Polymer Solution Thermodynamics: Flory-Huggins Theory of Polymer Solutions

[tetrakis]

Homework Statement

(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 (very) rusty - especially when it comes to partial derivatives.

Homework Equations

Gibbs Free Energy of Mixing (Flory-Huggins Theory):
ΔG^M=R·T[n₁·ln(ϕ₁)+n₂·ln(ϕ₂)+n₁·

·ϕ₂]

partial molar Gibbs free energy of dilution:
Δμ₁=R·T[ln(1-ϕ₂)+(1-1/r)·ϕ₂ +

·ϕ₂²]

where
K_B=Boltzmann Constant,
n_i=number of moles of i,
ϕ_i=volume fraction of i,

=Flory-Huggins Interaction Parameter

ϕ₁=N₁/N₀=N₁/(N₁+r·N₂)
ϕ₂=(r·N₂)/N₀=(r·N₂)/(N₁+r·N₂)
N₀=N₁+r·N₂

where
N₀ is the number of lattice sites
N₁ is the number of solvent molecules
N₂ is the number of polymer molecules, each occupying "r" lattice sites (or "r" segments)

and R=K_B·N_A
where
N_A= Avogadro's constant

The Attempt at a Solution

ΔG^M=R·T[n₁·ln(ϕ₁)+n₂·ln(ϕ₂)+n₁·

·ϕ₂]

applying R=K_B·N_A
ΔG^M=K_B·T[N₁·ln(ϕ₁)+N₂·ln(ϕ₂)+N₁·

·ϕ₂]

expressing ϕ₁ and ϕ₂ in terms of N₁ and N₂ gives
ΔG^M=K_B·T[N₁·ln(N₁/(N₁+r·N₂))+N₂·ln((r·N₂)/(N₁+r·N₂))+N₁·

·(r·N₂)/(N₁+r·N₂)]

Next step would be to take the partial derivative of the equation with respect to N₁, 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
Ψ₁=N₁·ln(N₁/(N₁+r·N₂))
Ψ₂=N₂·ln((r·N₂)/(N₁+r·N₂))
Ψ₃=N₁·

·(r·N₂)/(N₁+r·N₂)

taking the partial derivative
dΨ₁/dN₁=ln(N₁/(N₁+rN₂)+[1-(N₁+rN₁N₂)/(N₁+rN₂)]
dΨ₂/dN₁=-N₂[(1+rN₂)/(N₁+rN₂)]
dΨ₂/dN₁=

[(rN₂)/(N₁+rN₂)]+N₁

[(-rN₂)/(N₁+rN₂)²

which when I plug it all back in, gives me a big mess..

 

[Chestermiller]

Homework Statement

(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 (very) rusty - especially when it comes to partial derivatives.

Homework Equations

Gibbs Free Energy of Mixing (Flory-Huggins Theory):
ΔG^M=R·T[n₁·ln(ϕ₁)+n₂·ln(ϕ₂)+n₁·[PLAIN]https://upload.wikimedia.org/math/7/9/c/79c40dd2b43f3c03eaf88b5fc4c199b8.png·ϕ[SUB]2[/SUB]]

partial molar Gibbs free energy of dilution:
Δμ₁=R·T[ln(1-ϕ₂)+(1-1/r)·ϕ₂ +[PLAIN]https://upload.wikimedia.org/math/7/9/c/79c40dd2b43f3c03eaf88b5fc4c199b8.png·ϕ[SUB]2[/SUB][SUP]2[/SUP]]

where
K_B=Boltzmann Constant,
n_i=number of moles of i,
ϕ_i=volume fraction of i,
[PLAIN]https://upload.wikimedia.org/math/7/9/c/79c40dd2b43f3c03eaf88b5fc4c199b8.png=Flory-Huggins Interaction Parameter

