Introduction to Polymers: Mechanics of Materials (3rd Edition) – Robert J. Young – Complete Solution Manual Chapters 1–25 | Full Polymer Mechanics A+ Study Guide

Solution Manual For
Introduction to Polymers Mechanics of Materials 3rd Edition by Robert J. Young
Chapter 1-25
IntroductionExams serve as a fundamental tool in evaluating a student's understanding of a subject, particularly in fields as diverse as business, law, and mathematics. These disciplines




Part I
Problems
Concepts, Nomenclature
and
Synthesis of Polymers

, Chapter 1
Concepts and Nomenclature


1.1
Polystyrene has the repeat unit structure
IntroductionExams serve as a fundamental tool in evaluating a student's understanding of a subject, particularly in fields as diverse as business, law, and mathematics. These disciplines




Hence
M0 = (8 x 12 g mol-1) + (8 x 1 g mol-1) = 104 g mol-1
Neglecting contributions from end-groups (which is reasonable because M n is high), then xn
can be calculated using a rearranged form of Equation (1.1)
xn = M n / M0
89 440 g mol1
xn   860
104 g mol1

Molar-mass dispersity ĐM is defined by ĐM = M w / M n and so if ĐM = 1.5, then
M w = 1.5 x 89 440 g mol-1 = 134 160 g mol-1

1.2
The mean repeat unit molar mass of the copolymer M cop
0 is given by Equation (1.2), which
requires knowledge of the mole fraction Xj and the molar mass M 0j of each type j of repeat
unit. Since there are only two types of repeat unit, it is necessary to calculate Xj for only one
of the two repeat units (because XE + XVAc = 1). The two repeat units are of structure:




and so
M 0E = (2 x 12 g mol-1) + (4 x 1 g mol-1) = 28 g mol-1

M 0VAc = (4 x 12 g mol-1) + (6 x 1 g mol-1) + (2 x 16 g mol-1) = 86 g mol-1

, Thus
(100 12.9 g (100 g cop) g mol1
XE 
{(100 12.9 g (100 g cop)1) / 28 g mol1}{12.9 g (100 g cop) g mol1}
XE = 0.954
Hence
XVAc = 1 – 0.954 = 0.046
M 0cop can now be calculated from Equation (1.2)

M 0cop = (0.954 x 28 g mol-1) + (0.046 x 86 g mol-1)

M 0cop = 30.67 g mol-1
Finally, using a rearranged form of Equation (1.1)
cop 39 870 g mol1
xn   1300
30.67 g mol1


1.3
IntroductionExams serve as a fundamental tool in evaluating a student's understanding of a subject, particularly in fields as diverse as business, law, and mathematics. These disciplines

(a) M n and M w can be calculated using, respectively, Equations (1.4) and (1.9) with
N1 = N2 = N3 = N:
( N  10 000 g mol1)  ( N  30 000 g mol1)  ( N  100 000 g mol1)
Mn 
NNN
M n = 46 667 g mol-1
{N  (10 000 g mol1)2}{N  (30 000 g mol1)2}{N  (100 000 g mol1)2}
Mw 
( N  10 000 g mol1)  ( N  30 000 g mol1)  ( N  100 000 g mol1)
M w = 78 571 g mol-1
M w / M n = 1.68
(b) M n and M w can be calculated using, respectively, Equations (1.7) and (1.8) with w1 =
w2 = w3 = 13 :

1
Mn 
( 000 g mol1)  ( 000 g mol1)  ( 000 g mol1)
3 3 3
M n = 20 930 g mol-1
M  ( 1  10 000 g mol1)  ( 1  30 000 g mol1)  ( 1  100 000 g mol1)
w 3 3 3
M w = 46 667 g mol-1
M w / M n = 2.23

, (c) The calculations are carried out as in (b) but with w1 = 0.145, M1 = 10 000 g mol-1 and w2
= 0.855, M2 = 100 000 g mol-1 :
1
M 
n
(0. 000 g mol1)  (0. 000 g mol1)
M n = 43 384 g mol-1

M  (0.145  10 000 g mol1)  (0.855  100 000 g mol1)
w
M w = 86 950 g mol-1
M w / M n = 2.00
IntroductionExams serve as a fundamental tool in evaluating a student's understanding of a subject, particularly in fields as diverse as business, law, and mathematics. These disciplines



The calculations for mixtures (a) and (b) show that, compared to mixing by equal numbers of
molecules, mixing polymer samples with different molar masses by equal weight greatly
increases the number of molecules of low molar mass and so reduces M n and M w .

The calculations also highlight the inadequacies of using M w / M n to assess molar mass
distributions. The molar mass distribution for each mixture is multi-modal and this cannot be
interpreted from M w / M n . Furthermore, despite M w / M n = 2 for the mixture in (c), this
mixture does not have a molar mass distribution consistent with the most probable
distribution or other common distributions for which M w / M n = 2, thereby highlighting the
deficiencies in using M w / M n as the only means of assessing the functional form of molar
mass distributions.