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Unformatted text preview: Chapter 10, Problem 1
A highmolecularweight polyethylene has an average molecular weight of410.000 g."'mol. What is its average
degree of 1:3oly111erization'7J Chapter 10, Solution 1 To assess the average degree of polymerization (DP). the mass of the polyethylene mer is first calculated: (4
hydrogen atoms X 1 gs’mol) + (2 carbon atoms X 12 glmol) = 28 g.."mol.
The DP. defined in units of mers per mol. is: DP _ molecular weight of polymer (g 11101) H H
molecular weight of 111e1‘ (g mol 5" 1ne1‘) _ 410,000 g.:"' 11101  
28 g 1 mole" 111er H H
2 14’ 643 mers Polyethylene Chapter 10, Problem 2 If a type of polyethylene has an average degree of polymerization of 10.000. What is its average molecular weight? Chapter ‘10, Solution 2
MWav (polymer) 2 DP x MWav (1ne1‘) = (10, 000 mers)(28 g"(tnol‘tnerD
= 280, 000 glmol Chapter 10, Problem 6
A copolymet‘ consists of 70 wt 93 polystyrene and 30 wt % polyacrylonitt‘ile. Calculate the mole fraction of each
component in this material. Chapter ‘10, Solution 6
Using a basis of 100 g of copolymer. we have 70 g of polystyrene and 30 g of polyacrylonitl'ile. The con‘esponding
Il‘LLIIlbel‘ of moles of each component is: H H
  H H
ca 3/; C" "  
 (7% C a
H  
n H C E N n
Polystyrene Poly‘ac1ylonitrile M’ofes ofpohts{t'rene MWPolwym = 9 H atoms X 1 g.."mol  S C‘ atoms X 12 gfmol = 105 gr’mol 70 2
No. of moles of polystyrene in 100 g of copolymer = —f = 0.66? 11101
105 gs'mol 1140/95 ofpoh‘acrft'lortm'ile MWPAN = 3 H atoms x 1 gx’mol + 3 C.‘ atoms >\ 12 gt"1nol+ l N atom x 14 g.."'mol
= 53 gfmol
. . . . 30 a
No. of moles of polyacrylonltt‘lle In 100 g of copolymer = —r = 0.566 mol
53 g."mol Thus. the mole fractions of the components are calculated as: 0.66? 11101 —0.541
0.667 1nol+ 0.566 mol Mole fraction of polystyrene — 0.566 mol — 20.459
0.66? 1nol+0.566 11101 Mole fraction of polyacrylonitt‘ile = Chapter 10, Solution 8 From Example Problem 10.3. the molecular weights of PVC and PVA are 62.5. g.‘"mol and 86.0 g/mol. respectively.
The mole fractions of these polymers can be determined based upon the average molecular weight of the copolymer 1116f . T MW’mtmer) = fpvc MWTPW + fPVAh/IVVPVA = fP\.:C MVVPVC + (l —fp\_.vc )MWWA where the average molecular weight of the copolymer mer is,
MWav (polymer) _ l l. 000 gr'tnol = 73.33 g."'(mol‘ mer)
DP 1 50 mers MW’mtmer) = The mole fraction of PV C can be obtained by equating these two equations. Chapter 10, Problem 9 How much sulfur must be added to T0 g of butadiene rubber to crosslink 3 .0 percent of the mers? (Assume all
sulfur is used to crosslink the nlers and that only one sulfur atom is involved in each crosslinking bond.) Chapter 10, Solution 9 Assuming only one sulfur atom is involved in each crosslinking bond with butadiene. we know there is a oneto
one correspondence between the number of moles of sulfur and the number of moles of butadiene in the reaction
we are therefore able to assess the amount of sulfur required for 100% crosslinking. mass of S 100% = MWS X(No. of moles of S) = MWS x (No. of moles ofbutadiene) I mass of butadiene
3 MW (mer) butadiene The average molecular weight of the butadiene mer is: waulybumdim = 6 H atoms X 1 gr'mol + 4 C atoms X 12 glmol = 54 glmol Substituting. 0 s
54 gr'mol mass of Shem = (32 gr‘mol)[ ]= 41.48 g S     Thus for only 3 percent crosslinking. mass of S 3%: 0.03 x 41.48 g S 21.24 g S
' Polybutadiene Chapter 10, Problem 10
If 5 g of sulfur is added to 90 g of butadiene rubber. what is the IllaXiIllUlll fraction of the crosslink sites that can be connected? Chapter 10, Solution 10
Assuming one sulfur atom is involved in crosslinking each butadiene Iner. the fraction of crosslink sites is simply
the ratio of the moles of sulfur to the moles of butadiene. 5 a
No. ofntoles ofS : —% 20156111018
32 gr'mol 90g —. 21.66? 11101 butadiene
54 g.."1nol X0. of moles of butadiene = The fraction of crosslink sites is thus.