ϕ₁=N₁/N₀=N₁/(N₁+r·N₂)
ϕ₂=(r·N₂)/N₀=(r·N₂)/(N₁+r·N₂)
N₀=N₁+r·N₂

where
N₀ is the number of lattice sites
N₁ is the number of solvent molecules
N₂ is the number of polymer molecules, each occupying "r" lattice sites (or "r" segments)

and R=K_B·N_A
where
N_A= Avogadro's constant

The Attempt at a Solution

ΔG^M=R·T[n₁·ln(ϕ₁)+n₂·ln(ϕ₂)+n₁·[PLAIN]https://upload.wikimedia.org/math/7/9/c/79c40dd2b43f3c03eaf88b5fc4c199b8.png·ϕ[SUB]2[/SUB]]

applying R=K_B·N_A
ΔG^M=K_B·T[N₁·ln(ϕ₁)+N₂·ln(ϕ₂)+N₁·[PLAIN]https://upload.wikimedia.org/math/7/9/c/79c40dd2b43f3c03eaf88b5fc4c199b8.png·ϕ[SUB]2[/SUB]]

expressing ϕ₁ and ϕ₂ in terms of N₁ and N₂ gives
ΔG^M=K_B·T[N₁·ln(N₁/(N₁+r·N₂))+N₂·ln((r·N₂)/(N₁+r·N₂))+N₁·[PLAIN]https://upload.wikimedia.org/math/7/9/c/79c40dd2b43f3c03eaf88b5fc4c199b8.png·(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 N₁, 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
Ψ₁=N₁·ln(N₁/(N₁+r·N₂))
Ψ₂=N₂·ln((r·N₂)/(N₁+r·N₂))
Ψ₃=N₁·[PLAIN]https://upload.wikimedia.org/math/7/9/c/79c40dd2b43f3c03eaf88b5fc4c199b8.png·(r·N[SUB]2[/SUB])/(N[SUB]1[/SUB]+r·N[SUB]2[/SUB])

taking the partial derivative
dΨ₁/dN₁=ln(N₁/(N₁+rN₂)+[1-(N₁+rN₁N₂)/(N₁+rN₂)]
dΨ₂/dN₁=-N₂[(1+rN₂)/(N₁+rN₂)]
dΨ₃/dN₁=[PLAIN]https://upload.wikimedia.org/math/7/9/c/79c40dd2b43f3c03eaf88b5fc4c199b8.png[(rN[SUB]2[/SUB])/(N[SUB]1[/SUB]+rN[SUB]2[/SUB])]+N[SUB]1[/SUB][PLAIN]https://upload.wikimedia.org/math/7/9/c/79c40dd2b43f3c03eaf88b5fc4c199b8.png[(-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..

Start out with $\frac{\partial \psi _3}{\partial n_1}$. You need to reduce the rhs to the least common denominator. You almost have it. At least get that term.

Chet

 

[tetrakis]

Start out with $\frac{\partial \psi _3}{\partial n_1}$. You need to reduce the rhs to the least common denominator. You almost have it. At least get that term.

Chet

Hey Chet,
Thanks for the help, I've got the third term now. Am I even close with the other 2 terms? I've been playing around with them a bit more, but I can't quite seem to get it, especially the first term with the ln(1-ϕ₂)

 

[tetrakis]

for $\frac{\partial \psi _2}{\partial N_1}$ I'm taking $\frac{\partial N _2}{\partial N_1}$ and $\frac{\partial ln(rN_2)}{\partial N_1}$ as zero based on the assumption that it is similar to taking the derivative of a constant, would this be correct?

 

[Chestermiller]

for $\frac{\partial \psi _2}{\partial N_1}$ I'm taking $\frac{\partial N _2}{\partial N_1}$ and $\frac{\partial ln(rN_2)}{\partial N_1}$ as zero based on the assumption that it is similar to taking the derivative of a constant, would this be correct?

This is correct. But your partial derivatives of psi₁ and psi₂ with respect to n₁ are both incorrect. You need to be more careful. Try again.

Also note the phi₁ = 1 - phi₂

Chet

 

[tetrakis]

This is correct. But your partial derivatives of psi₁ and psi₂ with respect to n₁ are both incorrect. You need to be more careful. Try again.

Also note the phi₁ = 1 - phi₂

Chet

Thank you, I've gotten the correct equation now.