# solutions_2_14

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KE-100.3410 Polymer properties
Solution to exercise 2: Crystallinity
Exercise 2.1
Polyethylene is a crystalline polymer which forms orthorhombic unit cell, i.e. ===90ᵒC,
where  and are the angles between the b and c, a and c, and a and b axes.
a.
Calculate the density of each crystal when the values |a|, |b| and |c| are 736, 492, 254
and 2[-CH2-CH2-] units per cell.
b. The crystalline lamella of PE can at its thinnest consist of 22 carbons and at the
thickest 150 carbons. Calculate the melting temperature range for PE using

Thompson-Gibbs equation: Tm  Tm0 1 

2 
.
L  H m 
Tm = melting temperature of a lamellar with thickness L
T = melting temperature of an infinitely thick and complete crystallite (414.2K)
 = free surface energy per unit area (79 x 10 -3 J / m2)
Hm = Enthalpy change per volume (288 x 10 6 J / m3)
L = lamellae thickness,
( ) = typical values for PE
Solution 2.1
a) The volume for the orthorhombic unit of PE can be calculated:
V = 73649225410-36 m3 = 9,2010-29 m3
Weight of the two molecules in the unit is:
2
m  nM 
N
2
kg
M
 0.0280
 9.30 1026 kg
1
NA
mol
6.022 1023
mol
And thus the density is:  
m 9.30 1026 kg
kg

 1010 3
29 3
V 9.20 10 m
m
b)

2 
Thompson-Gibbs equation: Tm  Tm0 1 

 L  H m 
Lamella thickness can be calculated from the chain length of the monomer:
22 carbons (11 monomer units): L = 11254 pm =2.79 nm
Substituting the values gives:
Tm,min
J


2  79 103 2


m
 415K  1 
  330 K  57C
9
6 J
 2, 79 10 m  288 10 3 
m 

Tm,max
J


2  79 103 2


m
 415K  1 
  403K  130C
 19,1109 m  288 106 J3 
m 

The melting temperature range for PE is 57-130oC.
Exercise 2.2
Experiments were carried out to measure max,, the maximum crystallinity, in a series of
random copolymers of ethylene and propylene. Linear PE gave max= 95%. Introduction of 4
CH3 –groups per 100 main-chain carbon atoms reduced max to 50 % and adding 20 CH3 –
groups per 100 main chain carbons max=0%.
a)
Calculate the weight fraction of propylene in copolymer.
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b)
Calculate the glass transition temperatures
Tg (K) for these amorphous
ethylene/propylene copolymers using the equation below. w1 and w2 are the weight
proportions of comonomers and Tg,1 and Tg,2 are the glass transition temperatures for
equivalent homopolymers. PE Tg = -120&deg;C and PP Tg = -19&deg;C.
w
w
1
 1  2
T
T
T
g
g ,1
g ,2
(1)
Solution 2.2
Introducing 4 CH3 –groups per 100 main-chain carbon atoms means the main chain contains
50 monomer units. Thus there are 4 propene groups per 46 ethene units.
a)
Molecular weights of the repeating units:
M(C2H4) = 28.1 g/mol and M(C3H6) = 42.1 g/mol.
Weight fraction of propene in copolymer where there is 4 CH3-groups:
w1 
n1M (Pr.)
n1M (Pr.)  n2 M ( Et .)

4  42.1
4  42.1
g
mol
g
g
 46  28.1
mol
mol
 0.115  11.5wt.  %
Weight fraction of propene in copolymer where there is 20 CH3-groups:
w1 
20  42.1
g
mol
g
g
20  42.1
 30  28.1
mol
mol
 0.500  50.0wt.  %
b)
Using equation and solving for Tg:
w
w
1
 1  2
T
T
T
g
g ,1
g ,2
The glass transition temperature for copolymer containing 11.5 wt.% propene:
4
1
0.115
1  0.115


 Tg  160 K  113C
Tg  273  19  K  273  120  K
The glass transition temperature for copolymer containing 50 wt.% propene
Tg = -82&deg;C
Exercise 2.3
Impurities can affect the melting temperature Tm. When adding solvent or plasticizer to
polymer, the effect on the melting temperature can be calculated using the equation:
1
1
R Vu
 0
(v1 )
Tm Tm H f V1
where Tm0 is the melting temperature of pure polymer, R = 8.3145 J/(K mol) gas constant, Hf
melting enthalpy of pure polymer, Vu molar volume of the repeating unit in polymer, V1 molar
volume of plasticizer and v1 volume fraction of plasticizer.
Polydecamethyleneadiapate,   CO  (CH 2 )4  CO  O  (CH 2 )10  O n 
(density 0.99g/cm3) was plasticized using different amounts of dimethylformamide
(CH3)2NCHO, (density 0.9445g/cm3) and the following melting temperatures were observed:
v1
Tm (oC)
0.079 72.5
0.202 66.5
0.422 61.5
0.603 57.5
What is the melting temperature Tm0 and melting enthalpy H of the pure polymer?
Solution 2.3
Calculate the molecular weights and molar volume for the components; (C16H28O4) and
(C3H7NO):
M (C16 H 28O4 )  (16 12.011  28 1.008  4 15.999)
g
g
 284.396
mol
mol
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M (C3 H 7 NO)  (3 12.011  7 1.008  14.007  15.999)
V VM
M
Vu  2  2 2  2 
n2
m2
2
g
g
 73.1
mol
mol
g
3
mol  287.3 cm
g
mol
0.99 3
cm
284.396
g
M1
cm3
mol
V1 

