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# Polymer Chemistry. The Basic Concepts - Himenz P.C.

Himenz P.C. Polymer Chemistry. The Basic Concepts - Copyright, 1984. - 736 p. Previous << 1 .. 220 221 222 223 224 225 < 226 > 227 228 229 230 231 232 .. 312 >> Next desired result. A fair amount of algebra is required to convert this
result into a usable form. We outline the tidying up in the following
steps; the reader can supply the intervening steps:
516
Thermodynamics of Polymer Solutions
1. Substitute Eq. (8.32) for ńî in ?2:
zn2(z- 1)n2 (n-2>
n2
Ď [N- n(i- l)]n
N2 J NN2(n_1)
i= 1
2. Examine the product of terms in this expression, replacing N by nN/n:
6. Apply Sterling's approximation to the logarithm of ?2 and simplify:
7. Consider the above result for the case of N2 = 0, that is, for
pure solvent:
The expression in item (6) is proportional to the entropy of the
pure
solvent: St = 0.
8. If Nt = 0, that is, for pure polymer, the expression in item (6)
gives
S2 = kN2 [In z + (n - 2) In (z - 1) + (1 - n) + In n]
This is the entropy of the disordered polymer when the latter
entirely fills the lattice.
3. Write out a few terms of the product:
Ď
4. This is the same as
5. Substitute items (2) and (3) into item (1):
K1 ? KT - / v\_Ë \ w\ M _ y. . i n . p
+ N2 [In z + (n - 2) In (z - 1) + (1 - n) + In
n]
The Flory-Huggins Theory: The Entropy of Mixing
517
The entropy of the mixture minus the entropy of the pure components-
that is, item (6) minus items (7) and (8)-gives ASm according to this
model:
Multiplying and dividing the right-hand side by Nt + N2 and letting this
sum equal Avogadro's number of molecules gives ASm per mole of solution:
This result should be compared with Eq. (8.28) for the case of the ideal
mixture. It is reassuring to note that for n = 1, Eq. (8.36) reduces to
Eq. (8.28). Next let us consider whether a change of notation will
clarify Eq. (8.36) still more. Recognizing that the solvent, the repeat
unit, and the lattice site all have the same volume, we see that Nt/N is
the volume fraction occupied by the solvent in the mixture and nN2/N is
the volume fraction of the polymer. Letting ô{ be the volume fraction of
component i, we see that Eq. (8.36) becomes
This model then leads us through a thicket of statistical and algebraic
detail to the satisfying conclusion that going from small solute
molecules to polymeric solutes only requires the replacement of mole
fractions with volume fractions within the logarithms. Note that the mole
fraction weighting factors are unaffected.
Since the system was defined to be athermal, AGm = - T ASm, so
Since the 0's are fractions, the logarithms in Eq. (8.38) are less than
unity and AGm is negative for all concentrations. In the case of athermal
mixtures entropy considerations alone are sufficient to account for
polymer-solvent miscibility at all concentrations. Exactly the same is
true for ideal solutions. As a matter of fact, it is possible to regard
the expressions for ASm and AGm for ideal solutions as special cases of
Eqs. (8.37) and (8.38) for the situation where n happens to equal unity.
The following example compares values for ASm for ideal and Flory-Huggins
solutions to examine quantitatively the effect of variations in n on the
entropy of mixing.
+ N2 In
nN2
~N~
(8.35)
nN2
"fT

(8.36)
ASm = - R (xt In 0! + x2 In 02)
(8.37)
AGm = RT (xt In 0! + x2 In 02 )
(8.38)
518
Thermodynamics of Polymer Solutions
Example 8.1
Evaluate ASm for ideal solutions and for athermal solutions of polymers
having n values of 50, 100, and 500 by solving Eqs. (8.28) and (8.38) at
regular intervals of mole fraction. Compare these calculated quantities
by preparing a suitable plot of the results.
Solution
We express the calculated entropies of mixing in units of R. For ideal
solutions the values of ASm are evaluated directly from Eq. (8.28):
x2 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
0.9
x2 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2
0.1
ASm/R 0.325 0.500 0.611 0.673 0.693 0.673 0.611
0.500 0.325
For polymer solutions we seek the relationship between mole fraction and
volume fraction. Since 02/0t = nN2/Nt, N2/Nt = (1/n) (02/0i). Also,
Na _ N2/Nt _ (Öď)ô21ô1
*2 Nt + N2 1 + N2/Nt 1+(1/ď)02/0!
Remember that both x and 0 are fractional concentrations, so xt + x2 = 1
and 0i + 02 = 1. Therefore
x = (1/n) 0a /(1 - 03 )
*2 1 + (1/n) 02 /(1 - 02)
from which we find
<b = X2
2 (l/n) + x2(l - 1/n)
This result enables us to convert mole fractions to volume fractions.
Table 8.1 lists the corresponding values of ô{ and Xj for n = 50, 100,
and 500 as needed for the evaluation of ASm. With Xj's and the
corresponding 0.'s available, the
required values of ASmix /R are calculated by Eq. (8.38):
n = 50 1.71 2.10 2.19 2.14 1.97 1.75 1.49 1.06 0.62 Previous << 1 .. 220 221 222 223 224 225 < 226 > 227 228 229 230 231 232 .. 312 >> Next 