Handbook of Solvents - George Wypych - ChemTech - Ventech!

184 Christian Wohlfarth
sponding polymer concentrations. If, for example, one combines Δ mixG-ansatz by
Koningsveld and Kleintjens, 51 Equation [4.4.15], with the spinodal condition:
2
∂ ΔmixG = 0
[4.4.44]
∂ϕ ∂ϕ
i j
one obtains for a binary polymer solution:
α +
β( 1−γ)
( 1−
)
1 1
− −
3
γϕ ϕ rw 1−ϕ
2
2 2
= 0
[4.4.45]
where:
α acts as constant within a certain temperature range
β describes a temperature function like β = β0 + β1/T γ is also a constant within a certain temperature range.
rw mass average segment number, compare r in Equation (4.4.13)
ϕ 2 total volume fraction of the polymer
The adjustable parameters α, β, γ have to be fitted to spinodal data ϕ2 T spinodal and solvent activities can be calculated from the following relation:
where:
β( −γ)
( 1−γϕ
)
2
lna1 ln( 1 2) 1 2 2
rn 1 ⎛ ⎞
1
= − ϕ + ⎜ − ⎟ ϕ + αϕ +
⎝ ⎠
r n
2
2
ϕ [4.4.46]
number average segment number, compare r in Equation [4.4.13]
2
2
spinodal vs.
Gordon and coworkers 157-159 improved this method and developed the so-called PICS
(pulse-induced critical scattering) apparatus - details and history were summarized by
Galina et al. 160 PICS enables not only investigations within the metastable range, i.e., nearer
to the spinodal, but also of high-viscous solutions and polymer blends for determining
spinodal and binodal (cloud-point) curves. How to obtain solvent activities from demixing
equilibrium is explained in the text below.
4.4.3.2.3 X-ray scattering
X-ray scattering can be measured by the classical Kratky camera or more modern synchrotron
techniques. Technical details can be found in a number of books, e.g., those by Guinier
and Fouret, 161 Chen and Yip, 162 or Glatter and Kratky. 163
Small angle X-ray scattering (SAXS) can be used in analogy to light scattering
to measure second virial coefficients of binary polymer solutions. Zimm-diagrams can
be constructed following the same ways as in light scattering. This was demonstrated, for
example, in papers by Kirste and coworkers. 164-166 In analogy to Equation [4.4.43], one can
derive
where:
( ) / ( ) ( / )
2
Kc I θ= 0 = 1 M + 2Ac + 3Ac
+� [4.4.47]
2 2 2 2 3 2
c 2
mass by volume concentration c 2 =m 2/ν