CHAPTER 14POLYMER STRUCTURES PROBLEM SOLUTIONS Hydrocarbon Molecules Polymer Molecules The Chemistry 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) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 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)(16.00 g/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 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 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 42.08 g/mol m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.6 M n = ( DP )m = (26.3.14 g/mol) = 2. m = 8(AC) + 8(AH) = (8)(12. each repeat unit has eight carbons and eight hydrogens.000)(104. Solution (a) The repeat unit molecular weight of polystyrene is called for in this portion of the problem.008 g/mol) = 104.70 × 106 g/mol Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Since the degree of polymerization is 26.4 (a) Compute the repeat unit molecular weight of polystyrene.01 g/mol) + (8)(1. (b) Compute the number-average molecular weight for a polystyrene for which the degree of polymerization is 26. from Table 14.14. using Equation 14.000. . For polystyrene.000. Thus.14 g/mol (b) We are now asked to compute the number-average molecular weight. 20 0.07 0.05 0.000 28.16 3200 24.11 Solution (a) From the tabulated data.02 240 16. the number-average molecular weight.000 44.000 0.000 0. Molecular Weight Range (g/mol) xi wi 8.000–56.000 0.000 36.800 40. Range Mean Mi wi wiMi 8.000–24.880 48.000–24.000–24.10 2000 24.02 16.10 24.16 0.000 12. and (c) the degree of polymerization.000 0.000–40.000 0. we are asked to compute M n .000 0.24 6720 32.240 g/mol Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.000–32.000–32.30 40. Compute (a) the number-average molecular weight. .27 11.000–40.000 0.000–56. we are asked to compute M w .000 0.28 10.20 32.000 0.30 10.000 0.000–48.000 0.000 0.14. Molecular wt. This is carried out below.080 40.000 0.07 3640 Mn = ∑xM i i = 33.000 0.05 600 16.000 52.000–32. the weight-average molecular weight.5 The following table lists molecular weight data for a polypropylene material.000–16.000 12. (b) the weight-average molecular weight.20 5600 32.000–16. 040 g/mol (b) From the tabulated data.000 0.000–48.000 0.000–56. Molecular wt Range Mean Mi xi xiMi 8.000 0.24 0.27 48.000 44.11 5720 M w = ∑ wi M i = 36.000–48.000 0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.000 20.000–40.000–16.000 28.000 20.000 52.20 8800 48.28 0.000 36. 08 g/mol Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.6. which is possible using Equation 14. .(c) Now we are asked to compute the degree of polymerization. For polypropylene. the repeat unit molecular weight is just m = 3(AC) + 6(AH) = (3)(12.08 g/mol And DP = Mn 33.01 g/mol) + (6)(1.008 g/mol) = 42.040 g/mol = = 785 m 42. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 500 0. .000 0. we are asked to compute M n .550 75.700 105.000–30. Molecular wt.000–135.000 22.16 105.04 900 30.000 127.01 30. (c) If it is known that this material's degree of polymerization is 750.000–105.000 37. and (b) the weight-average molecular weight.000–105.000–30.000 112.04 45.3 is this polymer? Why? Molecular Weight Range g/mol 15.000–90.500 0.000 97.000–60.800 g/mol Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.000 0.6 Molecular weight data for some polymer are tabulated here.000–45.500 0.26 0.16 8400 60.000 xi 0.04 wi 0.500 0.05 Solution (a) From the tabulated data.16 0.500 0. which one of the polymers listed in Table 14.000–60.12 11.27 90.14.000 67.000–120.12 0.000 0.000 0. the number-average molecular weight.24 75. This is carried out below.000–45. Range Mean Mi xi xiMi 15.24 0.000–135.24 19.500 0.000 0.000 52.08 0.03 Mn = 3825 ∑ xi M i = 73.000–90.800 90.500 0.000 0.07 2625 45.11 60.08 9000 120.12 120.500 0.000–75.000 82.000–75.26 17.000 0.000–120.07 0.03 0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Compute (a) the number-average molecular weight. 49 g/mol Polytetrafluoroethylene—100.000 112.14 g/mol Poly(methyl methacrylate)—100.3 is this material? It is necessary to compute m in Equation 14.000 82.000–45.32 g/mol PET—192.27 g/mol Therefore.05 g/mol Poly(vinyl chloride)—62.11 5775 60.600 105.800 g/mol = = 98.450 g/mol (c) We are now asked if the degree of polymerization is 750. which of the polymers in Table 14.500 0.12 13.000–30.000–105.000 37.500 0.