much time to grow while those formed in the beginning will have become very large. Particles grow faster in the B process than in the M ADD and E ADD versions. Thus the time for disappearance of soap micelles will be greater in the M ADD than the B process. Since the rate of radical generation is the same in all cases, more micelles will be “stung” and more particles will be generated in the M ADD than B process. The average size of the particles will therefore be smaller in the M ADD version than in the batch process and the particle size distribution will also be somewhat broader because the formation of particles extended over a greater time interval.

Copyright © 2013 Elsevier Inc. All rights reserved.

Solutions Manual

12-4

12-5 (a)

([ ]

[ ])

[ ] [ ] at 50% conversion with no volume shrinkage. In an initial volume of 50 3 m there will be 16.67 m3 H2O and 33.33 m3 monomers. When the reaction has gone to 50% conversion the volume of the same mixture will have shrunk to a total of 16.67 (H2O) + 16.67 (unreacted monomer) + 16.67/1.27 (polymer) = 46.47 m3. Then the true monomer concentration in the effluent is (3 mol/l)(50/46.47) = 3.23 mol/l reaction mixture. ( [ ]

)

-

[ ]

(b) F(t) = fraction of exiting material which is in the reactor for a time < t (t = θ in this question).

For a perfectly stirred CSTR ( ) 12-6 The volume of the larger vessel in 8 times that of the smaller vessel. Thus the rate of generation of heat will be 8 times greater. But the wall area is only 4 times as great in the larger vessel. Transfer through the reactor walls can remove heat 4 times as fast as in the smaller vessel. The rate of heat removal in the smaller vessel was 1/8th the rate of the heat generation in the larger vessel and 80% of this heat was taken out through the reactor walls. Thus the rate of heat removal through the walls in the larger vessel is 4(0.8)(Q/8) = 0.4Q, where Q is the rate of heat generated in the larger reactor. By difference, 60% of the process heat in the larger reactor must be removed by taking off and condensing propylene vapor (cf. eq. (12-15).

Copyright © 2013 Elsevier Inc. All rights reserved.

Rudin & Choi: The Elements of Polymer Science & Engineering, 3rd Edition Chapter 13 Solutions Manual

13-1 There are three pathways to producing biopolymers: 1) extraction and modification of natural polymers from biomass 2) polymerization of bio-based monomers 3) extraction of polymers produced in microorganisms.

13-2 Both starch and cellulose are made up of the same glucose units but in the case of starch, the glucose units are linked by α-1,4 linkage while in the case of cellulose, β-1,4 linkage.

13-3 Natural rubber consists of 97% cis-1,4-polyisoprene. The other 3% contains various proteins that would trigger allergies. Most of the natural rubber products are crosslinked.

Therefore, they are not biodegradable.

13-4 When natural rubber is highly crosslinked, the material will become very rigid and hence not suitable for tire applications.

Copyright © 2013 Elsevier Inc. All rights reserved.