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# Material Science

In: Science

Submitted By ikmal
Words 1463
Pages 6
HW 9
Problem 14.5
a. To Find:
(a) The number-average molecular weight
(b) The weight-average molecular weight
(c) The degree of polymerization for the given polypropylene material
b. Given:
Molecular Weight
Range (g/mol)

xi

wi

8,000–16,000

0.05

0.02

16,000–24,000

0.16

0.10

24,000–32,000

0.24

0.20

32,000–40,000

0.28

0.30

40,000–48,000

0.20

0.27

48,000–56,000

0.07

0.11

c. Assumptions:
The given data is accurate; the material (polypropylene) is pure.
d. Solution:
(a) Number-average molecular weight:
Molecular wt.
Range

Mean Mi

8,000-16,000
16,000-24,000
24,000-32,000

12,000
20,000
28,000

32,000-40,000
40,000-48,000
48,000-56,000

36,000
44,000
52,000

xi
0.05
0.16
0.24

xiMi
600
3200
6720

0.28
10,080
0.20
8800
0.07
3640
____________________________
Mn =



 xi M i = 33,040 g/mol

(b) Weight-average molecular weight:
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

wi

wiMi

12,000
20,000
28,000
36,000
44,000
52,000

0.02
0.10
0.20
0.30
0.27
0.11

240
2000
5600
10,800
11,880
5720

___________________________
M w = wi M i = 36,240 g/mol
(c) Degree of polymerization:


For polypropylene, the repeat unit molecular weight, m = 3(AC) + 6(AH) = (3)(12.01 g/mol) + (6)(1.008 g/mol)
= 42.08 g/mol

DP =



Mn
33,040 g/mol
=
= 785 m 42.08 g/mol

(a) g/mol
(b) 36240 g/mol 
(c) 


Problem 14.8
a. To Find:
(a) Wt% of Chlorine to be added
(b) How this chlorinated poly-ethylene differs from PVC
b. Given:
High-density poly-ethylene is chlorinated such that Cl atoms replace 5% of the original H atoms.
c. Assumptions:
(i)
(ii)
(iii)
(iv)

All the Chlorine added is utilized in replacing the existing H atoms – no Cl atom stays unutilized. Exactly 5 % of the original H atoms are replaced by Cl atoms.
The original material is pure poly-ethylene.
No other atom species (which could potentially replace H atoms) is added.

d. Solution:

(a) Consider 50 carbon atoms in the high density poly-ethylene. These correspond to 100 possible side-bonding sites. Initially, all 100 sites are occupied by H. Post-chlorination, 95 of these sites are occupied by hydrogen and 5 sites are occupied by Cl.
Mass of 50 carbon atoms,mC = 50(AC) = (50)(12.01 g/mol) = 600.5 g
Mass of 95 H atoms, mH = 95(AH) = (95)(1.008 g/mol) = 95.76 g
Mass of 95 Cl atoms, mCl = 5(ACl) = (5)(35.45 g/mol) = 177.25 g
Using a modified form of Equation 4.3, concentration of chlorine, CCl is:
CCl =

mCl mC + mH + mCl

x 100 =

177.25 g
 100 = 20.3 wt%
600.5 g + 95.76 g + 177.25 g

20.3 wt%




(b) Compared to chlorinated poly-ethylene, in poly(vinyl chloride):
(1) 25% of the side-bonding sites are substituted with Cl
(2) Substitution is probably less random

Problem 14.17
a. To Find:
The number-average molecular weight of a random nitrile rubber [poly(acrylonitrile-butadiene) copolymer]. b. Given:
1. Fraction of butadiene repeat units, fBu = 0.30 (=> fraction of acrylonitrile repeat units, fAc
= 0.70)
2. Degree of polymerization = 2000
c. Assumptions:
Given data is accurate; the material is pure.
d. Solution:
The two repeat units in this co-polymer are acrylonitrile and butadiene.
From Table 14.5, the acrylonitrile repeat unit contains 3 Carbon atoms, 1 Nitrogen atom and 3
Hydrogen atoms. Thus, the molecular weight of the acrylonitrile repeat unit 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 repeat unit contains 4 Carbon atoms and 6 Hydrogen atoms. Thus, the molecular weight of the butadiene repeat unit is: mBu = 4(AC) + 6(AH) = (4)(12.01 g/mol) + (6)(1.008 g/mol) = 54.09 g/mol
From Equation 14.7, the average repeat unit molecular weight is: m = f AcmAc + fBu mBu = (0.70)(53.06 g/mol) + (0.30)(54.09 g/mol) = 53.37 g/mol
From Equation 14.6, the number-average molecular weight is:



M n = m (DP) = (53.37 g/mol)(2000) = 106,740 g/mol



106740 g/mol

Problem 14.19
a. To Find:
(a) Ratio of butadiene to styrene repeat units in the given co-polymer
(b) Type of co-polymer
b. Given:
1. Number-average molecular weight of copolymer = 350000 g/mol
2. Degree of polymerization = 4425
c. Assumptions:
(i)
(ii)

Given data is accurate.
The material is pure.

d. Solution:
(a) From Equation 14.6, the average repeat unit molecular weight of the copolymer, m , is calculated as: m =



