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CHAPTER 4
POLYMER STRUCTURES
PROBLEM SOLUTIONS
Hydrocarbon Molecules Polymer Molecules The Chemistry of Polymer Molecules
4.1 The mer structures called for are sketched below. (a) Polychlorotrifluoroethylene
(b) Poly(vinyl alcohol)
Molecular Weight
4.2 Mer weights for several polymers are asked for in this problem.
(a) For polytetrafluoroethylene, each mer unit consists of two carbons and four fluorines (Table 4.3). If AC and AF represent the atomic weights of carbon and fluorine, respectively, then m = 2(AC) + 4(AF)
= (2)(12.01 g/mol) + (4)(19.00 g/mol) = 100.02 g/mol
(b) For poly(methyl methacrylate), from Table 4.3, each mer 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
(c) For nylon 6,6, from Table 4.3, each mer unit has twelve carbons, twenty-two hydrogens, two nitrogens, and two oxygens. Thus, m = 12(AC) + 22(AH) + 2(AN) + 2(AO)
= (12)(12.01 g/mol) + (22)(1.008 g/mol) + (2)(14.01 g/mol) + (2)(16.00 g/mol)
= 226.32 g/mol
(d) For poly(ethylene terephthalate), from Table 4.3, each mer 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
4.3 We are asked to compute the number-average degree of polymerization for polypropylene, given that the number-average molecular weight is 1,000,000 g/mol. The mer 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
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If we let nn represent the number-average degree of polymerization, then from Equation 4.4a
Mn m 106 g/mol = 23, 700 42.08 g/mol
nn =
=
4.4 (a) The mer molecular weight of polystyrene is called for in this portion of the problem. For polystyrene, from Table 4.3, each mer 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 weight-average molecular weight. Since the weight-average degree of polymerization, nw, is 25,000, using Equation 4.4b
M w = nw m = (25, 000)(104.14 g/mol) = 2.60 x 106 g/mol 4.5 (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 8,000-16,000 16,000-24,000 24,000-32,000 32,000-40,000 40,000-48,000 48,000-56,000
Mean Mi 12,000 20,000 28,000 36,000 44,000 52,000
xi
0.05 0.16 0.24 0.28 0.20 0.07
xiMi
600 3200 6720 10,080 8800 3640
____________________________ Mn = xi M i = 33, 040 g/mol
(b) From the tabulated data, we are asked to compute M w , the weight-average molecular weight.
Molecular wt.
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Range 8,000-16,000 16,000-24,000 24,000-32,000 32,000-40,000 40,000-48,000 48,000-56,000
Mean Mi 12,000 20,000 28,000 36,000 44,000 52,000
wi
0.02 0.10 0.20 0.30 0.27 0.11
wiMi
240 2000 5600 10,800 11,880 5720
___________________________ Mw = wi M i = 36, 240 g/mol
(c) Now we are asked to compute nn (the number-average degree of polymerization), using Equation 4.4a. For polypropylene,
m = 3(AC) + 6(AH)
= (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol
And
nn =
Mn m
=
33, 040 g/mol = 785 42.08 g/mol
(d) And, finally, we are asked to compute nw, the weight-average degree of polymerization, as
nw =
Mw m
=
36, 240 g/mol = 861 42.08 g/mol
4.6 (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 8,000-20,000 20,000-32,000 32,000-44,000 44,000-56,000 56,000-68,000
Mean Mi 14,000 26,000 38,000 50,000 62,000
xi
0.05 0.15 0.21 0.28 0.18
xi M i
700 3900 7980 14,000 11,160
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68,000-80,000 80,000-92,000
74,000 86,000
0.10 0.03
7400 2580 _________________________
M n = x i M i = 47, 720 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 8,000-20,000 20,000-32,000 32,000-44,000 44,000-56,000 56,000-68,000 68,000-80,000 80,000-92,000
Mean Mi 14,000 26,000 38,000 50,000 62,000 74,000 86,000
wi
0.02 0.08 0.17 0.29 0.23 0.16 0.05
wiMi
280 2080 6460 14,500 14,260 11,840 4300 _________________________
M w = w i M i = 53, 720 g/mol
(c) We are now asked if the number-average degree of polymerization is 477, which of the polymers in Table 4.3 is this material? It is necessary to compute m in Equation 4.4a as
m =
Mn nn
=
47, 720 g/mol = 100.04 g/mol 477
The mer molecular weights of the polymers listed in Table 4.3 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
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PET--192.16 g/mol Polycarbonate--254.27 g/mol Therefore, polytetrafluoroethylene is the material since its mer molecular weight is closest to that calculated above. (d) The weight-average degree of polymerization may be calculated using Equation 4.4b, since M w and m were computed in portions (b) and (c) of this problem. Thus Mw m 53, 720 g/mol = 537 100.04 g/mol
nw =
=
4.7 This problem asks if it is possible to have a poly(vinyl chloride) homopolymer with the given molecular weight data and a number-average degree of polymerization of 1120. The appropriate data are given below along with a computation of the number-average molecular weight.
