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Handbook of Solvents - George Wypych - ChemTech - Ventech!

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152 Christian Wohlfarth<br />

derivative <strong>of</strong> the Gibbs free energy <strong>of</strong> mixing with respect to the amount <strong>of</strong> substance <strong>of</strong> the<br />

solvent:<br />

0 ∂ Δ<br />

Δμ1<br />

= μ1 − μ1<br />

=<br />

∂ 1<br />

⎛ n ⎞ mixG<br />

⎜<br />

⎟<br />

⎝ n ⎠<br />

T, p, nj≠1 [4.4.11]<br />

where:<br />

ni amount <strong>of</strong> substance (moles) <strong>of</strong> component i<br />

n total amount <strong>of</strong> substance (moles) <strong>of</strong> the mixture: n = Σni ΔmixG molar Gibbs free energy <strong>of</strong> mixing.<br />

For a truly binary polymer solutions, the classical Flory-Huggins theory leads to: 46,47<br />

or<br />

where:<br />

ΔmixG / RT = x lnϕ + x lnϕ<br />

+ gx ϕ [4.4.12a]<br />

1 1 2 2 1 2<br />

( )<br />

ΔmixG / RTV = ( x / V )ln ϕ + x / V lnϕ<br />

+ BRTϕ<br />

ϕ [4.4.12b]<br />

1 1 1 2 2 2 1 2<br />

xi mole fraction <strong>of</strong> component i<br />

ϕi volume fraction <strong>of</strong> component i<br />

g integral polymer-solvent interaction function that refers to the interaction <strong>of</strong> a solvent<br />

molecule with a polymer segment, the size <strong>of</strong> which is defined by the molar volume <strong>of</strong><br />

the solvent V1 B interaction energy-density parameter that does not depend on the definition <strong>of</strong> a<br />

segment but is related to g and the molar volume <strong>of</strong> a segment Vseg by B = RTg/Vseg V molar volume <strong>of</strong> the mixture, i.e. the binary polymer solution<br />

Vi molar volume <strong>of</strong> component i<br />

The first two terms <strong>of</strong> Equation [4.4.12] are named combinatorial part <strong>of</strong><br />

Δ mixG, the third one is then a residual Gibbs free energy <strong>of</strong> mixing. Applying Equation<br />

[4.4.11] to [4.4.12], one obtains:<br />

or<br />

or<br />

where:<br />

2<br />

Δμ 1 ( 1 ϕ 2) 1 ϕ 2 χϕ 2<br />

1 ⎛ ⎞<br />

/ RT = ln − + ⎜ − ⎟ + [4.4.13a]<br />

⎝ r ⎠<br />

⎡<br />

⎛ ⎞ ⎤<br />

χ = ⎢ μ − ( 1−ϕ ) −⎜1− ⎟ϕ<br />

⎥ ϕ<br />

⎣<br />

⎝ ⎠ ⎦<br />

1<br />

/ RT ln /<br />

r<br />

2<br />

Δ 1 2 2 2<br />

⎡<br />

⎛ ⎞ ⎤<br />

χ = ⎢lna<br />

−ln ( 1−ϕ ) −⎜1− ⎟ϕ<br />

⎥/<br />

ϕ<br />

⎣<br />

⎝ r ⎠ ⎦<br />

1<br />

1 2 2<br />

2<br />

2<br />

[4.4.13b]<br />

[4.4.13c]<br />

r ratio <strong>of</strong> molar volumes V 2/V 1, equal to the number <strong>of</strong> segments if V seg =V 1

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