8.4: Flory–Huggins Model of Polymer Solutions
- Page ID
- 294309
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Let’s being by defining the variables for the lattice:
- \(M\): total number of lattice cells
- \(N_P\): number of polymer molecules
- \(n\): number of beads per polymer
- \(N_S\): number of solvent cells
- \(nN_P\) = total number of polymer beads
The total number of lattice sites is then composed of the fraction of sites occupied by polymer beads and the remaining sites, which we consider occupied by solvent:
\[M = nN_P + N_S\nonumber\]
Volume fractions of solvent and polymer:
\[\phi_S = \dfrac{N_S}{M} \ \ \ \ \phi_P = \dfrac{nN_P}{M} \ \ \ \ \phi_S + \phi_P = 1 \nonumber\]
The mole fraction of polymer:
\[x_P = \dfrac{N_P}{N_S + N_P}\nonumber\]
\(x_P\) is small even if the volume fraction is high.
Excluded Volume for Single Polymer Chain
Generally, excluded volume is difficult to account for if you don’t want to elaborate configurations explicitly, as in self-avoiding walks. However, there is a mean field approach we can use to account for excluded volume.
A better estimate for chain configurations that partially accounts for excluded volume:
Large \(n\):
\[\Omega_P \approx \dfrac{(z - 1)^{n - 1}}{M} \dfrac{M!}{(M - n)!} \nonumber \]
Entropy of Multiple Polymer Chains
For \(N_P\) chains, we count growth of chains by adding beads one at a time to all growing chains simultaneously.
1) First bead. The number of ways for placing the \(1^{\text{st}}\) bead for all chains:
2) Place the second bead on all chains. We assume the solution is dilute and neglect collisions between chains.
3) For placing the \(n^{\text{th}}\) bead on \(N_P\) growing chains. Here we neglect collisions between site \(i\) and sites \(>(i+4)\), which is the smallest separation that one can clash on a cubic lattice.
\[V^{(n)} = \left (\dfrac{z - 1}{M} \right )^{N_P(N - 1)} \dfrac{(M - N_P)!}{(M - n \cdot N_P)!} \nonumber\]
4) Total number of configurations of \(N_P\) chains with \(n\) beads:
Entropy of Polymer Solution
Entropy of polymer/solvent mixture:
\[S_{\text{mix}} = k_B \ln \Omega_P\nonumber\]
Calculate entropy of mixing:
The pure polymer has many possible entangled configurations \(\Omega_P^0\), and therefore a lot of configurational entropy: \(S_{\text{polymer}}^0\). But we can calculate \(\Omega_P^0\) just by using the formula for \(\Omega_P\) with the number of cells set to the number of polymer beads \(M = nN_P\).
\[\Omega_P^0 = \left (\dfrac{z - 1}{N_P \cdot n} \right )^{N_P(n - 1)} \dfrac{(N_P \cdot n)!}{N_P!} \nonumber\]
\[\dfrac{\Omega_P}{\Omega_P^0} = \left (\dfrac{N_P \cdot n}{M} \right )^{N_P (n - 1)} \dfrac{M!}{N_S!(N_P \cdot n)!}\]
Since \(\Delta S_{\text{mix}} = k_B \ln \dfrac{\Omega_P}{\Omega_P^0}\)
\[\begin{array} {rcl} {\Delta S_{\text{mix}}} & = & {-k_B N_S \ln \left (\dfrac{N_S}{M} \right) - k_B N_P \ln \left (\dfrac{N_P \cdot n}{M} \right)} \\ {} & = & {-Mk_B \left (\phi_S \ln \phi_S + \dfrac{\phi_P}{n} \ln \phi_P \right )} \end{array} \nonumber\]
where the volume fractions are:
\[\phi_S = \dfrac{N_S}{M} \ \ \ \ \ \ \ \ \ \phi_P = \dfrac{nN_P}{M} = 1 - \phi_S \nonumber\]
Note for \(n = 1\), we have original lattice model of fluid.