(b) From the tabulated data.6—226. the weight-average molecular weight.08 g/mol Polystyrene—104.16 g/mol Nylon 6.500 120.4 g/mol DP 750 The repeat unit molecular weights of the polymers listed in Table 14.16 15. This determination is performed as follows: Molecular wt. .6 as m= Mn 73.24 16.500 0. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.000 127.500 0.000 22.05 Mw = 6375 ∑ wi M i = 81.01 225 30.000–135.3 are as follows: Polyethylene—28. Range Mean Mi wi wiMi 15.275 90.500 0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.500 0.000 52.000–75.000–60.11 g/mol Phenol-formaldehyde—133.000–120.04 1500 45.000–90.500 0.27 22.000 97.200 75.16 g/mol Polycarbonate—254. polytetrafluoroethylene is the material since its repeat unit molecular weight is closest to that calculated above.000 67. we are asked to compute M w .500 0.02 g/mol Polypropylene—42. 000 0.29 0.10 80.000–20. Molecular wt.000 0.21 7980 44.16 0.720 g/mol For PMMA.17 0.000 26.03 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 530.000 38.000–44.000 0.000–80.000–44.18 11.000–68. Thus.000 0. and two oxygens.000–80. The appropriate data are given below along with a computation of the number-average molecular weight.000 0.008 g/mol) + (2)(16.10 7400 80.000–56.000 0.000 0. Range Mean Mi xi xiMi 8.05 0.000 0.14.160 68.28 56.11 g/mol Now.000–32.000 wi 0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.03 Mn = 2580 ∑ xi M i = 47.00 g/mol) = 100.000–68.000 0.000 62.6 as Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.000–32.15 32.000 0.02 xi 0.3.000–56.05 20. m = 5(AC) + 8(AH) + 2(AO) = (5)(12.000 0. each repeat unit has five carbons. from Table 14.000 0.18 68.000 56.7 Is it possible to have a poly(methyl methacrylate) homopolymer with the following molecular weight data and a of polymerization of 530? Why or why not? Molecular Weight Range (g/mol) 8.000–20.01 g/mol) + (8)(1.000 0. . eight hydrogens.08 0.000–92.000 50.000 86.000–92.21 44.15 3900 32.000 14.000 74.05 700 20.23 0. we will compute the degree of polymerization using Equation 14.28 14. . Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.DP = Mn 47.11 g/mol Thus.720 g/mol = = 477 m 100. such a homopolymer is not possible since the calculated degree of polymerization is 477 (and not 530). 8 High-density polyethylene may be chlorinated by inducing the random substitution of chlorine atoms for hydrogen. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. mH = 92(AH) = (92)(1. the mass of these 50 carbon atoms.6 g × 100 = 29. for hydrogen and chlorine. we are asked to determine the weight percent of chlorine added for 8% Cl substitution of all original hydrogen atoms. in that.01 g/mol) = 600. . Thus. is just mC = 50(AC) = (50)(12. (a) Determine the concentration of Cl (in wt%) that must be added if this substitution occurs for 8% of all the original hydrogen atoms. (1) 25% of the sidebonding sites are substituted with Cl.5 g Likewise.73 g mCl = 8(ACl) = (8)(35. and (2) the substitution is probably much less random.0 wt% 600.3 as CCl = = mCl × 100 mC + mH + mCl 283.73g + 283.14. Consider 50 carbon atoms. mC. Ninety-five are occupied by hydrogen and five are occupied by Cl. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. CCl. the concentration of chlorine. is determined using a modified form of Equation 4.008 g/mol) = 92. there are 100 possible side-bonding sites. (b) In what ways does this chlorinated polyethylene differ from poly(vinyl chloride)? Solution (a) For chlorinated polyethylene.6 g (b) Chlorinated polyethylene differs from poly(vinyl chloride).5g + 92.6 g Thus.45 g/mol) = 283. for PVC. 154 nm and θ = 109° (Section 14. from Equation 14.Molecular Shape 14. as follows: ⎛θ⎞ L = Nd sin ⎜ ⎟ ⎝ 2⎠ (14. there are two C—C chain bonds per repeat unit. N. DP. is just (2)(5000) = 10. It is necessary to calculate the degree of polymerization. d = 0. compute average values of L and r for this material. using Equation 14. For polytetrafluoroethylene. Solution