0.156
1.66? X100% 2 9.36% Fraction of crosslink sites 2 Chapter 10, Problem 15 What weight percent sulfur must be added to polybutadiene to crosslink 20 percent of the possible crosslink sites”? Chapter 10, Solution 15 Using a basis of 100 g and recalling that the molecular weight ofpolybutadiene is 54.0 g/mol. the number of nloles
of rubber is: 100g — 218511101
54.0 gs‘mol No. of moles of polybutadiene = For 20 percent crosslinking. the mass of sulfur required is.
mass of 8120.3,“ 2 (0.20)(1.85 11101)(32 gfinol) =1 1.84 g And the weight percent of sulfur required for this reaction is 11.84g —x100% : 10.59%
100 g + 11.84 g W’t % S : Chapter 10, Problem 18 A stress of 9.0 MPa is applied to an elastomeric material at a constant stress at 20°C. After 25 days. the stress
decreases to 6.0 MPa. (a) What is the relaxation time I for this material? (b) What will be the stress after 50 days'? Chapter 10, Solution 18 (a) The decrease in stress is given by 0' = 0' 08—“ r . Thus. —.’ —25 days I _ _
111(0 00) 111(6 MPa MPa) — 61.7 (1 ays
(b) For a relaxation time constant of 61.7' days. the stress after 50 days will be. a = (9.0 NIPakfﬁo W355” ‘1“) = 4.0 MPa Chapter 10, Problem 20 A stress of 1000 psi is applied to an elastonier at 2?°C‘. and after 25 days the stress is reduced to 750 by stress
relaxation. When the temperature is raised to 50°C. the stress is reduced from 1100 to 400 psi in 30 days. Calculate
the activation energy for this relaxation process using an Arrheniustype rate equation. Chapter 10, Solution 20 First. the relaxation time must be calculated for each set of conditions: _ I
I] — I. — f1?“ _ — 86.90 days
111(0' f 0'0) 111050 p31 1000 p31)
_ _3 .v .
r3 — r 0 day” — 29.66 days 111(0' HOG) 111(400 psit’l 100 psi)
Since the Alrheniustype equation uses absolute telnperattu‘e. we must next convert the temperatures to Kelvin:
T1 = 2? °+ 2T3 “’2 300 K; T: = 50 0+ 2—13 °= 323 K . Applying the Alrlleniustype rate equation. . — IRE" . . .
1 .I' I = C9 Q .. we obtain URL'0 equations 111 “V0 unknowns. _ (we—0.5m: 1 1
— = (.9 9" H1 and —
’51 Is Dividing these equations. we obtain a single equation with one unknown — Q — l l
Ezexp _Q ———
7:1 R 11 T2 Substituting the values provided and the calculated time constants. 29.66%“ —Q 1 1
86.90 ‘13 8.314Jr’1novi 300K 323K [—8.314 It'niol‘K] [29.66 4 I 231653 J."'1n01=37.65 ka’mol
2.3?36X10 I'K 86.90 Chapter 10, Problem 22 A polymeric material has a relaxation time of 100 days at 2?°C when a stress of 6.0 MPa is applied. (a) How many
days will be required to decrease the stress to 4.2 MPa? (b) “That is the relaxation tiine at 40°C if the activation
energy for this process is 25 Lil"11101? Chapter 10, Solution 22 (a) The decrease in stress is given by O' : 008—!" r . Thus.
r=—t [111(0' 00)]: —(100 days)[ln(4.2 MPa 6 MPa)] = 35.7 days . . . a — ER?" . . .
(b) By applying the Arrheniustype rate equation. l.‘ I : ('9 Q . for two sets 0t general conditions
and dividing the resulting equations. we obtain — l l
L2:exp _Q _,_
Ii R Ti T2 Substituting the values provided and the T1 calculated above. 1:2 _ex 25,000 J l 1
100 ‘13 8.314Jf1nol‘K 300K 313K 1:9 =65.95 days ...
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 Spring '08
 wholedepartment

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