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Materials & technology 112

Question For each of the following pairs of polymers, decide which is more likely to have the greater tensile strength, and then give reasons foryour choice:
(a) Lightly cross-linked polyethylene; network Bakelite
A network Bakelite has a greater tensile strength than a lightly cross-linked polyethylene. 1 Crosslinking hinder the re-organization of molecules during crystallization; hence, a lightly cross-linked polyethylene has a lower degree of crystallinity as compared to its linear form. A decreased in crystallinity makes crosslinked polymers less rigid and weaker. On the other hand, a network of Bakelite, though has a highly crosslinked structure, often produce hard, impervious, black, and tough solids2.
(b) High density polyethylene; low density polyethylene
High density polyethylene has high levels of crystallization. 3 The crystallization process is less likely to hinder because of low level branching in its structure. Crystallization markedly enhances strength, rigidity, and opacity of a polymer. High density polyethylene, then, is the stiffest of all types of polyethylene due to its high degree of crystallinity. Thereby, high density polyethylene has a greater tensile strength than low density polyethylene.
(c) 95% crystalline and linear PTFE; 95% crystalline and branched PTFE
Branching tend to impede crystallization, making a polymer less rigid, more easily to deform, and weaker4. Thus, a 95% crystalline and linear polytetrafluoroethylene (PTFE) has a greater tensile strength than a 95% crystalline and branched PTFE.
Question 2:
Molecular weight data for some polymer are tabulated below. 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 polymerisation is 477, which one of the polymers listed in Table 14. 3 of the lecture notes is this polymer? Why?
Number-Average Molecular Weight (Mn)
Mass per molecule, Mi (g/mol)
Number fraction, Xi
XiMi (g/mol)
14, 000
0. 05
700
26, 000
0. 15
3, 900
38, 000
0. 21
7, 980
50, 000
0. 28
14, 000
62, 000
0. 18
11, 160
74, 000
0. 10
7, 400
86, 000
0. 03
2, 580
Mn
47, 720
Mn = ∑ XiMi
= 47, 720 g/mol
Weight-Average Molecular Weight
Mass per molecule, Mi (g/mol)
Weight fraction, Wi
WiMi (g/mol)
14, 000
0. 02
280
26, 000
0. 08
2, 080
38, 000
0. 17
6, 460
50, 000
0. 29
14, 500
62, 000
0. 23
14, 260
74, 000
0. 16
11, 840
86, 000
0. 05
4, 300
Mw
53, 720
Mw = ∑ WiMi
= 53, 720 g/mol
nn = number-average degree of polymerization
m = molecular weight of a mer unit
nn = Mn / m
nn = 477 Mn = 47, 720 g/mol
nn = 47, 720 g/mol
477
nn = 100. 042 g/mol
This value is closer to the molecular weight of the mer unit of PTFE/Teflon.
Question 3:
The density and associated % crystallinity for two polypropylene materials are as follows:
 (g/cm3)
Crystallinity (%)
0. 904
62. 8
0. 895
54. 4
(a) Compute the densities of totally crystalline and totally amorphous polypropylene.
Crystallinity = C ρ = density
C = % crystallinity / 100
= ρc (ρs – ρa)
ρs (ρc – ρa)
by rearrangement:
ρc ( C ρs- ρs) + ρc ρa – C ρs ρa = 0
where:
ρc and ρa are unknown
Since ρc and C are specified in the problem, two equations can further be derived:
ρc ( C1 ρs1- ρs1) + ρc ρa – C1 ρs1 ρa = 0
ρc ( C2 ρs2- ρs2) + ρc ρa – C2 ρs2 ρa = 0
where:
ρs1 = 0. 904 g/cm3ρs2 = 0. 895 g/cm3
C1 = 0. 628C2 = 0. 544
ρa = ρs1 ρs2 (C1 – C2)
C1 ρs1 – C2 ρs2
= (0. 904 g/cm3) (0. 895 g/cm3) (0. 628- 0. 544)­­__
(0. 628) (0. 904 g/cm3) – (0. 544) (0. 895 g/cm3)
ρa = 0. 841 g/cm3
ρc = ___ρs1 ρs2 (C2 – C1)___
ρs2 (C2 – 1) – ρs1 (C1-1)
= ­_____(0. 904 g/cm3) (0. 895 g/cm3) (0. 544 – 0. 628)­­_____
0. 895 g/cm3 (0. 544 – 1. 0) – (0. 904 g/cm3) (0. 628 – 1. 0)
ρc = 0. 946 g/cm3
(b) Determine the density of a specimen having 74. 6% crystallinity.
ρs = density of specimen
ρs = ____- ρc ρa______
C (ρc – ρa) – ρc
= __________- (0. 946 g/cm3) (0. 841 g/cm3)________
(0. 746) (0. 946 g/cm3 – 0. 841 g/cm3) – 0. 946 g/cm3
ρs = 0. 917 g/cm3
Question 4:
Carbon dioxide diffuses through a high density polyethylene (HDPE) sheet 50 mm thick at a rate of 2. 2  10-8 (cm3 STP)/cm2-s at 325 K. The pressures of carbon dioxide at the two faces are 4000 kPa and 2500 kPa, which are maintained constant. Assuming conditions of steady state, what is the permeability coefficient at 325 K?
50 mm = 5 cm
Q / t = PA (P1-P2) / l
Q – quantity of permeantP – permeabilityA – area
t – timeP1 – P2 – drop in pressure
2. 21 x 10-8 cm3/ cm2-s = P (14. 8038 atm)
5 cm
P = 7. 4643 x 10-9 _cm3-cm_
cm2-s-atm
Bibliography
Peacock, Andrew J. and Calhoun, Allison. Polymer Chemistry: Properties and Applications. Munich: Hanser, 2006.
Robello, Douglas. Formaldehyde and Related Polymers. Synthetic Polymer Chemistry. http://chem. chem. rochester. edu/~chem424/novolak. htm (accessed August 17, 2010).
Sperling, Leslie Howard. Introduction to Physical Polymer Science. Hoboken, NJ: Wiley, 2006.

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