Materials &amp

Technology 112Question 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 BakeliteA 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 polyethyleneHigh 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 PTFEBranching 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)14,0000.0570026,0000.153,90038,0000.217,98050,0000.2814,00062,0000.1811,16074,0000.107,40086,0000.032,580Mn 47,720Mn = ∑ XiMi= 47,720 g/molWeight-Average Molecular WeightMass per molecule, Mi (g/mol)Weight fraction, Wi WiMi (g/mol)14,0000.0228026,0000.082,08038,0000.176,46050,0000.2914,50062,0000.2314,26074,0000.1611,84086,0000.054,300Mw53,720Mw = ∑ WiMi= 53,720 g/molnn = number-average degree of polymerizationm = molecular weight of a mer unitnn = Mn / mnn = 477 Mn = 47, 720 g/molnn = 47, 720 g/mol 477nn = 100.042 g/molThis 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 (%)0.90462.80.89554.4(a) Compute the densities of totally crystalline and totally amorphous polypropylene.Crystallinity = C ρ = densityC = % crystallinity / 100= ρc (ρs – ρa) ρs (ρc – ρa) by rearrangement:ρc ( C ρs- ρs) + ρc ρa – C ρs ρa = 0where:ρc and ρa are unknownSince ρc and C are specified in the problem, two equations can further be derived:ρc ( C1 ρs1- ρs1) + ρc ρa – C1 ρs1 ρa = 0ρc ( C2 ρs2- ρs2) + ρc ρa – C2 ρs2 ρa = 0where: ρs1 = 0.904 g/cm3 ρs2 = 0.895 g/cm3 C1 = 0.628 C2 = 0.544ρa = 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/cm3ρc = ___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 specimenρs = ____- 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/cm3Question 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 cmQ / t = PA (P1-P2) / lQ – quantity of permeant P – permeability A – areat – time P1 – P2 – drop in pressure2.21 x 10-8 cm3/ cm2-s = P (14.8038 atm) 5 cmP = 7. 4643 x 10-9 _cm3-cm_ cm2-s-atmBibliographyPeacock, Andrew J. and Calhoun, Allison. Polymer Chemistry: Properties and Applications. Munich: Hanser, 2006.Robello, Douglas. Formaldehyde and Related Polymers. Synthetic Polymer Chemistry. (accessed August 17, 2010). Sperling, Leslie Howard. Introduction to Physical Polymer Science. Hoboken, NJ: Wiley, 2006.