Unformatted text preview: EMA 3066, Exam 1 January 29, 2009 Name:______________________________
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1. Nylon 66 and PET have roughly the same glass transition temperature, while the
polymer shown below has a much lower glass transition. Explain how the structure of
each of these polymers causes the Tg’s to behave in this way.
O ( O ( CH2 ) O
6 C O ( CH2 ) 4 C ) EMA 3066, Exam 1 January 29, 2009 2. Calculate Mn, Mw, and PDI for the following sample of PVC:
mol% of chains
4000 EMA 3066, Exam 1 January 29, 2009 3. The graph below shows a plot of reduced viscosity versus concentration for a sample
of polystyrene measured in toluene at 25º C. From this data determine the molecular
weight. The Mark-Houwink-Sakurada constants are K = 7.5x10-5 dL/g and a = 0.75 inherent viscosity (dL/g) 2.5 2 1.5 1
0 0.05 0.1 0.15 0.2 concentration (g/dL) 0.25 0.3 0.35 EMA 3066, Exam 1 January 29, 2009 4.
a. Polystyrene with Mv of 350,000 g/mol is placed into cyclohexane at 34.5˚ C. These are
theta conditions for PS. Under these conditions, the Mark-Houwink-Sakurada constants
are K = 8.46x10-4 dL/g and a = 0.50. Calculate:
i. The unperturbed end-to-end distance, assuming a freely jointed chain. Note: The length
of a C-C bond is 1.54 Å.
ii. The real end-to-end distance under these conditions.
iii. The characteristic ratio of PS.
b. Using the Mark-Houwink-Sakurada constants given in problem 3, calculate the chain
expansion factor for PS in toluene at 25˚ C. EMA 3066, Exam 1 January 29, 2009 5. Calculate the solubility parameters for each of the following polymers using the
parameters of Hoy, and identify a good solvent for each. A table with solvent solubility
parameters is provided at the end of the exam.
a. Nylon 6, = 1.15 g/cm3
b. polystyrene, = 1.1 g/cm3 EMA 3066, Exam 1 January 29, 2009 6. Based on the lattice theory of mixing, how do you expect the entropy of mixing of a
polymer and a solvent to vary with molecular weight of the polymer? Explain your
answer. EMA 3066, Exam 1 January 29, 2009
EMA 3066 Exam 1 Equation Sheet Mn NiMi Ni N i M i2 w i M i
Mw NiMi w i
Mn RT lim
M n c 0 c r = / s = t / ts
sp = (-s) / s = r –1
red = sp / c
inh = (ln r) / c lim red lim inh
c 0 c 0 red k ' 2 c
inh k" 2 c KM a
c a a wt %crystallinity c s
x100 s c a /2
∑ = 2.1x1021 dL/mol-cm3
R = 8.31 J/mol-K EMA 3066, Exam 1 January 29, 2009 EMA 3066, Exam 1 January 29, 2009 ...
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- Polymer, Glass, Trigraph, Ema, Nylon, Glass transition