Uploaded by riva novikov

Question number 3 and 5 - solvent polymer solution

advertisement
Question number #5
The vapor pressure of the solvent in a polymer solution relative to the vapor pressure of the pure solvent may, to a first
approximation, be equated to the activity of the solvent, which in turn is related to the chemical potential by:
ln π‘Žπ‘  =
πœ‡π‘  +πœ‡π‘ 0
𝑅𝑇
For the system polystyrene/cyclohexane the following values of ln as were obtained at 34°C (Krigbaum and Geymer,
JACS, 81, 1859 (1959)):
(a) Calculate c for each value of Φp (assume M is large).
(b) Plot a graph of χ vs Φp and comment on the result.
ln π‘Žπ‘  =
πœ‡π‘  + πœ‡π‘ 0
1
1
= ln πœ™π‘  + πœ™π‘ ∗ (1 + ) + π‘‹πœ™π‘2 = 𝑖𝑓 𝑉 (π‘šπ‘œπ‘™π‘Žπ‘Ÿ π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›)π‘™π‘Žπ‘Ÿπ‘”π‘’ → (1 + ) ≅ 1
𝑅𝑇
𝑉
𝑉
Values of ln π‘Žπ‘  were obtained at 34 [∁°]
π‘π‘Žπ‘™
𝑅 − π‘’π‘›π‘–π‘£π‘’π‘Ÿπ‘ π‘Žπ‘™ π‘”π‘Žπ‘  π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ = 1.987 [
] ; 𝑇 − π‘‘π‘’π‘šπ‘ = 34 + 273.15 = 302.15 [𝐾]
π‘šπ‘œπ‘™ − 𝐾
ln π‘Žπ‘  =
πœ‡π‘  + πœ‡π‘ 0
= ln πœ™π‘  + πœ™π‘ + π‘‹πœ™π‘2
𝑅𝑇
π‘™π‘›π‘Žπ‘  − ln πœ™π‘  − πœ™π‘
X=
=[
πœ™π‘2
ln (
π‘Žπ‘›
) − πœ™π‘
1 − πœ™π‘
]
πœ™π‘2
Φp [volum fraction]
0.343
0.388
0.435
0.485
0.543
0.637
0.69
0.768
0.818
0.902
ln (as)
-0.004
-0.0042
-0.013
-0.018
-0.024
-0.063
-0.089
-0.151
-0.232
-0.46
as
0.991
0.990
0.971
0.959
0.946
0.865
0.815
0.706
0.586
0.347
1-Φ
0.657
0.612
0.565
0.515
0.457
0.363
0.31
0.232
0.182
0.098
ln(as/1-Φ)-Φ
0.067861
0.093352
0.105996
0.137142
0.184810
0.231290
0.276253
0.345328
0.351549
0.361599
χ polystyrene/cyclohexane
0.577
0.620
0.560
0.583
0.627
0.570
0.580
0.585
0.525
0.444
Minimum in energy occurs at F3 when the representative points on the free energy curve are joined by the lowest straight line.
In this case, have phase separation into compositions cA and cB, with phase's  and .
For compositions
c < cA have  phase
ο€  cA < c < cB have  + 
cB<c have just .
Proportions of  and  given by Lever rule:
For c<cA A dissolves B
For cB<c B dissolves A
cA and cB define solubility limits.
Question number #3
The values of the solubility parameter, δs , for certain solvent in (cal/cm3 ) 1/2 are given as follows:
Calculate the value of χ for solutions of polystyrene in these solvents (use a “fudge factor” of 0.34). Indicate which
solutions are likely to be single phase and which are likely to be phase separated. Use the program “solubility
parameter” to calculate the solubility parameter of polystyrene from group contributions. Now do the same for
poly(butadiene). Note that the solubility parameters of random copolymers can be calculated using δ = φAδA + φBδB).
Calculate the solubility parameter of a 50/50 styrene/butadiene random copolymer Now calculate the value of χ for
solutions of this copolymer in the solvents listed above. Compare the solubility of polystyrene with that of the random
copolymer.
