Ch14 Solution Manual Material Science and Engineering 8th Edition PDF

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CHAPTER 14

POLYMER STRUCTURES

PROBLEM SOLUTIONS

Hydrocarbon Molecules Polymer Molecules The Che mistry of Polymer Molecules 14.1 On the basis of the structures presented in this chapter, sketch repeat unit structures for the following polymers: (a) polychlorotrifluoroethylene, and (b) poly(vinyl alcohol). Solution The repeat unit structures called for are sketched below. (a) Polychlorotrifluoroethylene

(b) Poly(vinyl alcohol)

Molecular Weight 14.2 Compute repeat unit molecular weights for the following: (a) poly(vinyl chloride), (b) poly(ethylene terephthalate), (c) polycarbonate, and (d) polydimethylsiloxane. Solution (a) For poly(vinyl chloride), each repeat unit consists of two carbons, three hydrogens, and one chlorine (Table 14.3). If AC, AH and ACl represent the atomic weights of carbon, hydrogen, and chlorine, respectively, then

m = 2(AC ) + 3(AH) + (ACl)

= (2)(12.01 g/mol) + (3)(1.008 g/mol) + 35.45 g/mol = 62.49 g/mol

(b) For poly(ethylene terephthalate), from Table 14.3, each repeat unit has ten carbons, eight hydrogens, and four oxygens. Thus, m = 10(AC) + 8(AH ) + 4(AO)

= (10)(12.01 g/mol) + (8)(1.008 g/mol) + (4)(16.00g/mol) = 192.16 g/mol

(c) For polycarbonate, from Table 14.3, each repeat unit has sixteen carbons, fourteen hydrogens, and three oxygens. Thus, m = 16(AC) + 14(AH) + 3(AO)

= (16)(12.01 g/mol) + (14)(1.008 g/mol) + (3)(16.00 g/mol)

= 254.27 g/mol

(d) For polydimethylsiloxane, from Table 14.5, each repeat unit has two carbons, six hydrogens, one silicon and one oxygen. Thus, m = 2(AC) + 6(AH) + (ASi) + (AO)

= (2)(12.01 g/mol) + (6)(1.008 g/mol) + (28.09 g/mol) + (16.00 g/mol) = 74.16 g/mol

14.3 The number-average molecular weight of a polypropylene is 1,000,000 g/mol. Compute the degree of polymerization. Solution We are asked to compute the degree of polymerization for polypropylene, given that the number-average molecular weight is 1,000,000 g/mol. The repeat unit molecular weight of polypropylene is just m = 3(AC) + 6(AH)

= (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol

Now it is possible to compute the degree of polymerization using Equation 14.6 as

DP =

M n 1, 000, 000 g/mol = = 23, 760 m 42.08 g/mol

14.4 (a) Compute the repeat unit molecular weight of polystyrene. (b) Compute the number-average molecular weight for a polystyrene for which the degree of polymerization is 25,000. Solution (a) The repeat unit molecular weight of polystyrene is called for in this portion of the problem. For polystyrene, from Table 14.3, each repeat unit has eight carbons and eight hydrogens. Thus, m = 8(AC) + 8(AH)

= (8)(12.01 g/mol) + (8)(1.008 g/mol) = 104.14 g/mol

(b) We are now asked to compute the number-average molecular weight. Since the degree of polymerization is 25,000, using Equation 14.6

M n = (DP)m = (25, 000)(104.14 g/mol)

= 2.60 ´ 10 6 g/mol

14.5 Below, molecular weight data for a polypropylene material are tabulated. Compute (a) the numberaverage molecular weight, (b) the weight-average molecular weight, and (c) the degree of polymerization. Molecular Weight Range (g/mol)

xi

wi

8,000–16,000

0.05

0.02

16,000– 24,000

0.16

0.10

24,000– 32,000

0.24

0.20

32,000– 40,000

0.28

0.30

40,000– 48,000

0.20

0.27

48,000–56,000

0.07

0.11

Solution (a) From the tabulated data, we are asked to compute M n , the number-average molecular weight. This is carried out below.

Molecular wt Range

Mean Mi

xi

xi Mi

8,000-16,000

12,000

0.05

600

16,000-24,000

20,000

0.16

3200

24,000-32,000

28,000

0.24

6720

32,000-40,000

36,000

0.28

10,080

40,000-48,000

44,000

0.20

8800

48,000-56,000

52,000

0.07

3640

____________________________ Mn = xi M i = 33, 040 g/mol

å

(b) From the tabulated data, we are asked to compute M w, the weight-average molecular weight.