Molecular wt. Range 8,000-20,000 20,000-32,000 32,000-44,000 44,000-56,000 56,000-68,000 68,000-80,000 80,000-92,000
Mean Mi 14,000 26,000 38,000 50,000 62,000 74,000 86,000
xi 0.05 0.15 0.21 0.28 0.18 0.10 0.03
xi M i 700 3900 7980 14,000 11,160 7440 2580 _________________________
Mw =
xi M i
= 47, 720 g/mol
For PVC, from Table 4.3, each mer unit has two carbons, three hydrogens, and one chlorine. Thus, m = 2(AC) + 3(AH) + (ACl)
= (2)(12.01 g/mol) + (3)(1.008 g/mol) + (35.45 g/mol) = 62.49 g/mol Now, we will compute nn using Equation 4.4a as
nn =
Mn 47, 720 g/mol = = 764 62.49 g/mol m
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Thus, such a homopolymer is not possible since the calculated nn is 764 not 1120.
4.8 (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 carbon atoms; there are 100 possible side-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 just
CCl =
177.25 g x 100 = 20.3 wt% 600.5 g + 95.76 g + 177.25 g
(b) Chlorinated polyethylene differs from poly(vinyl chloride), in that, for PVC, (1) 25% of the sidebonding sites are substituted with Cl, and (2) the substitution is probably much less random.
Molecular Shape
4.9 This problem first of all asks for us to calculate, using Equation 4.7, the average total chain length, L, for a linear polyethylene polymer having a number-average molecular weight of 300,000 g/mol. It is necessary to calculate the number-average degree of polymerization, nn, using Equation 4.4a. For polyethylene, from Table 4.3, each mer unit has two carbons and four hydrogens. Thus, m = 2(AC) + 4(AH)
= (2)(12.01 g/mol) + (4)(1.008 g/mol) = 28.05 g/mol
and
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nn =
Mn m
=
300, 000 g/mol = 10, 695 28.05 g/mol
which is the number of mer units along an average chain. Since there are two carbon atoms per mer unit, there are two C--C chain bonds per mer, which means that the total number of chain bonds in the molecule, N, is just (2)(10,695) = 21,390 bonds. Furthermore, assume that for single carbon-carbon bonds, d = 0.154 nm and = 109 (Section 4.4); therefore, from Equation 4.7
L = Nd sin 2
109 = (21, 390)(0.154 nm) sin = 2682 nm 2
It is now possible to calculate the average chain end-to-end distance, r, using Equation 4.8 as r = d N = (0.154 nm) 21, 390 = 22.5 nm
4.10 (a) This portion of the problem asks for us to calculate the number-average molecular weight for a linear polytetrafluoroethylene for which L in Equation 4.7 is 2000 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 = 109. Thus L
d sin 2
N =
=
2000 nm = 15, 900 109 (0.154 nm) sin 2
Since there are two C--C bonds per PTFE mer unit, there is an average of N/2 or 15,900/2 = 7950 mer units per chain, which is also the number-average degree of polymerization, nn. In order to compute the value of M n using Equation 4.4a, we must first determine m for PTFE. Each PTFE mer unit consists of two carbon and four fluorine atoms, thus
m = 2(AC) + 4(AF)
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= (2)(12.01 g/mol) + (4)(19.00 g/mol) = 100.02 g/mol
Therefore
M n = nn m = (7950)(100.02 g/mol) = 795, 000 g/mol
(b) Next, we are to determine the number-average molecular weight for r = 15 nm. Solving for N from Equation 4.8 leads to (15 nm) 2 = = 9490 d2 (0.154 nm) 2
r2
N =
which is the total number of bonds per average molecule. Since there are two C--C bonds per mer unit, then nn =
N/2 = 9490/2 = 4745. Now, from Equation 4.4a M n = nn m = (4745)(100.02 g/mol) = 474, 600 g/mol
Molecular Configurations
4.11 We are asked to sketch portions of a linear polypropylene molecule for different configurations (using two-dimensional schematic sketches). (a) Syndiotactic polypropylene