Polystyrene
𝛿𝑃𝑆
π‘π‘Žπ‘™ 1/2
= 9.5 [ 3 ]
π‘π‘š
Polybutadiene
𝛿𝑃𝐡
π‘π‘Žπ‘™ 1/2
= 8.1 [ 3 ]
π‘π‘š
Styrene butadiene rubber (SBR)
𝛿𝑆𝐡𝑅 = 0.5 ∗ 9.5 + 0.5 ∗ 8.1
π‘π‘Žπ‘™ 1/2
= 8.8 [ 3 ]
π‘π‘š
πœ’ = πœ’π» + πœ’π‘† =
𝑉𝑆
2
∗ (𝛿𝑠 − 𝛿𝑝 ) + 0.34
𝑅𝑇
π‘π‘š3
𝑉𝑠 − π‘šπ‘œπ‘™π‘Žπ‘Ÿ π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘œπ‘“ π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘ [
]
π‘šπ‘œπ‘™
𝑅 − π‘’π‘›π‘–π‘£π‘’π‘Ÿπ‘ π‘Žπ‘™ π‘”π‘Žπ‘  π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘ = 1.987 [
π‘π‘Žπ‘™
] ; 𝑇 − π‘‘π‘’π‘šπ‘ = 298 [𝐾]
π‘šπ‘œπ‘™ − 𝐾
𝑖𝑓 πœ’ < 0.5 → π‘ π‘œπ‘™π‘£π‘’π‘›π‘‘ − π‘π‘œπ‘™π‘¦π‘šπ‘’π‘Ÿ π‘šπ‘–π‘ π‘π‘–π‘π‘–π‘™π‘–π‘‘π‘¦
As much as solubility parameter of the solvent and polymer close / identical, (𝛿𝑆 − 𝛿𝑃 )2 ≅ 0 will be miscible in all
proportions.
Polystyrene:
Solvent
δs [cal/cm3]^1/2 δps [cal/cm3]^1/2 Vs [cm3/mol]
n-Hexane
7.2
9.5
132
Carbon tetrachloride
8.6
9.5
97
Benzene
9.2
9.5
89
Acetone
9.7
9.5
73.3
Methanol
14.5
9.5
40.4
R [cal/mol-K]
1.987
1.987
1.987
1.987
1.987
T [K]
298
298
298
298
298
χ [interaction parameter]
1.519
0.473
0.354
0.345
2.046
Polybutadiene:
Solvent
n-Hexane
Carbon tetrachloride
Benzene
Acetone
Methanol
δs [cal/cm3]^1/2
7.2
8.6
9.2
9.7
14.5
δps [cal/cm3]^1/2 Vs [cm3/mol] R [cal/mol-K]
8.1
132
1.987
8.1
97
1.987
8.1
89
1.987
8.1
73.3
1.987
8.1
40.4
1.987
T [K]
298
298
298
298
298
χ [interaction parameter]
0.521
0.381
0.522
0.657
3.135
Styrene butadiene rubber (SBR):
Solvent
n-Hexane
Carbon tetrachloride
Benzene
Acetone
Methanol
δs [cal/cm3]^1/2
7.2
8.6
9.2
9.7
14.5
δStyrene [cal/cm3]^1/2 δbutadiene [cal/cm3]^1/2
8.1
9.5
8.1
9.5
8.1
9.5
8.1
9.5
8.1
9.5
δSBR [cal/cm3]^1/2
8.8
8.8
8.8
8.8
8.8
Vs [cm3/mol] R [cal/mol-K]
132
1.987
97
1.987
89
1.987
73.3
1.987
40.4
1.987
T [K]
298
298
298
298
298
χ [interaction parameter]
0.911
0.347
0.364
0.440
2.557
solvents
n-Hexane
Carbon tetrachloride
Benzene
Acetone
Methanol
Polystyrene
Phase separation
critical point
Miscible mixture
Miscible mixture
Phase separation
Polybutadiene
critical point
Miscible mixture
critical point
Phase separation
Phase separation
Styrene butadiene rubber (SBR)
Phase separation
Miscible mixture
Miscible mixture
Miscible mixture
Phase separation
Styrene butadiene rubber (SBR) with Carbon tetrachloride solvent will be miscible, contrary to Polystyrene. This can be
seen by solubility parameter of solvent and polymer, solubility parameter of SBR (8.8) more close to solubility
parameter of Carbon tetrachloride (8.6), than solubility parameter of PS (9.5).
Download