Chemistry Reference
In-Depth Information
(a) Calculate the actual DP of PE and PS.
(b) Calculate
ΔS
m
,
ΔH
m
, and
ΔG
m
.
(c) Calculate
χ
critical
.
Solution
(a) PE: 32.74 cm
3
/mol
DPPE
1000
-
5
PS:
84:16cm
3
=mol
-
eachunitoccupies
84:16
1000
2:57
5
389
32:74
5
2:57latticesites:'DP
PS
5
(b)
0
@
1
A
0:4
1000
ln 0:4
1
0:6
1000
ln 0:6
ΔS
m
52
R
5
0:0056 J=molK
ΔH
m
5
χRTφ
1
φ
2
5
0:1
3
8:314
3
300
3
0:4
3
0:6
5
59:86 J=mol
ΔG
m
5
59.86
2
300(0.0056)
5
58.2 J/mol
2
1000
1000
(c)
χ
critical
5
2
1
1
5
0:002
1
p
p
EXAMPLE 5-2
According to the FloryHuggins lattice theory, the activity coefficient of a solvent (i.e., γ
1
)in
a solvent (1)-polymer (2) mixture is given by the following equation:
ð1
2
φ
1
Þ
1
χð1
2
φ
1
Þ
lnγ
1
5
ln
φ
1
1
m
2
1
2
x
1
1
(a) For a solvent at infinite dilution in a polymer with infinite molecular weight, show that χ
is given by the following equation:
1
1
ln
φ
1
x
1
χ
5
lnγ
1
2
(b) Both
x
1
for a particular solvent-polymer system are measured to be 0.001 and
0.02, respectively; if the interaction energy of the components in this system follows
the geometric mean rule (i.e.,
φ
1
and
can be calculated using the individual components'
Hildebrand solubility parameters), show that the minimum possible
χ
γ
1
for such a system
is 0.135.
(c) At 100
C, γ
1
and the reference volume based upon which χ is calculated are deter-
mined to be 0.2 and 55 cm
3
/mol, respectively; calculate the difference in the solubility
parameters of the solvent and polymer (cal/cm
3
)
1/2
. R
5
1.987 cal/mol/K.
Solution
(a)
ð1
2
φ
1
Þ
1
χð1
2
φ
1
Þ
lnγ
1
5
ln
φ
1
1
m
2
x
1
1
1
2
Infinite dilution
-
φ
1
B0
Infinite molecular weight
-
1/m
-
0
