# Chapter 14: Polymer Structure¶

### Example 14.1 Page No 499¶

In :
Mi1=7500          #g/mol , Molecular wt
Mi2=12500
Mi3=17500
Mi4=22500
Mi5=27500
Mi6=32500
Mi7=37500
xi1=0.05           #Number fraction
xi2=0.16
xi3=0.22
xi4=0.27
xi5=0.2
xi6=0.08
xi7=0.02

xM1=Mi1*xi1
xM2=Mi2*xi2
xM3=Mi3*xi3
xM4=Mi4*xi4
xM5=Mi5*xi5
xM6=Mi6*xi6
xM7=Mi7*xi7
xM=xM1+xM2+xM3+xM4+xM5+xM6+xM7

nC=2          #no of carbon atoms in repeat unit
nH=3          #no of hydrogen atoms in repeat unit
nCl=1         #no of chlorine atoms in repeat unit
MwC=12.01     #Molecular wt of carbon
MwH=1.01       #Molecular wt of Hydogen
MwCl=35.45     #Molecular wt of  chlorine
m=nC*MwC+nH*MwH+nCl*MwCl     #Total wt for PVC
DP=xM/m

wi1=0.02           #weight fraction
wi2=0.1
wi3=0.18
wi4=0.29
wi5=0.26
wi6=0.13
wi7=0.02

wM1=Mi1*wi1
wM2=Mi2*wi2
wM3=Mi3*wi3
wM4=Mi4*wi4
wM5=Mi5*wi5
wM6=Mi6*wi6
wM7=Mi7*wi7
wM=wM1+wM2+wM3+wM4+wM5+wM6+wM7

print"(a)The no Av Molecular wt is",xM,"g/mol"
print"(b)Degree of Polymerisation is",round(DP,0)
print"(c)The weight  Av Molecular wt is",wM,"g/mol"

(a)The no Av Molecular wt is 21150.0 g/mol
(b)Degree of Polymerisation is 338.0
(c)The weight  Av Molecular wt is 23200.0 g/mol


### Example 14.2 Page No 511¶

In :
Ac=12.01       #in g/mol  Molecular weight of Carbon
Ah=1.008       #in g/mol  molecular weight of hydrogen
a=7.41*10**-8  #in cm
b=4.94*10**-8  #in cm
c=2.55*10**-8  #in cm
Na=6.023*10**23

Vc=a*b*c
n=2
A=(2*Ac)+(4*Ah)
density_c=n*A/(Vc*Na)

density_a=0.870  # in g/cm**3
density_s=0.925  # in g/cm**3
pc=density_c*(density_s-density_a)*100/(density_s*(density_c-density_a))

print"(a)Density is",round(density_c,3),"g/cm**3"
print"(b)percentage crystallinity is",round(pc,1),"%"

(a)Density is 0.998 g/cm**3
(b)percentage crystallinity is 46.4 %


### Example 14.3 Page No 516¶

In :
P1=400000.0       # in Pa  Pressure inside the bottle
P2=400.0          # in Pa  Pressure outside the bottle
Pm=0.23*10**-13  #Solubility Coefficient
dx=0.05          # in cm Thickness of wall

J=(-Pm*(P2-P1)/dx)

A=500           #surface area of bottle in cm**2
V_lose=750.0    #cm**3 STP
V=J*A
t=V_lose/round(V,5)

print"Diffusion flux is ",round(J,9),"cm**3 STP/cm**2-s"
print"Time to escape is ",round(t/(3600*24),1),"days"

Diffusion flux is  1.84e-07 cm**3 STP/cm**2-s
Time to escape is  96.5 days

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