(b) Atactic polypropylene
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(c) Isotactic polypropylene
4.12 This problem asks for us to sketch cis and trans structures for butadiene and chloroprene. (a) The structure for cis polybutadiene is
The structure of trans butadiene is
(b) The structure of cis chloroprene is
The structure of trans chloroprene is
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Thermoplastic and Thermosetting Polymers
4.13 This question asks for comparisons of thermoplastic and thermosetting polymers. (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.
4.14 (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
4.15 This problem asks for sketches of the mer structures for several alternating copolymers. (a) For poly(ethylene-propylene)
(b) For poly(butadiene-styrene)
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(c) For poly(isobutylene-isoprene)
4.16 For a poly(styrene-butadiene) alternating copolymer with a number-average molecular weight of 1,350,000 g/mol, we are asked to determine the average number of styrene and butadiene mer units per molecule. Since it is an alternating copolymer, the number of both types of mer units will the be same. Therefore, consider them as a single mer unit, and determine the number-average degree of polymerization. For the styrene mer, there are eight carbon atoms and eight hydrogen atoms, while the butadiene mer consists of four carbon atoms and six hydrogen atoms. Therefore, the styrene-butadiene combined mer weight is just
m = 12(AC) + 14(AH)
= (12)(12.01 g/mol) + (14)(1.008 g/mol) = 158.23 g/mol
From Equation 4.4a, the number-average degree of polymerization is just
Mn m
nn =
=
135, 000 g/mol = 8530 158.23 g/mol
Thus, there is an average of 8530 of both mer types per molecule.
4.17 This problem asks for us to calculate the number-average molecular weight of a random nitrile [poly(acrylonitrile-butadiene) copolymer]. For the acrylonitrile mer there are three carbon, one nitrogen, and three hydrogen atoms. Thus, its mer molecular weight is
mAc = 3(AC) + (AN) + 3(AH)
= (3)(12.01 g/mol) + 14.01 g/mol + (3)(1.008 g/mol) = 53.06 g/mol
The butadiene mer is composed of four carbon and six hydrogen atoms. Thus, its mer molecular weight is
mBu = 4(AC) + 6(AH)
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= (4)(12.01 g/mol) + (6)(1.008 g/mol) = 54.09 g/mol
From Equation 4.5, the average mer molecular weight is just
m = fAc mAc + fBu mBu
= (0.70)(53.06 g/mol) + (0.30)(54.09 g/mol) = 53.37 g/mol Since nn = 2000 (as stated in the problem), M n may be computed using Equation 4.4a as
M n = m nn = (53.37 g/mol)(2000) = 106, 740 g/mol
4.18 For an alternating copolymer that has a number-average molecular weight of 100,000 g/mol and a number-average degree of polymerization of 2210, we are to determine one of the mer types if the other type is ethylene. It is first necessary to calculate m using Equation 4.4a as
Mn nn
m =
=
100, 000 g/mol = 42.25 g/mol 2210
Since this is an alternating copolymer we know that chain fraction of each mer type is 0.5; that is fe = fx = 0.5, fe and fx being, respectively, the chain fractions of the ethylene and unknown mers. Also, the mer molecular weight for ethylene is
ms = 2(AC) + 4(AH)
= 2(12.01 g/mol) + 4(1.008 g/mol) = 28.05 g/mol Now, using Equation 4.5, it is possible to calculate the mer weight of the unknown mer type, mx. Thus
m - feme fx
mx =
=
45.25 g/mol - (0.5)(28.05 g/mol) = 62.45 g/mol 0.5
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Finally, it is necessary to calculate the mer molecular weights for each of the possible other mer types. These are calculated below:
mstyrene = 8(AC) + 8(AH) = 8(12.01 g/mol) + 8(1.008 g/mol) = 104.16 g/mol mpropylene = 3(AC) + 6(AH) = 3(12.01 g/mol) + 6(1.008 g/mol) = 42.08 g/mol mTFE = 2(AC) + 4(AF) = 2(12.01 g/mol) + 4(19.00 g/mol) = 100.02 g/mol mVC = 2(AC) + 3(AH) + (ACl) = 2(12.01 g/mol) + 3(1.008 g/mol) + 35.45 g/mol = 62.49 g/mol
Therefore, vinyl chloride is the other mer type since its m value is almost the same as the calculated mx.
4.19 (a) This portion of the problem asks us to determine the ratio of butadiene to acrylonitrile mers in a copolymer having a weight-average molecular weight of 250,000 g/mol and a weight-average degree of polymerization of 4640. It first becomes necessary to calculate the average mer molecular weight of the copolymer,
m , using Equation 4.4b as Mw nw
250, 000 g/mol = 53.88 g/mol 4640
m =
=
If we designate fb as the chain fraction of butadiene mers, since the copolymer consists of only two mer types, the chain fraction of acrylontrile mers fa is just 1 fb. Now, Equation 4.5 for this copolymer may be written in the form
m = f b mb + fa ma = f b mb + (1 - f b ) ma
in which mb and ma are the mer molecular weights for butadiene and acrylontrile, respectively. These values are calculated as follows:
mb = 4(AC) + 6(AH) = 4(12.01 g/mol) + 6(1.008 g/mol) = 54.09 g/mol ma = 3(AC) + 3(AH) + (AN) = 3(12.01 g/mol) + 3(1.008 g/mol) + (14.01 g/mol)
= 53.06 g/mol. Solving for fb in the above expression yields
fb =
m - ma
b
m - ma
=
53.88 g/mol - 53.06 g/mol = 0.80 54.09 g/mol - 53.06 g/mol
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Furthermore, fa = 1 fb = 1 0.80 = 0.20; or the ratio is just
fb fa
=
0.80 = 4.0 0.20
(b) Of the possible copolymers, the only one for which there is a restriction on the ratio of mer types is alternating; the ratio must be 1:1. Therefore, on the basis of the result in part (a), the possibilities for this copolymer are random, graft, and block.
4.20 For a copolymer consisting of 60 wt% ethylene and 40 wt% propylene, we are asked to determine the fraction of both mer types. In 100 g of this material, there are 60 g of ethylene and 40 g of propylene. The ethylene (C2H4) molecular weight is
m(ethylene) = 2(AC) + 4(AH)
= (2)(12.01 g/mol) + (4)(1.008 g/mol) = 28.05 g/mol The propylene (C3H6) molecular weight is
m(propylene) = 3(AC) + 6(AH)
= (3)(12.01 g/mol) + (6)(1.008 g/mol) = 42.08 g/mol
Therefore, in 100 g of this material, there are 60 g = 2.14 mol of ethylene 28.05 g / mol
and 40 g = 0.95 mol of propylene 42.08 g / mol
Thus, the fraction of the ethylene mer, f(ethylene), is just 2.14 mol = 0.69 2.14 mol + 0.95 mol
f (ethylene) =
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Likewise, 0.95 mol = 0.31 2.14 mol + 0.95 mol
f (propylene) =
4.21 For a random poly(isobutylene-isoprene) copolymer in which M w = 200,000 g/mol and nw = 3000, we are asked to compute the fractions of isobutylene and isoprene mers. From Table 4.5, the isobutylene mer has four carbon and eight hydrogen atoms. Thus,
mib = (4)(12.01 g/mol) + (8)(1.008 g/mol) = 56.10 g/mol
Also, from Table 4.5, the isoprene mer has five carbon and eight hydrogen atoms, and
mip = (5)(12.01 g/mol) + (8)(1.008 g/mol) = 68.11 g/mol
From Equation 4.5
m = fibmib + fipmip
Now, let x = fib, such that
m = 56.10x + (68.11)(1 - x) since fib + fip = 1. Also, from Equation 4.4b Mw m
nw = Or
3000 =
200, 000 g / mol [56.10x + 68.11(1 - x)] g / mol
Solving for x leads to x = fib = f(isobutylene) = 0.12. Also,
f(isoprene) = 1 x = 1 0.12 = 0.88
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Polymer Crystallinity
4.22 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.