This problem first of all asks for us to calculate. from Table 14.01 g/mol) + (4)(19.000 bonds. which means that the total number of chain bonds in the molecule.4). .00 g/mol) = 100.000 g/mol. the average total chain length.11 ⎛θ⎞ L = Nd sin ⎜ ⎟ ⎝ 2⎠ Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. the total extended chain length L depends on the bond length between chain atoms d. for a linear polytetrafluoroethylene polymer having a number-average molecular weight of 500. the total number of bonds in the molecule N. using Equation 14.9 For a linear freely rotating polymer molecule. Thus. Furthermore.000 g/mol. m = 2(AC) + 4(AF) = (2)(12. each repeat unit has two carbons and four flourines.3.000 g/mol = = 5000 m 100. and the angle between adjacent backbone chain atoms θ.6.6 is equal to r = d N (14. Since there are two carbon atoms per repeat unit. therefore. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.11) Furthermore.02 g/mol and DP = Mn 500. L. the average end-to-end distance r for a randomly winding polymer molecule in Figure 14.12) A linear polytetrafluoroethylene has a number-average molecular weight of 500.11.02 g/mol which is the number of repeat units along an average chain. assume that for single carbon-carbon bonds. 4 nm Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. .154 nm) 10. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.⎡ ⎛ 109° ⎞⎤ = (10.12 as r = d N = (0.000)(0. r.154 nm) ⎢sin ⎜ ⎟⎥ = 1254 nm ⎣ ⎝ 2 ⎠⎦ It is now possible to calculate the average chain end-to-end distance.000 = 15. using Equation 14. (b) the number-average molecular weight for r = 30 nm. It is first necessary to compute the value of N using this equation.05 g/mol) = 532. and θ = 109°. . which is also the degree of polymerization.950 2 d (0.12).369 repeat units per chain. Now.14. then DP = N/2 = 37. there is an average of N/2 or 20.975.738 ⎛ 109° ⎞ (0.05 g/mol Therefore M n = ( DP )m = (10. Each polyethylene repeat unit consists of two carbon and four hydrogen atoms.154 nm.975)(28.11) and average chain endto-end distance r (Equation 14. Thus N = = L ⎛θ⎞ d sin ⎜ ⎟ ⎝ 2⎠ 2.154 nm) 2 which is the total number of bonds per average molecule.249 g/mol Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. for the C—C chain bond.600 nm. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. from Equation 14.6 M n = (DP )m = (18. DP. we are to determine the number-average molecular weight for r = 30 nm. Solving for N from Equation 14.008 g/mol) = 28. d = 0. for a linear polyethylene determine the following: (a) the number-average molecular weight for L = 2.600 nm.950/2 = 18.738/2 = 10.154 nm) sin ⎜ ⎝ 2 ⎟⎠ Since there are two C—C bonds per polyethylene repeat unit. where. thus m = 2(AC) + 4(AH) = (2)(12.600 nm = 20.369)(28.12 leads to N = r2 (30 nm) 2 = = 37. 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 14.01 g/mol) + (4)(1.850 g/mol (b) Next.10 Using the definitions for total chain molecule length L (Equation 14.6. Since there are two C—C bonds per repeat unit. In order to compute the value of M n using Equation 14.05 g/mol) = 290. we must first determine m for polyethylene.11 is 2. (a) Syndiotactic polystyrene (b) Atactic polystyrene (c) Isotactic polystyrene Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. (b) atactic. .11 Sketch portions of a linear polystyrene molecule that are (a) syndiotactic.Molecular Configurations 14. Use two-dimensional schematics per footnote 8 of this chapter. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Solution We are asked to sketch portions of a linear polystyrene molecule for different configurations (using two-dimensional schematic sketches). and (c) isotactic. 14.5) is The structure of trans butadiene is (b) The structure of cis chloroprene (Table 14. Use two-dimensional schematics per footnote 11 of this chapter.12 Sketch cis and trans structures for (a) butadiene. Solution This problem asks for us to sketch cis and trans structures for butadiene and chloroprene. (a) The structure for cis polybutadiene (Table 14. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. .5) is The structure of trans chloroprene is Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. and (b) chloroprene. while for thermosetting polymers. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.13 Make comparisons of thermoplastic and thermosetting polymers (a) on the basis of mechanical characteristics upon heating. the structures will normally be network or crosslinked. Solution (a) Thermoplastic polymers soften when heated and harden when cooled. harden upon heating. whereas thermosetting polymers. (b) Thermoplastic polymers have linear and branched structures.Thermoplastic and Thermosetting Polymers 14. while further heating will not lead to softening. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. . and (b) according to possible molecular structures. 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. may be remolded. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. will soften when reheated. it is possible to grind up and reuse polypropylene since it is a thermoplastic polymer. (b) Yes.14. and. . therefore. is not amenable to remolding. thus. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (a) For poly(butadiene-chloroprene) (b) For poly(styrene-methyl methacrylate) (c) For poly(acrylonitrile-vinyl chloride) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.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).Copolymers 14. . Solution This problem asks for sketches of the repeat unit structures for several alternating copolymers. 16 The number-average molecular weight of a poly(styrene-butadiene) alternating copolymer is 1.350.23 g/mol From Equation 14. and determine the number-average degree of polymerization. Therefore.008 g/mol) = 158. there is an average of 8.14. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.530 m 158. Solution Since it is an alternating copolymer.000 g/mol.01 g/mol) + (14)(1. consider them as a single repeat unit.6. while the butadiene repeat consists of four carbon atoms and six hydrogen atoms. Therefore. the number of both types of repeat units will be the same.23 g/mol Thus. For the styrene repeat unit. there are eight carbon atoms and eight hydrogen atoms. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. the degree of polymerization is just DP = Mn 1.350. . determine the average number of styrene and butadiene repeat units per molecule. the styrene-butadiene combined repeat unit weight is just m = 12(AC) + 14(AH) = (12)(12.000 g/mol = = 8.530 of both repeat unit types per molecule. and three hydrogen atoms. .7.30. its repeat unit molecular weight is mAc = 3(AC) + (AN) + 3(AH) = (3)(12.37 g/mol)(2500) = 133.01 g/mol) + (6)(1. Thus. M n may be computed using Equation 14.01 g/mol) + 14.06 g/mol The butadiene repeat unit is composed of four carbon and six hydrogen atoms.14. the average repeat unit molecular weight is just m = f Ac mAc + f Bu mBu = (0.425g/mol Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. its repeat unit molecular weight is mBu = 4(AC) + 6(AH) = (4)(12. Thus.008 g/mol) = 53.17 Calculate the number-average molecular weight of a random nitrile rubber [poly(acrylonitrilebutadiene) copolymer] in which the fraction of butadiene repeat units is 0.30)(54.6 as M n = m( DP ) = (53.37 g/mol Since DP = 2500 (as stated in the problem).06 g/mol) + (0.008 g/mol) = 54.09 g/mol From Equation 14. Solution This problem asks for us to calculate the number-average molecular weight of a random nitrile rubber copolymer. For the acrylonitrile repeat unit there are three carbon.01 g/mol + (3)(1. assume that this concentration corresponds to a degree of polymerization of 2500. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.09 g/mol) = 53. one nitrogen.70)(53. Also. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. propylene is the other repeat unit type since its m value is almost the same as the calculated m x. which of ethylene.14 g/mol Now.7.01 g/mol) + 6(1.008 g/mol) + 35. fs and fx being.008 g/mol) = 104.01 g/mol) + 4(19.5)(104.000 g/mol and a degree of polymerization of 3500.01 g/mol) + 4(1. it is necessary to calculate the repeat unit molecular weights for each of the possible other repeat unit types. tetrafluoroethylene.18 An alternating copolymer is known to have a number-average molecular weight of 250. we are to determine one of the repeat unit types if the other is styrene. mx.43g/mol DP 3500 Since this is an alternating copolymer we know that chain fraction of each repeat unit type is 0. It is first necessary to calculate m using Equation 14. propylene.08 g/mol mTFE = 2(AC) + 4(AF) = 2(12.008 g/mol) = 42. These are calculated below: methylene = 2(AC) + 4(AH) = 2(12.00 g/mol) = 100.5 Finally.14.05 g/mol mpropylene = 3(AC) + 6(AH) = 3(12. If one of the repeat units is styrene.01 g/mol) + 8(1. .02 g/mol mVC = 2(AC) + 3(AH) + (ACl) = 2(12.45 g/mol = 62.01 g/mol) + 3(1. Thus mx = = m − f s ms fx 71.5. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.14 g/mol) = 38.5. using Equation 14. respectively. that is fs = fx = 0. the chain fractions of the styrene and unknown repeat units.000 g/mol = = 71.6 as m= M n 250.43g/mol − (0.49 g/mol Therefore. it is possible to calculate the repeat unit weight of the unknown repeat unit type.008 g/mol) = 28. the repeat unit molecular weight for styrene is ms = 8(AC) + 8(AH) = 8(12.000 g/mol and a degree of polymerization of 3500. and vinyl chloride is the other repeat unit? Why? Solution For an alternating copolymer which has a number-average molecular weight of 250.72 g/mol 0. 008 g/mol) = 54.01 g/mol) + 8(1.53 = 0.09 g/mol − 104.14 g/mol Solving for fb in the above expression yields fb = m − ms 77. considering the following possibilities: random.7 for this copolymer may be written in the form m = f b mb + f s ms = f b mb + (1 − f b )ms in which mb and ms are the repeat unit molecular weights for butadiene and styrene.14 g/mol = = 0. respectively. m .000 g/mol = = 77.14.78g/mol DP 4500 If we designate fb as the chain fraction of butadiene repeat units. (b) Which type(s) of copolymer(s) will this copolymer be.008 g/mol) = 104. These values are calculated as follows: mb = 4(AC) + 6(AH) = 4(12. and block? Why? Solution (a) This portion of the problem asks us to determine the ratio of butadiene to styrene repeat units in a copolymer having a weight-average molecular weight of 350. since the copolymer consists of only two repeat unit types. Equation 14. or the ratio is just f b 0. using Equation 14.14 g/mol Furthermore.19 (a) Determine the ratio of butadiene to styrene repeat units in a copolymer having a numberaverage molecular weight of 350.09 g/mol ms = 8(AC) + 8(AH) = 8(12. graft. the chain fraction of styrene repeat units fs is just 1 – fb.53 mb − ms 54.47. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.53 = = 1. fs = 1 – fb = 1 – 0. It first becomes necessary to calculate the average repeat unit molecular weight of the copolymer.6 as m= M n 350. Now.000 g/mol and a degree of polymerization of 4500. alternating.000 g/mol and degree of polymerization of 4500.78g/mol − 104.1 f s 0. .01 g/mol) + 6(1.47 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Therefore. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. the only one for which there is a restriction on the ratio of repeat unit types is alternating. . Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. the possibilities for this are random. the ratio must be 1:1. graft and block.(b) Of the possible copolymers. on the basis of the result in part (a). 14 mol of ethylene 28.20 Crosslinked copolymers consisting of 60 wt% ethylene and 40 wt% propylene may have elastic properties similar to those for natural rubber.008 g/mol) = 42.95 mol Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.14 mol + 0.08 g/mol Therefore. f (propylene) = 0. The ethylene (C2H4) molecular weight is m(ethylene) = 2(AC) + 4(AH) = (2)(12. In 100 g of this material.31 2. there are 60 g = 2. we are asked to determine the fraction of both repeat unit types. Solution For a copolymer consisting of 60 wt% ethylene and 40 wt% propylene.08 g/mol Thus. in 100 g of this material. the fraction of the ethylene repeat unit.01 g/mol) + (6)(1.95 mol = 0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. there are 60 g of ethylene and 40 g of propylene. is just f (ethylene) = 2.14 mol + 0.05 g/mol and 40 g = 