Molecular wt. Range

Mean Mi

wi

wiMi

8,000-16,000

12,000

0.02

240

16,000-24,000

20,000

0.10

2000

24,000-32,000

28,000

0.20

5600

32,000-40,000

36,000

0.30

10,800

40,000-48,000

44,000

0.27

11,880

48,000-56,000

52,000

0.11

5720

___________________________ wi M i = 36, 240 g/mol Mw =

å

(c) Now we are asked to compute the degree of polymerization, which is possible using Equation 14.6. For polypropylene, the repeat unit molecular weight is just m = 3(AC) + 6(AH)

= (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol

And

DP =

Mn 33, 040 g/mol = = 785 m 42.08 g/mol

14.6 Molecular weight data for some polymer are tabulated here. Compute (a) the number-average molecular weight, and (b) the weight-average molecular weight. (c) If it is known that this material's degree of polymerization is 710, which one of the polymers listed in Table 14.3 is this polymer? Why? Molecular Weight Range g/mol 15,000–30,000

xi 0.04

wi 0.01

30,000– 45,000

0.07

0.04

45,000– 60,000

0.16

0.11

60,000– 75,000

0.26

0.24

75,000– 90,000

0.24

0.27

90,000–105,000

0.12

0.16

105,000–120,000

0.08

0.12

120,000–135,000

0.03

0.05

Solution (a) From the tabulated data, we are asked to compute M n , the number -average molecular weight. This is carried out below.

Molecular wt. Range

Mean Mi

xi

15,000-30,000

22,500

0.04

900

30,000-45,000

37,500

0.07

2625

45,000-60,000

52,500

0.16

8400

60,000-75,000

67,500

0.26

17,550

75,000-90,000

82,500

0.24

19,800

90,000-105,000

97,500

0.12

11,700

105,000-120,000

112,500

0.08

9000

120,000-135,000

127,500

0.03 3825 _________________________

xiMi

Mn =

å xi M i

= 73,800 g/mol

(b) From the tabulated data, we are asked to compute M w, the weight -average molecular weight. This determination is performed as follows:

Molecular wt. Range

Mean Mi

wi

wiMi

15,000-30,000

22,500

0.01

225

30,000-45,000

37,500

0.04

1500

45,000-60,000

52,500

0.11

5775

60,000-75,000

67,500

0.24

16,200

75,000-90,000

82,500

0.27

22,275

90,000-105,000

97,500

0.16

15,600

105,000-120,000

112,500

0.12

13,500

120,000-135,000

127,500

0.05 6375 _________________________

M w = å wi M i = 81, 450 g/mol

(c) We are now asked if the degree of polymerization is 710, which of the polymers in Table 14.3 is this material? It is necessary to compute m in Equation 14.6 as

m =

73, 800 g/mol Mn = = 103.94 g/mol DP 710

The repeat unit molecular weights of the polymers listed in Table 14.3 are as follows:

Polyethylene--28.05 g/mol Poly(vinyl chloride)--62.49 g/mol Polytetrafluoroethylene--100.02 g/mol Polypropylene--42.08 g/mol Polystyrene--104.14 g/mol Poly(methyl methacrylate)--100.11 g/mol Phenol-formaldehyde--133.16 g/mol Nylon 6,6--226.32 g/mol PET--192.16 g/mol Polycarbonate--254.27 g/mol Therefore, polystyrene is the material since its repeat unit molecular weight is closest to that calculated above.

14.7 Is it possible to have a poly(methyl methacrylate) homopolymer with the following molecular weight data and a of polymerization of 527? Why or why not? Molecular Weight Range (g/mol) 8,000– 20,000

wi 0.02

xi 0.05

20,000–32,000

0.08

0.15

32,000–44,000

0.17

0.21

44,000–56,000

0.29

0.28

56,000–68,000

0.23

0.18

68,000–80,000

0.16

0.10

80,000–92,000

0.05

0.03

Solution This problem asks if it is possible to have a poly(methyl methacrylate) homopolymer with the given molecular weight data and a degree of polymerization of 527. The appropriate data are given below along with a computation of the number-average molecular weight.

