Question

2.22 Use the group contribution method to calculate the solubility parameter of linear polyethylene (PE) and poly(vinyl chloride) (PVC) given the following molar attraction constants:

F(-CH2) = 135 cal^(1/2) / cm^(3/2)

F (-CH)= 60 cal^(1/2) / cm^(3/2)

And F(-Cl )= 230 cal^(1/2) / cm^(3/2 )

The densities of PE and PVC are 0.86 and 1.39 g/cm3, respectively. Compare the values of δ you obtain with the observed values of 16.2 and 19.8 MPa1/2 for these two polymers.

Answer 1

δ = [ρ*∑E] / M

where

δ =Solubility parameter

ρ = Density

∑E = Sum of the molar attraction constants

M = Molecular weight of the monomer

Solubility parameter of PE, δ = [ρ*∑E] / M = [0.86*(135 + 135)] / 28 = 8.3 cal^1/2 cm^-3/2

= 8.3*2.045 MPa^1/2 = 17.0 MPa^1/2

Solubility parameter of PE, δ = 17.0 MPa^1/2

Solubility parameter of PVC, δ = [ρ*∑E] / M = [1.39*(135 + 60 + 230)] / 62.5 = 9.5 cal^1/2 cm^-3/2

= 9.5*2.045 MPa^1/2 = 19.3 MPa^1/2

Solubility parameter of PVC, δ = 19.3 MPa^1/2

Comparison:

The obtained solubility parameter of PE (17.0 MPa^1/2) is more than the observed value of 16.2 MPa^1/2

The obtained solubility parameter of PVC (19.3 MPa^1/2) is less than the observed value of 19.8 MPa^1/2