0. f(ethylene). .14 mol = 0.69 2.95 mol Likewise. For a copolymer of this composition.14.95 mol of propylene 42.01 g/mol) + (4)(1. determine the fraction of both repeat unit types.05 g/mol The propylene (C3H6) molecular weight is m(propylene) = 3(AC) + 6(AH) = (3)(12.008 g/mol) = 28. mib = (4)(12.5.10 x + 68.88 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.21 A random poly(isobutylene-isoprene) copolymer has a number-average molecular weight of 200.11 g/mol From Equation 14.14.000 g/mol and DP = 3000.12 = 0.10 g/mol Also. the isobutylene repeat unit has four carbon and eight hydrogen atoms.5. . we are asked to compute the fractions of isobutylene and isoprene repeat units. f(isoprene) = 1 – x = 1 – 0.000 g/mol and a degree of polymerization of 3000.01 g/mol) + (8)(1.12.11)(1 − x) since fib + fip = 1. and mip = (5)(12. let x = fib. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. from Equation 14.10x + (68.11(1 − x)] g/mol Solving for x leads to x = fib = f(isobutylene) = 0. such that m = 56.008 g/mol) = 68.6 DP = Mn m Or 3000 = 200. From Table 14. Thus.01 g/mol) + (8)(1. Also. from Table 14. Also. 000 g/mol [56. the isoprene repeat unit has five carbon and eight hydrogen atoms. Compute the fraction of isobutylene and isoprene repeat units in this copolymer.008 g/mol) = 56.7 m = f ib mib + f ip mip Now. Solution For a random poly(isobutylene-isoprene) copolymer in which M n = 200. Polymer Crystallinity 14. . Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Solution The tendency of a polymer to crystallize decreases with increasing molecular weight because as the chains become longer it is more difficult for all regions along adjacent chains to align so as to produce the ordered atomic array.22 Explain briefly why the tendency of a polymer to crystallize decreases with increasing molecular weight. (b) No. linear and isotactic polystyrene. The alternating poly(styrene-ethylene) copolymer is more likely to crystallize. (c) Yes.23 For each of the following pairs of polymers. (a) Linear and syndiotactic poly(vinyl chloride). the phenyl side-group for polystyrene is bulkier than the Cl sidegroup for poly(vinyl chloride). (d) Alternating poly(styrene-ethylene) copolymer. linear and heavily crosslinked cis-isoprene. The repeat unit structure for polypropylene is chemically more complicated than is the repeat unit structure for polyethylene. random poly(vinyl chloridetetrafluoroethylene) copolymer. for these two polymers it is possible to decide. lightly branched isotactic polypropylene. Solution (a) Yes. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. (c) Linear polyethylene. . then state why. (b) Network phenol-formaldehyde. Furthermore. (2) if it is possible. do the following: (1) state whether it is possible to determine whether one polymer is more likely to crystallize than the other. The linear polyethylene is more likely to crystallize. and (3) if it is not possible to decide. note which is the more likely and then cite reason(s) for your choice. it is possible to decide for these two polymers. The linear and syndiotactic poly(vinyl chloride) is more likely to crystallize. Alternating copolymers crystallize more easily than do random copolymers. Syndiotactic and isotactic isomers are equally likely to crystallize. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. it is not possible to decide for these two polymers.14. branched structures are less likely to crystallize than are linear structures. Both heavily crosslinked and network polymers are not likely to crystallize. it is possible to decide for these two copolymers. (d) Yes. 078 nm)(0. Before this can be carried out we must first calculate VC.008 g/mol) = 42.5 leads to n = ρVC NA A Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.01 g/mol) + 6(1.3.62°) = 0. Also. an expression for the volume of its unit cell. in which we solve for n.24 The density of totally crystalline polypropylene at room temperature is 0. Solution For this problem we are given the density of polypropylene (0.62° c = 0. and A the repeat unit molecular weight. and the lattice parameters.946 g/cm3).666 nm α = 90° b = 2.8869 nm3 = 8.14.650 nm γ = 90° If the volume of a monoclinic unit cell.650 nm) sin (99. and are asked to determine the number of repeat units per unit cell. solving for n from Equation 3. This computation necessitates the use of Equation 3. . at room temperature the unit cell for this material is monoclinic with lattice parameters a = 0. the unit cell volume. from which the value of A may be determined as follows: A = 3(AC) + 6(AH) = 3(12.078 nm β = 99.5. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.869 × 10−22 cm3 The repeat unit for polypropylene is shown in Table 14. is a function of these lattice parameters as Vmono = abc sin β determine the number of repeat units per unit cell.08 g/mol Finally. For VC VC = abc sin β = (0. Vmono.666 nm)(2.946 g/cm3. 08 g/mol = 12.0 repeat unit/unit cell Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.946 g/cm3 )(8.869 × 10 −22 cm 3 /unit cell)(6.022 × 1023 repeat units/mol) 42. .= (0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Solving the above two equations for ρa and ρc leads to ρa = = ρs1 ρs 2 (C1 − C2 ) C1 ρs1 − C2 ρs 2 (2.8 let C = % crystallinity . (b) Determine the percent crystallinity of a specimen having a density of 2.215 g/cm3) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.144 g/cm3)( 2.215 g/cm3)(0. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.144 51.513 − 0.000 g/cm3 (0.215 74.3 2.14.144 g/cm3.742) ( 2.144 g/cm 3) − (0.742) = 2. Since two values of ρs and C are specified in the problem statement. 100 such that C = ρc ( ρ s − ρ a ) ρ s ( ρc − ρ a ) Rearrangement of this expression leads to ρc (C ρs − ρs ) + ρc ρa − C ρs ρa = 0 in which ρc and ρa are the variables for which solutions are to be found.513) (2. . Solution (a) We are asked to compute the densities of totally crystalline and totally amorphous polytetrafluoroethylene (ρc and ρa from Equation 14. From Equation 14.25 The density and associated percent crystallinity for two polytetrafluoroethylene materials are as follows: ρ (g/cm3) crystallinity (%) 2.8). ρs2 = 2.215 g/cm3. two equations may be constructed as follows: ρc (C1 ρs1 − ρs1 ) + ρc ρa − C1 ρs1 ρa = 0 ρc (C2 ρs 2 − ρs 2 ) + ρc ρa − C2 ρs 2 ρa = 0 In which ρs1 = 2.26 g/cm3.513.742. C1 = 0. and C2 = 0.2 (a) Compute the densities of totally crystalline and totally amorphous polytetrafluoroethylene. 260 g/cm3)( 2. using Equation 14.513) = 2.742 − 0.215 g/cm3 )(0.301 g/cm3)( 2.215 g/cm3 )(0.000 g/cm3) = 87.513 − 1) (b) Now we are to determine the % crystallinity for ρs = 2.144 g/cm3 )( 2. .742 − 1) − (2.26 g/cm3.260 g/cm3 (2.And ρc = = ρs1 ρs 2 (C2 − C1 ) ρs 2 (C2 − 1) − ρs1 (C1 − 1) (2.9% Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.301 g/cm3 × 100 − 2.301 g/cm3 (2.144 g/cm3 )(0.8 % crystallinity = = ρc ( ρ s − ρ a ) ρ s ( ρc − ρ a ) (2.000 g/cm3 ) × 100 − 2. Again. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 152 g/cm3) (0.188 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Solution (a) We are asked to compute the densities of totally crystalline and totally amorphous nylon 6.6 materials are as follows: ρ (g/cm3) crystallinity (%) 1.188 67.188 g/cm3) − (0.152 g/cm3 ) (1. such that 100 C = ρc ( ρ s − ρ a ) ρ s ( ρc − ρ a ) Rearrangement of this expression leads to ρc (C ρs − ρs ) + ρc ρa − C ρs ρa = 0 in which ρc and ρa are the variables for which solutions are to be found. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. C1 = 0.673) (1.091 g/cm3 (0.188 g/cm3.673.8). and C2 = 0.7 (a) Compute the densities of totally crystalline and totally amorphous nylon 6.8 let C = . (b) Determine the density of a specimen having 55.3 1. .26 The density and associated percent crystallinity for two nylon 6. From Equation 14.14.437. Solving the above two equations for ρa and ρc leads to ρa = = ρs1 ρs 2 (C1 − C2 ) C1 ρs1 − C2 ρs 2 g/cm3)(1. two equations may be constructed as follows: ρc (C1 ρs1 − ρs1 ) + ρc ρa − C1ρs1ρa = 0 ρc (C2 ρs 2 − ρs 2 ) + ρc ρa − C2 ρs 2 ρa = 0 In which ρs1 = 1.437) (1.437) = 1.152 43.4% crystallinity.152 g/cm3.673− 0. ρs2 = 1.6 (ρc % crystallinity and ρa from Equation 14. Since two values of ρs and C are specified in the problem.6. 