Molecular wt. Range

Mean Mi

xi

xiMi

8,000-20,000

14,000

0.05

700

20,000-32,000

26,000

0.15

3900

32,000-44,000

38,000

0.21

7980

44,000-56,000

50,000

0.28

14,000

56,000-68,000

62,000

0.18

11,160

68,000-80,000

74,000

0.10

7400

80,000-92,000

86,000

0.03 2580 _________________________

Mn =

å xi M i =

47, 720 g/mol

For PMMA, from Table 14.3, each repeat unit has five carbons, eight hydrogens, and two oxygens. Thus, m = 5(AC) + 8(AH) + 2(AO)

= (5)(12.01 g/mol) + (8)(1.008 g/mol) +(2)(16.00 g/mol) = 100.11 g/mol

Now, we will compute the degree of polymerization using Equation 14.6 as

DP =

Mn 47, 720 g/mol = 477 = 100.11 g/mol m

Thus, such a homopolymer is not possible since the calculated degree of polymerization is 477 (and not 527).

14.8 High-density polyethylene may be chlorinated by inducing the random substitution of chlorine atoms for hydrogen. (a) Determine the concentration of Cl (in wt%) that must be added if this substitution occurs for 5% of all the original hydrogen atoms. (b) In what ways does this chlorinated polyethylene differ from poly(vinyl chloride)? Solution (a) For chlorinated polyethylene, we are asked to determine the weight percent of chlorine added for 5% Cl substitution of all original hydrogen atoms. Consider 50 caUERQ DW RP V W KHUHDUH

SRVVLEOHVLGH-bonding sites.

Ninety-five are occupied by hydrogen and five are occupied by Cl. Thus, the mass of these 50 carbon atoms, mC, is just mC = 50(AC) = (50)(12.01 g/mol) = 600.5 g

Likewise, for hydrogen and chlorine, mH = 95(AH ) = (95)(1.008 g/mol) = 95.76 g mCl = 5(ACl) = (5)(35.45 g/mol) = 177.25 g Thus, the concentration of chlorine, CCl, is determined using a modified form of Equation 4.3 as

CCl =

=

mCl x 100 mC + mH + mCl

177.25 g 600.5 g + 95.76 g + 177.25 g

´ 100 = 20.3 wt%

(b) Chlorinated polyethylene differs from poly(vinyl chloride), in that, for PVC, (1) 25% of the side-bonding sites are substituted with Cl, and (2) the substitution is probably much less random.

Molecular Shape 14.9 For a linear polymer molecule, the total chain length L depends on the bond length between chain atoms d, the total number of bonds in the molecule N, and the angle between adjacent backbone chain atoms θ, as follows:

æqö L = Nd sin ç ÷ è2 ø

(14.11)

Furthermore, the average end-to-end distance for a series of polymer molecules r in Figure 14.6 is equal to

r = d

N

A linear polytetrafluoroethylene has a number-DYHUDJ H P ROHFXODU Z HLJ KWRI

(14.12)

J P RO FRP SXW H DYHUDJ H

values of L and r for this material. Solution This problem first of all asks for us to calculate, using Equation 14.11, the average total chain length, L, for a linear polytetrafluoroethylene polymer having a number-average molecular weight of 500,000 g/mol. It is necessary to calculate the degree of polymerization, DP, using Equation 14.6. For polytetrafluoroethylene, from Table 14.3, each repeat unit has two carbons and four flourines. Thus, m = 2(AC) + 4(AF)

= (2)(12.01 g/mol) + (4)(19.00 g/mol) = 100.02 g/mol

and

DP =

Mn 500, 000 g/mol = 5000 = 100.02 g/mol m

which is the number of repeat units along an average chain. Since there are two carbon atoms per repeat unit, there are two C—C chain bonds per repeat unit, which means that the total number of chain bonds in the molecule, N, is just (2)(5000) = 10,000 bonds. Furthermore, assume that for single carbon-carbon bonds, d= 0.154 nm and q = 109° 6HFW LRQ

W KHUHIRUH IURP ( TXDW LRQ

æqö L = Nd sin ç ÷ è2 ø

é æ109° öù = (10, 000)(0.154 nm) êsin ç ÷ú = 1254 nm ë è 2 øû

It is now possible to calculate the average chain end-to-end distance, r , using Equation 14.12 as

r = d N = (0.154 nm)

10, 000 = 15.4 nm

14.10 Using the definitions for total chain molecule length, L (Equation 14.11) and average chain end-toend distance r (Equation 14.12), for a linear polyethylene determine: (a) the number-average molecular weigKWIRU/

QP

(b) the number-average molecular weight for r = 20 nm. Solution (a) This portion of the problem asks for us to calculate the number-average molecular weight for a linear polyethylene for which L in Equation 14.11 is 2500 nm. It is first necessary to compute the value of N using this equation, where, for the C—C chain bond, d = 0.154 nm, and q = 109°. Thus

