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Polymer Solution Thermodynamics: Flory-Huggins Theory of Polymer Solutions
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[QUOTE="tetrakis, post: 4868138, member: 495486"] [h2]Homework Statement [/h2] (Note: This isn't an assignment problem, more a curiosity about the derivation of an equation - hopefully it is still posted in the right forum..) I have working through the derivation for the partial molar Gibbs free energy of mixing from the Flory-Huggins expression for the Gibbs free energy of mixing, however my math skills are ([B]very[/B]) rusty - especially when it comes to partial derivatives. [h2]Homework Equations[/h2] Gibbs Free Energy of Mixing (Flory-Huggins Theory): ΔG[SUP]M[/SUP]=R·T[n[SUB]1[/SUB]·ln(ϕ[SUB]1[/SUB])+n[SUB]2[/SUB]·ln(ϕ[SUB]2[/SUB])+n[SUB]1[/SUB]·[ATTACH=full]195962[/ATTACH]·ϕ[SUB]2[/SUB]] partial molar Gibbs free energy of dilution: Δμ[SUB]1[/SUB]=R·T[ln(1-ϕ[SUB]2[/SUB])+(1-1/r)·ϕ[SUB]2[/SUB] +[ATTACH=full]195963[/ATTACH]·ϕ[SUB]2[/SUB][SUP]2[/SUP]] where K[SUB]B[/SUB]=Boltzmann Constant, n[SUB]i[/SUB]=number of moles of i, ϕ[SUB]i[/SUB]=volume fraction of i, [ATTACH=full]195964[/ATTACH]=Flory-Huggins Interaction Parameter ϕ[SUB]1[/SUB]=N[SUB]1[/SUB]/N[SUB]0[/SUB]=N[SUB]1[/SUB]/(N[SUB]1[/SUB]+r·N[SUB]2[/SUB]) ϕ[SUB]2[/SUB]=(r·N[SUB]2[/SUB])/N[SUB]0[/SUB]=(r·N[SUB]2[/SUB])/(N[SUB]1[/SUB]+r·N[SUB]2[/SUB]) N[SUB]0[/SUB]=N[SUB]1[/SUB]+r·N[SUB]2[/SUB] where N[SUB]0[/SUB] is the number of lattice sites N[SUB]1[/SUB] is the number of solvent molecules N[SUB]2[/SUB] is the number of polymer molecules, each occupying "r" lattice sites (or "r" segments) and R=K[SUB]B[/SUB]·N[SUB]A[/SUB] where N[SUB]A[/SUB]= Avogadro's constant [h2]The Attempt at a Solution[/h2] ΔG[SUP]M[/SUP]=R·T[n[SUB]1[/SUB]·ln(ϕ[SUB]1[/SUB])+n[SUB]2[/SUB]·ln(ϕ[SUB]2[/SUB])+n[SUB]1[/SUB]·[ATTACH=full]195965[/ATTACH]·ϕ[SUB]2[/SUB]] applying R=K[SUB]B[/SUB]·N[SUB]A[/SUB] ΔG[SUP]M[/SUP]=K[SUB]B[/SUB]·T[N[SUB]1[/SUB]·ln(ϕ[SUB]1[/SUB])+N[SUB]2[/SUB]·ln(ϕ[SUB]2[/SUB])+N[SUB]1[/SUB]·[ATTACH=full]195966[/ATTACH]·ϕ[SUB]2[/SUB]] expressing ϕ[SUB]1[/SUB] and ϕ[SUB]2[/SUB] in terms of N[SUB]1[/SUB] and N[SUB]2[/SUB] gives ΔG[SUP]M[/SUP]=K[SUB]B[/SUB]·T[N[SUB]1[/SUB]·ln(N[SUB]1[/SUB]/(N[SUB]1[/SUB]+r·N[SUB]2[/SUB]))+N[SUB]2[/SUB]·ln((r·N[SUB]2[/SUB])/(N[SUB]1[/SUB]+r·N[SUB]2[/SUB]))+N[SUB]1[/SUB]·[ATTACH=full]195967[/ATTACH]·(r·N[SUB]2[/SUB])/(N[SUB]1[/SUB]+r·N[SUB]2[/SUB])] Next step would be to take the partial derivative of the equation with respect to N[SUB]1[/SUB], I've tried many times but I cannot get anything close to the equation for the partial molar Gibbs free energy of dilution.. I broke it up, letting Ψ[SUB]1[/SUB]=N[SUB]1[/SUB]·ln(N[SUB]1[/SUB]/(N[SUB]1[/SUB]+r·N[SUB]2[/SUB])) Ψ[SUB]2[/SUB]=N[SUB]2[/SUB]·ln((r·N[SUB]2[/SUB])/(N[SUB]1[/SUB]+r·N[SUB]2[/SUB])) Ψ[SUB]3[/SUB]=N[SUB]1[/SUB]·[ATTACH=full]195968[/ATTACH]·(r·N[SUB]2[/SUB])/(N[SUB]1[/SUB]+r·N[SUB]2[/SUB]) taking the partial derivative dΨ[SUB]1[/SUB]/dN[SUB]1[/SUB]=ln(N[SUB]1[/SUB]/(N[SUB]1[/SUB]+rN[SUB]2[/SUB])+[1-(N[SUB]1[/SUB]+rN[SUB]1[/SUB]N[SUB]2[/SUB])/(N[SUB]1[/SUB]+rN[SUB]2[/SUB])] dΨ[SUB]2[/SUB]/dN[SUB]1[/SUB]=-N[SUB]2[/SUB][(1+rN[SUB]2[/SUB])/(N[SUB]1[/SUB]+rN[SUB]2[/SUB])] dΨ[SUB]2[/SUB]/dN[SUB]1[/SUB]=[ATTACH=full]195969[/ATTACH][(rN[SUB]2[/SUB])/(N[SUB]1[/SUB]+rN[SUB]2[/SUB])]+N[SUB]1[/SUB][ATTACH=full]195970[/ATTACH][(-rN[SUB]2[/SUB])/(N[SUB]1[/SUB]+rN[SUB]2[/SUB])[SUP]2[/SUP] which when I plug it all back in, gives me a big mess.. [/QUOTE]
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Polymer Solution Thermodynamics: Flory-Huggins Theory of Polymer Solutions
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