8 and substitution for ρa and ρc which were computed in part (a) yields ρs = = − ρc ρ a C ( ρc − ρ a ) − ρc − (1.170 g/cm3 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Solving for ρs from Equation 14.554) (1.091 g/cm3 ) (0.673 −1) (1.152 g/cm3) (0. .242 g/cm3 = 1.152 g/cm3) (0.242 g/cm3 − 1.242 g/cm3)(1.091 g/cm3 ) − 1.And ρc = = ρs1 ρs 2 (C2 − C1 ) ρs 2 (C2 − 1) − ρs1 (C1 − 1) g/cm3)(1.188 g/cm3) (0.673) = 1.242 g/cm3 (1.437 − 0.437 −1) − (1. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.188 (b) Now we are asked to determine the density of a specimen having 55.4% crystallinity. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Thus.9 J = PM P −P ∆P = PM 2 1 ∆x ∆x and taking P1 = 1 kPa (1.000 Pa) and P2 = 10 kPa (10.000 Pa) we get ⎡ (cm3 STP)(cm) ⎤ ⎛ 10. The permeability coefficient of H2O through PP is given in Table 14.9.Diffusion in Polymeric Materials 14. The pressures of H2O at the two faces are 1 kPa and 10 kPa.6 as 38 × 10−13 (cm3 STP) cm / cm 2 ⋅ s ⋅Pa.2 cm cm 2 ⋅ s ⋅ Pa ⎦ ⎝ ⎣ = 1. which are maintained constant. 000 Pa − 1.000 Pa ⎞ = ⎢38 × 10−13 ⎟⎠ ⎥ ⎜ 0. from Equation 14. Assuming conditions of steady state. .27 Consider the diffusion of water vapor through a polypropylene (PP) sheet 2 mm thick. In order to solve this problem it is necessary to employ Equation 14. what is the diffusion flux [in (cm3 STP) / cm 2 ⋅ s ] at 298 K? Solution This is a permeability problem in which we are asked to compute the diffusion flux of water vapor through a 2-mm thick sheet of polypropylene.71 × 10−7 (cm3 STP) cm 2 ⋅ s Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. 000 Pa) = 3.000.2 × 10−13 (cm3 STP)(cm) cm 2 ⋅ s ⋅ Pa Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.000 Pa).28 Argon diffuses through a high density polyethylene (HDPE) sheet 40 mm thick at a rate of 4.9 in order to solve this problem.0 10 (4 cm) × ⎢ cm 2 ⋅ s ⎥⎦ ⎣ PM = (7.14. The pressures of argon at the two faces are 7000 kPa. what is the permeability coefficient at 330 K? Solution This problem asks us to compute the permeability coefficient for argon through high density polyethylene at 330 K given a steady-state permeability situation.000.000. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Assuming conditions of steady state. . which are maintained constant. Rearranging this expression and solving for the permeability coefficient gives PM = J ∆x J ∆x = ∆P P2 − P1 Taking P1 = 2000 kPa (2.000 Pa) and P2 = 7000 kPa (7. and 2000 kPa. 000 Pa − 2.0 × 10–7 (cm3 STP) / cm 2 ⋅ s at 330 K. It is necessary for us to use Equation 14. 000. the permeability coefficient of Ar through HDPE is equal to 3 ⎡ −7 (cm STP) ⎤ 4. Compute the diffusion flux [in (cm3 STP) / cm 2 ⋅ s ] at 360 K.000 Pa ⎞ ⎟⎠ 3. assume a condition of steady state diffusion Solution This problem asks that we compute the diffusion flux at 360 K for hydrogen in poly(dimethyl siloxane) (PDMSO). For this diffusion system PM 0 = 1. .31J/mol ⋅ K)(360 K) ⎦ ⎣ = 1.9. Consider the diffusion of hydrogen through a poly(dimethyl siloxane) (PDMSO) sheet 30 mm thick. which are maintained constant. It is first necessary to compute the value of the permeability coefficient at 360 K.47 × 10−7 (cm3 STP) cm 2 ⋅ s Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted.49 × 10−10 (cm3 STP)(cm) cm 2 ⋅ s ⋅ Pa And. we get ⎡ ⎡ ⎤ (cm3 STP)(cm) ⎤ 13.45 × 10 −8 (cm 3 STP)(cm) / cm 2 ⋅ s ⋅ Pa Qp = 13. the diffusion flux is equal to J = PM = 1. as follows: ⎛ Q ⎞ p ⎟ PM = PM exp ⎜⎜− ⎟ 0 RT ⎠ ⎝ And.7 kJ/mol Also. 700 J/mol PM = ⎢1.45 × 10 −8 ⎥ exp ⎢ − ⎥ 2 cm ⋅ s ⋅ Pa ⎦ ⎣ (8.29 The permeability coefficient of a type of small gas molecule in a polymer is dependent on absolute temperature according to the following equation: ⎛ Qp ⎞ PM = PM 0 exp ⎜− ⎟ ⎝ RT ⎠ where PM 0 and Qp are constants for a given gas–polymer pair. 000 Pa − 1.0 cm cm 2 ⋅ s ⋅ Pa ⎜⎝ = 4. incorporating values provided for the constants PM 0 and Qp.14. The hydrogen pressures at the two faces are 10 kPa and 1 kPa.49 × 10-10 P −P ∆P = PM 2 1 ∆x ∆x (cm3 STP)(cm) ⎛ 10. using Equation 14. The temperature dependence of PM is given in the problem statement. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.