Question

Using the definitions for total chain molecule length, L (Equation 4.9) and average chain end-to-end distance r (Equation 4.10), determine the following for a linear polyethylene: (a) the number-average molecular weight for L = 2500 nm; (b) the number-average molecular weight for r = 20 nm. please show equations used and calculations thank you

Answer 1

(a) Let us find the degree of polymerization, DP,

DP = N/2

N = L/d.sin(theta/2)

with,

d being C-C bond length = 0.154 nm

theta = 109 degree

L = 2500 nm

we get,

N = 2500/(0.154)sin(109/2) = 19940

DP = 19940/2 = 9970

So, number average molecular weight Mn becomes,

Mn = DP x molar mass of one ethylene unit

      = 9970 x 28.05 g/mol = 279,664 g/mol [or 280,000 g/mol]

(b) For r = 20 nm

N = r^2/d^2

   = (20)^2/(0.154)^2 = 16900

Degree of polymerization DP,

DP = 16900/2 = 8450

So, number average molecular weight Mn,

Mn = 8450 x 28.05 g/mol = 237,022 g/mol [or 237,000 g/mol]