Mn
350,000 g/mol
=
= 79.10 g/mol
DP
4425



From Table 14.5, the butadiene repeat unit contains 4 Carbon atoms and 6 Hydrogen atoms.
Thus, the molecular weight of the butadiene repeat unit is : mb = 4(AC) + 6(AH) = 4(12.01 g/mol) + 6(1.008 g/mol) = 54.09 g/mol
The styrene repeat unit contains 8 Carbon atoms and 8 Hydrogen atoms. Thus, the molecular weight of the styrene repeat unit is: ms = 8(AC) + 8(AH) = 8(12.01 g/mol) + 8(1.008 g/mol) = 104.14 g/mol
Let fb be the chain fraction of butadiene repeat units. Since there are only two repeat unit types in the copolymer, the chain fraction of styrene repeat units fs is 1 – fb. Equation 14.7 may be written as:

m = f bmb + f s ms = fbmb + (1  f b )ms





fb =

m  ms
79.10 g/mol  104.14 g/mol
=
= 0.50 mb  ms
54.09 g/mol  104.14 g/mol

fs = 1 – fb = 1 – 0.50 = 0.50

Hence, the ratio of butadiene to styrene repeat units =

fb
0.50
=
= 1.0 (or 1:1) fs 0.50

(b) The ratio of the the two units in the co-polymer is 1:1 in case of alternating co-polymers.
There is no fixed ratio prescribed for random, graft and block co-polymers (1:1 could be a possible ratio for each of these types!) 
Therefore, with the ratio of butadiene to styrene repeat units obtained in part (a), this copolymer be of alternating, random, graft and block types.

(a) 1 (or 1:1)
(b) Alternating, random, graft and block types

Problem 14.25
a. To Find:
(a) Densities of totally crystalline and totally amorphous polytetrafluoroethylene (PTFE).
3
(b) % Crystallinity of a specimen having a density of 2.26 g/cm .
b. Given: ρ (g/cm3)

crystallinity (%)

2.144

51.3

2.215

74.2

c. Assumptions:
Given data is accurate; material is pure.
d. Solution:
(a) From Equation 14.8: %Crystallinity =

c ( s   a )
*100
 s (c   a )

…(1)

Where c is the density of totally crystalline PTFE; a is the density of totally amorphous PTFE and s is the density of the given specimen.

We can obtain two equations from the two sets of data given and solve for a and c.
(A) For s = 2.144 and % crystallinity = 51.3

 c (2.144   a )
*100
2.144(  c   a )
 c (2.144   a )
=> 51.3*2.144/100 =
(c   a )
51.3 =

…(2)

(B) For s = 2.215 and % crystallinity = 74.2

 c (2.215   a )
*100
2.215(  c   a )
 c (2.215   a )
74.2 * 2.215 / 100 =
(c   a )

74.2 =

…(3)

Dividing (2) by (3) :
51.3*2.144 / (74.2*2.215) =

(2.144   a )
(2.215   a )

(2.144   a )
=> 1.482 – 0.669a = 2.144 - a => 0.331 a = 0.662
(2.215   a )

0.669 =

a = 2.000 g/ cm

3

Substituting the value of a in (2) :
51.3*2.144/100 =

 c (2.144  2.000)
(  c  2.000)

 c (0.144)
=> 1.1 c – 2.2 = 0.144 c => 0.956 c = 2.2
(  c  2.000)

1.1 =

c = 2.301 g/ cm

3

3

(b) Substituting the values of c and a in (1) for s = 2.26 g/cm :
% Crystallinity =

(2.301 g/cm3)( 2.260 g/cm3
(2.260 g/cm3)( 2.301 g/cm3

 2.000 g/cm3)
 100 = 87.9
 2.000 g/cm3)


3

3

(a) a = 2.000 g/ cm ; c = 2.301 g/ cm
(b) 87.9 % Crystallinity

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#### Mats

...MATS1101 Materials Strand/MATS9520 INTRODUCTION TO MATERIALS ENGINEERING Dislocations Ashby & Jones, Volume 1, Chapter 9 Ideal Strength of Materials Last week we saw that the slope of the interatomic force-distance curve at equilibrium spacing is proportional to Young's Modulus, E. Interatomic forces become negligible for r > 2r0. F ~2Fmax Fmax Attraction 0 Repulsion r0 2r0 r0 r0 ~1.25r0 2r0 r Figure 8 Interatomic force-distance curve The maximum in the force-distance curve occurs at ~1.25r0, (where F = Fmax). If applied stress is greater than Fmax per bond, bonds between atoms are broken and fracture occurs. Ideal strength, , corresponds to bond rupture at Fmax. Calculation of ideal strength:    Slope E   0.25r0   r0 2r0 r0 Figure 9 Ideal Strength  ~ From the force-distance curve, where r = 1.25r0,  = 0.25 and   2  is ideal strength).  Copyright School of Materials Science and Engineering, UNSW, 2012. E ~ 2 0.25 ~  E 8 ~  E 15 E 15 A better estimate using interatomic potential gives Glasses and some ceramics have a yield strength of For other ceramics and polymers, For metals, y   y  10 1  E 15 E 15  y  10 1  10 5   Actual vs Ideal Strength 10-1 10-2 10-3 Cement (nonreinforced Ceramics Silica glass Diamond Soda glass SiC Al2O3, Si3N4 MgO, ice Alkali halides Metals Polymers Low density PE Epoxies PP, PMMA High density PE......

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