N =

=

L æq ö d sin ç ÷ è2ø

2500 nm æ109° ö (0.154 nm) sin ç ÷ è 2 ø

= 19, 940

Since there are two C—C bonds per polyethylene repeat unit, there is an average of N/2 or 19,940/2 = 9970 repeat units per chain, which is also the degree of polymerization, DP. In order to compute the value of M n using Equation 14.6, we must first determine m for polyethylene. Each polyethylene repeat unit consists of two carbon and four hydrogen atoms, thus m = 2(AC) + 4(AH)

= (2)(12.01 g/mol) + (4)(1.008 g/mol) = 28.05 g/mol

Therefore

M n = (DP ) m = (9970)(28.05 g/mol)= 280,000 g/mo

(b) Next, we are to determine the number-average molecular weight for r = 20 nm. Solving for N from Equation 14.12 leads to

N =

(20 nm) 2 r2 = = 16, 900 (0.154 nm) 2 d2

which is the total number of bonds per average molecule. Since there are two C—C bonds per repeat unit, then DP = N/2 = 16,900/2 = 8450. Now, from Equation 14.6

M n = (DP ) m = (8450)(28.05 g/mol)= 237,000 g/mo

Molecular Configurations 14.11 Sketch portions of a linear polystyrene molecule that are (a) syndiotactic, (b) atactic, and (c) isotactic. Use two-dimensional schematics per footnote 8 of this chapter. Solution We are asked to sketch portions of a linear polystyrene molecule for different configurations (using twodimensional schematic sketches). (a) Syndiotactic polystyrene

(b) Atactic polystyrene

(c) Isotactic polystyrene

14.12

Sketch cis and trans structures for (a) butadiene, and (b) chloroprene. Use two-dimensional

schematics per footnote 11 of this chapter. Solution This problem asks for us to sketch cis and trans structures for butadiene and chloroprene. (a) The structure for cis polybutadiene (Table 14.5) is

The structure of trans butadiene is

(b) The structure of cis chloroprene (Table 14.5) is

The structure of trans chloroprene is

The rmoplastic and Thermosetting Polymers 14.13 Make comparisons of thermoplastic and thermosetting polymers (a) on the basis of mechanical characteristics upon heating, and (b) according to possible molecular structures. Solution (a) Thermoplastic polymers soften when heated and harden when cooled, whereas thermosetting polymers, harden upon heating, while further heating will not lead to softening. (b) Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked.

14.14 (a) Is it possible to grind up and reuse phenol-formaldehyde? Why or why not? (b) Is it possible to grind up and reuse polypropylene? Why or why not? Solution (a) It is not possible to grind up and reuse phenol-formaldehyde because it is a network thermoset polymer and, therefore, is not amenable to remolding. (b) Yes, it is possible to grind up and reuse polypropylene since it is a thermoplastic polymer, will soften when reheated, and, thus, may be remolded.

Copolymers 14.15 Sketch the repeat structure for each of the following alternating copolymers: (a) poly(butadienechloroprene), (b) poly(styrene-methyl methacrylate), and (c) poly(acrylonitrile-vinyl chloride). Solution This problem asks for sketches of the repeat unit structures for several alternating copolymers. (a) For poly(butadiene-chloroprene)

(b) For poly(styrene-methyl methacrylate)

(c) For poly(acrylonitrile-vinyl chloride)

14.16

The number-average molecular weight of a poly(styrene-butadiene) alternating copolymer is

J P RO GHW HUP LQHW KHDYHUDJ HQXP EHURI VW yrene and butadiene repeat units per molecule. Solution Since it is an alternating copolymer, the number of both types of repeat units will be the same. Therefore, consider them as a single repeat unit, and determine the number-average degree of polymerization. For the styrene repeat unit, there are eight carbon atoms and eight hydrogen atoms, while the butadiene repeat consists of four carbon atoms and six hydrogen atoms. Therefore, the styrene-butadiene combined repeat unit weight is just m = 12(AC) + 14(AH)

= (12)(12.01 g/mol) + (14)(1.008 g/mol) = 158.23 g/mol

From Equation 14.6, the degree of polymerization is just

DP =

Mn 1, 350, 000 g/mol = = 8530 m 158.23 g/mol

Thus, there is an average of 8530 of both repeat unit types per molecule.

14.17 Calculate the number-average molecular weight of a random nitrile rubber [poly(acrylonitrileEXW DGLHQH FRSRO\ P HU@LQ ZKLFK W KH IUDFW LRQ RI EXW DGLHQH UHSHDWXQLW V LV

DVVXP H W KDWW KLV FRQFHQW UDW LRQ

corresponds to a degree of polymerization of 2000. Solution This problem asks for us to calculate the number-average molecular weight of a random nitrile rubber cop...