Question 6 of 7: Copolymer Molecular Weight; Modulus–Temperature Behaviour of Amorphous and Semicrystalline Polymers

Question 6: Copolymer Molecular Weight; Modulus–Temperature Behaviour of Amorphous and Semicrystalline Polymers (supplementary)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (a) An ethylene–propylene copolymer made from $1$ kg C$_2$H$_4$ and $3$ kg C$_3$H$_6$, degree of polymerization $\overline{DP}=4500$.

Find. (a) The (number-average) molecular weight of the copolymer. (b) A schematic modulus–temperature curve for amorphous polyethylene and how crystallinity/cross-linking change it.

Approach

(a) Convert the two feed masses to moles of each mer using the page-1 atomic masses, form the mole fractions in the chain, take the mole-fraction-weighted average repeat-unit mass, then multiply by $\overline{DP}$.

1. (a) Repeat-unit molar masses. Using $M_C=12.01$, $M_H=1.01$ g/mol: $$M_{\text{ethylene}} = 2(12.01)+4(1.01) = 28.06\ \text{g/mol}, \qquad M_{\text{propylene}} = 3(12.01)+6(1.01) = 42.09\ \text{g/mol}$$
2. (a) Moles of each mer reacted. $$n_{\text{eth}} = \frac{1000\ \text{g}}{28.06\ \text{g/mol}} = 35.64\ \text{mol}, \qquad n_{\text{prop}} = \frac{3000\ \text{g}}{42.09\ \text{g/mol}} = 71.28\ \text{mol}$$ (the mole ratio comes out exactly $2\!:\!1$ propylene:ethylene).
3. (a) Mole fractions and average repeat-unit mass. $$f_{\text{eth}} = \frac{35.64}{35.64+71.28} = \frac{1}{3}, \qquad f_{\text{prop}} = \frac{2}{3}$$ $$\overline{M}_{\text{repeat}} = f_{\text{eth}}M_{\text{eth}} + f_{\text{prop}}M_{\text{prop}} = \tfrac13(28.06)+\tfrac23(42.09) = 37.41\ \text{g/mol}$$
4. (a) Molecular weight from the degree of polymerization. $$\overline{M} = \overline{DP}\times\overline{M}_{\text{repeat}} = 4500\times37.41$$ $$\boxed{\overline{M} \approx 1.684\times10^{5}\ \text{g/mol}\ \ (168{,}300\ \text{g/mol})}$$

(b) Amorphous polyethylene — modulus vs temperature. A fully amorphous, linear (uncross-linked) polymer shows a high, roughly temperature-independent glassy modulus ($\sim10^9$ Pa) at low $T$; a sharp, roughly one-decade drop in modulus at the glass transition $T_g$ as segmental (chain) motion becomes possible; a rubbery plateau just above $T_g$ where chain entanglements still resist flow; and, since there is no crystallinity and no cross-linking to prevent it, a further drop into viscous flow at higher $T$ as entanglements release and the whole chain can move — there is no true melting point $T_m$ for a purely amorphous polymer, only this terminal flow region. For polyethylene, $T_g$ lies far below room temperature ($\approx-100^{\circ}$C to $-120^{\circ}$C), and the $T_m$ marked on the sketch ($\approx115$–$135^{\circ}$C for PE) is where the crystallites melt once any crystallinity is present; for the fully amorphous chain the modulus simply falls into flow in that region. (i) Increasing crystallinity raises the rubbery-plateau modulus (crystallites act as physical cross-links/reinforcement) and introduces (or raises/sharpens) a genuine melting transition $T_m$ where the crystallites themselves melt. (ii) Increasing cross-linking raises the rubbery-plateau modulus even further and, at sufficient cross-link density, eliminates the terminal flow region entirely — a heavily-cross-linked (thermoset) network cannot flow at any temperature short of decomposition, since the covalent cross-links are permanent.