 77.4
1 0.9445 g
mol
3
cm
73.095
Plotting the melting temperature as a function of weight fractions of solvent the cross point of
R Vu
1
y-axis is 0 and the slope is
Tm
H f V1
The melting temperature and the melting enthalpy of the pure polymer can then be solved
from the plot Tm0:
3.05E-03
1/Tm (K-1)
3.00E-03
2.95E-03
2.90E-03
2.85E-03
0
0.1
0.2
0.3
y = 2.40E-04x + 2.88E-03
R&sup2; = 9.73E-01
0.4
v1
0.5
0.6
0.7
6
1
 2.88 103 K 1
0
Tm
 Tm0 
1
 346.74 K  73.6o C
3
1
2.88 10 K
The slope enables the calculation of Hf for the polymer
R Vu
 2.40 104 K 1
H f V1
 H f 
RVu
V1  2.40 104 K 1
J
cm3
 287.3
K  mol
mol  128 kJ

3
cm
mol
77.4
 2.40 104 K 1
mol
8.3145
Exercise 2.4*
20 wt.% of a styrene oligomer having a number-average degree of polymerization of 7 was
mixed with a commercial polystyrene sample having a number-average molecular weight of
100 000 g/mol. What is the Tg of the styrene oligomer? What is the Tg of the mixture? FoxFlory parameters for polystyrene are Tg = 373 K, K = 1.2105.
Solution 2.4*
The number-average molecular weight of the oligomer can be calculated from molecular
weight of the repeating unit in the polymer or oligomer and the number-average degree of
polymerization. Molecular weight of styrene is 104 g/mol and the number-average degree of
polymerization of the oligomer is 7:
M n  M 0 X n  104
g
g
 7  728
mol
mol
The Fox-Flory equation can be used to calculate Tg of the oligomer:
K
1.2 105
Tg ,oligomer  T 
 373K 
K  208K   65o C
728
Mn

g
To obtain the Tg of the mixture, the inverse rule of mixtures can be applied (Fox equation).
The Tg of the polystyrene is needed to calculate in order to use Fox equation:
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Tg ,PS  Tg 
K
1.2 105
 373K 
K  374 K  101o C
100000
Mn
1
w
w
0.2
0.8
1
 1  2 

 0.003101
Tg Tg ,1 Tg , 2 208K 374 K
K
 Tg ,mixture 
1
K  323K  50o C
0.003091
Exercise 2.5*
The following empirical equation can be applied when calculating the glass transition
temperature Tg for copolymers and blends
w
w
1
 1  2
Tg Tg ,1 Tg ,2
where w1 and w2 are the weight fractions of polymers and Tg,1 and Tg,2 are glass transition
temperatures of the polymers. The effect of plasticizer on polymer melting temperature Tm
can be estimated with the equation:
1
1
R Vu
 0
(v1  12v12 )
Tm Tm H f Vm,1
where Tm0 is the melting temperature of pure polymer, R = 8.3145 J/(K mol) gas constant, Hf
melting enthalpy of pure polymer, Vu molar volume of the repeating unit in polymer, Vm,1
molar volume of the plasticizer, v1 volume fraction of plasticizer and 12 is interaction
parameter for polymer-plasticizer.
How much plasticizer (Tg = -80oC) should be added to the Nylon 66 (C12H22O2N2) sample to
obtain Tg of 25oC? What would be the melting temperature; Vm,1 = 200cm3/mol and  = 0.40,
for Nylon66 Hu = 195.6 J/g, = 1.088 g/cm3, Tm0 = 265oC, Tg = 50 oC. Density of
plasticizer= 1.05 g/cm3.
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Solution 2.5
The weight fraction of polymer is w2  1  w1 and then weight fraction of the plasticizer can
be solved from the equation:
w
w
w 1  w1
1
 1  2  1 
Tg Tg ,1 Tg ,2 Tg ,1 Tg ,2
Tg ,1Tg ,2
 w1 
Tg
 Tg ,1
Tg ,2  Tg ,1
193.15K  323.15K
 193.15K
298.15
K

 0.1246
323.15K  193.15K
So the amount of the plasticizer should be 12 wt.%.
In order to calculate the melting temperature for the sample, one needs know the volume
fraction of the plasticizer v1:
v1 

m1
w1m


V1 1
w w   w2  2 


 1  1 1 1
m
m
V
1
g
g
 0.8754 1.088 3 )
3
cm
cm  0.1285
g
1.05 3
cm
0.1246  (0.1246 1.05
Molar volume of the repeating unit (C12H22O2N2) of Nylon 66 ([-NH-(CH2)6-NH-CO(CH2)4CO-]n):
g
3
mol  208 cm
Vu 

 2 1.088 g
mol
cm3
M0
226
The melting temperature of the polymer is:
1
1
R Vu
 0
(v1  12v12 )
Tm Tm H f Vm,1
9
 Tm 

1
1
R Vu

(v1  12 v12 )
0
Tm H f Vm ,1
1
J
cm3
8.3145
 208.0147
 (0.1285  0.4  0.12852 )
1
K  mol
mol

J
g
cm3
538.15K
195.6  226.32
 200
g
mol
mol
 531.34 K  258.19o C
Melting temperature is lowered from 265oC to 258 oC, when 12-wt.% of plasticizer is added.
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