Handbook of Solvents - George Wypych - ChemTech - Ventech!

320 V V Tereshatov, V Yu. Senichev, E Ya Denisyuk
Figure 6.2.3. Dependence of equilibrium swelling of the
crosslinked elastomer SCEPT-40 on the concentration
of component 2 in the mixtures: 1-heptane-toluene,
2-heptane-amyl acetate, 3-heptane-ethyl acetate.
[Adapted, by permission, from V. V. Tereshatov,
M. I. Balashova, A. I. Gemuev, Prediction and regulating
of properties of polymeric materials, Ural
Branch of AS USSR Press, Sverdlovsk, 1989, p. 3.]
Neglecting the change of volume on mixing, the solubility parameter of the mixture of
two liquids is represented by:
δ12 = δ1ϕ1+ δ2ϕ2 where:
ϕ1 and ϕ 2 volume fractions of components 1 and 2
More exact evaluation of the δ12 value is possible if the experimental data on enthalpy of
mixing, ΔH, of components of SL are taken into account: 4
2 2 ( 1 2 2 / 12)
δ = δ ϕ + δ ϕ −ΔH V
12 1
12 /
Figure 6.2.4. Dependence of equilibrium swelling of
the crosslinked elastomer SCN-26 on the concentration
of component 2 in the mixtures: 1-toluene-acetone,
2-ethyl acetate-dioxane, 3-ethyl acetate-acetone.
[Adapted, by permission, from V. V. Tereshatov,
M. I. Balashova, A. I. Gemuev, Prediction and regulating
of properties of polymeric materials, Ural
Branch of AS USSR Press, Sverdlovsk, 1989, p. 3.]
With a change in δ 1 and δ 2 parameters, the SL composition has the maximum equilibrium
swelling which corresponds to shifts in the field of composition of the liquid phase.
The δ 12 parameter is close or equal to the value of the solubility parameter of polymer. Such
a simplified approach to the extreme swelling of polymers in liquid mixtures frequently
works very well in practice. If there is a maximum on the curve of swelling in SL, then the
swelling has an extreme character (11 cases out of 12) (Figures 6.2.1-6.2.4).
The parameter of interaction, χ 123 , between polymer and a two-component liquid can
be used as a co-solvency criterion for linear polymers (or criterion of extreme swelling),
more general, than the equality (δ 12 = δ 3): 5
χ123 = χ13 ϕ1+ χ23 ϕ2− ϕ1ϕ2χ12 [6.2.1]