4.23 For each of four pairs of polymers, we are asked to (1) state whether it is possible to decide which is more likely to crystallize; (2) if so, which is more likely and why; and (3) it is not possible to decide then why. (a) No, it is not possible to decide for these two polymers. On the basis of tacticity, the isotactic PP is more likely to crystallize than the atactic PVC. On the other hand, with regard to side-group bulkiness, the PVC is more likely to crystallize. (b) Yes, it is possible to decide for these two copolymers. The linear and syndiotactic polypropylene is more likely to crystallize than crosslinked cis-isoprene since linear polymers are more likely to crystallize than crosslinked ones. (c) Yes, it is possible to decide for these two polymers. The linear and isotactic polystyrene is more likely to crystallize than network phenol-formaldehyde; network polymers rarely crystallize, whereas isotactic ones crystallize relatively easily. (d) Yes, it is possible to decide for these two copolymers. The block poly(acrylonitrile-isoprene)
copolymer is more likely to crystallize than the graft poly(chloroprene-isobutylene) copolymer. Block copolymers crystallize more easily than graft ones.
4.24 Given that polyethylene has an orthorhombic unit cell with two equivalent mer units, we are asked to compute the density of totally crystalline polyethylene. In order to solve this problem it is necessary to employ Equation 3.5, in which n represents the number of mer units within the unit cell (n = 2), and A is the mer molecular weight, which for polyethylene is A = 2(AC) + 4(AH)
= (2)(12.01 g/mol) + (4)(1.008 g/mol) = 28.05 g/mol Also, VC is the unit cell volume, which is just the product of the three unit cell edge lengths in Figure 4.10. Thus, nA V N
C
=
A
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=
(2 mers/uc)(28.05 g/mol)
( 7.41 x 10 -8 cm)( 4.94 x 10 -8 cm)( 2.55 x 10-8 cm)/(unit cell)( 6.023 x 10 23 mers/mol)
= 0.998 g/cm3
4.25 For this problem we are given the density of nylon 6,6 (1.213 g/cm3), an expression for the volume of its unit cell, and the lattice parameters, and are asked to determine the number of mer units per unit cell. This computation necessitates the use of Equation 3.5, in which we solve for n. Before this can be carried out we must first calculate VC, the unit cell volume, and A the mer molecular weight. For VC
V = abc 1 - cos2 - cos2 - cos 2 + 2 cos cos cos C
= (0.497)(0.547)(1.729) 1 - 0.441 - 0.054 - 0.213 + 2 (0.664)(0.232)(0.462)
= 0.3098 nm3 = 3.098 x 10-22 cm3
The mer unit for nylon 6,6 is shown in Table 4.3, from which the value of A may be determined as follows: A = 12(AC) + 22(AH) + 2(AO) + 2(AN)
= 12(12.01 g/mol) + 22(1.008 g/mol) + 2(16.00 g/mol) + 2(14.01 g/mol)
= 226.32 g/mol
Finally, solving for n from Equation 3.5 leads to
VC N A A
n =
=
(1.213 g/cm 3 )( 3.098 x 10 -22
cm 3 /unit cell)( 6.023 x 10 23 mers/mol)
