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

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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 ñî in ?2:

zn2(z- 1)n2 (n-2>

n2

Ï [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:

Ï

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 _ (Öï)ô21ô1

*2 Nt + N2 1 + N2/Nt 1+(1/ï)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

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