226.32 g/mol
= 1 mer/unit cell
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4.26 (a) We are asked to compute the densities of totally crystalline and totally amorphous polyethylene % crystallinity , such that (c and a from Equation 4.6). From Equation 4.6 let C = 100
c ( s - a ) s ( c - a )
C =
Rearrangement of this expression leads to
c (C s - s ) + c a - C s a = 0
in which c and a are the variables for which solutions are to be found. Since two values of s and C are specified in the problem statement, two equations may be constructed as follows:
c (C1 s1 - s1 ) + c a - C1 s1 a = 0 c (C 2 s2 - s2 ) + c a - C 2 s2 a = 0
In which s1 = 0.965 g/cm3, s2 = 0.925 g/cm3, C1 = 0.768, and C2 = 0.464. Solving the above two equations for
a and c leads to s1 s2 (C1 - C 2 ) C - C
1 s1 2 s2
a =
=
(0.965 g/cm 3 )( 0.925 g/cm 3 )(0.768 - 0.464) = 0.870 g/cm 3 (0.768) ( 0.965 g/cm 3 ) - (0.464) ( 0.925 g/cm 3 )
And
c =
s1 s2 (C 2 - C1 ) s2 (C2 - 1 ) - s1 (C1 - 1)
=
( 0.965 g/cm 3 )( 0.925 g/cm 3 )( 0.464 - 0.768) = 0.998 g/cm 3 ( 0.925 g/cm 3 ) (0.464 - 1.0) - (0.965 g/cm 3 )(0.768 - 1.0)
(b) Now we are to determine the % crystallinity for s = 0.950 g/cm3. Again, using Equation 4.6
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% crystallinity =
c ( s - a ) x 100 s ( c - a ) - 0.870 g/cm 3 ) - 0.870 g/cm 3 )
=
( 0.998 g/cm 3 )( 0.950 g/cm 3 ( 0.950 g/cm 3 )( 0.998 g/cm 3
= 65.7%
x 100
4.27 (a) We are asked to compute the densities of totally crystalline and totally amorphous polypropylene % crystallinity , such that (c and a from Equation 4.6). From Equation 4.6 let C = 100 c ( s - a ) s ( c - a )
C =
Rearrangement of this expression leads to c (C s - s ) + c a - C s a = 0 in which c and a are the variables for which solutions are to be found. Since two values of s and C are specified in the problem, two equations may be constructed as follows: c (C1 s1 - s1 ) + c a - C1 s1 a = 0 c (C 2 s 2 - s2 ) + c a - C 2 s2 a = 0 In which s1 = 0.904 g/cm3, s2 = 0.895 g/cm3, C1 = 0.628, and C2 = 0.544. Solving the above two equations for a and c leads to
s1 s2 (C1 - C 2 ) C1 s1 - C 2 s2
a =
=
( 0.904
g / cm 3 )( 0.895 g / cm 3 ) (0.628 - 0.544)
(0.628) (0.904 g / cm 3 ) - (0.544) (0.895 g / cm 3 )
= 0.841 g/cm 3
And
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c =
s1 s2 (C 2 - C1 )
s2
(C2
- 1) -
s1
(C1
- 1)
=
(0.904 ( 0.895
g / cm 3 )( 0.895 g / cm 3 ) (0.544 - 0.628)
g / cm 3 ) ( 0.544 - 1.0) - ( 0.904 g / cm 3 ) (0.628 - 1.0)
= 0.946 g/cm 3
(b) Now we are asked to determine the density of a specimen having 74.6% crystallinity. Solving for s from Equation 4.6 and substitution for a and c which were computed in part (a) yields - c a C ( c - a ) - c
s =
=
- ( 0.946 g / cm 3 )( 0.841 g / cm 3 )
( 0.746) (0.946 g / cm 3 - 0.841 g / cm 3 ) - 0.946 g / cm 3
= 0.917 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. 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.