Question 5 of 8: Melt Index and PTFE Chain Statistics

Question 5: Melt Index and PTFE Chain Statistics (20 marks)

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.

(a)(i) Melt index vs. molecular weight. The melt index varies inversely with molecular weight: higher-molecular-weight polymer melts are more viscous (melt viscosity scales roughly as $\eta\propto \bar M_w^{3.4}$ above the entanglement threshold), so under the same extrusion pressure and temperature a high-MW polymer flows through the standard die more slowly, giving a lower melt index (fewer grams extruded in 10 minutes). Conversely, low-MW polymer melts have lower viscosity and a higher melt index.

(a)(ii) Why $\bar M_w$, not $\bar M_n$. Melt flow behaviour is controlled by chain entanglement, and long chains contribute disproportionately to entanglement density and melt viscosity — a small number of very long chains can dominate the flow resistance even if they are a small fraction of the total chain count. $\bar M_w$ (which weights each chain by its own mass, so longer chains count more) tracks this entanglement-dominated viscosity far better than $\bar M_n$ (a simple chain-count average, which is skewed toward the much more numerous short chains and is insensitive to the high-MW tail that actually controls melt flow).

Given (b). Degree of polymerization $\overline{DP}=8000$; PTFE repeat unit $-\text{CF}_2-\text{CF}_2-$ (C$_2$F$_4$); sample mass $=1200$ g; atomic masses (from page 1) $M_C=12.0$, $M_F=19.0$ g/mol; $N_A=6.02\times10^{23}$ mol$^{-1}$.

Find (b). (i) Molecular weight of each chain. (ii) Total number of chains in 1200 g.

Approach

The repeat-unit molar mass follows directly from the PTFE formula and the given atomic masses; multiplying by the degree of polymerization gives the (monodisperse) chain molecular weight, and dividing the sample mass by that chain weight (then multiplying by $N_A$) gives the number of chains.

1. Repeat-unit molar mass. C$_2$F$_4$: $$M_0=2(12.0)+4(19.0)=\boxed{100.0\ \text{g/mol}}.$$
2. Chain molecular weight. $$\overline{M}=\overline{DP}\times M_0=8000\times100.0=\boxed{800{,}000\ \text{g/mol}}.$$
3. Number of chains in 1200 g. Moles of chains $=1200/800{,}000=1.500\times10^{-3}$ mol; number of chains $$N_{chains}=1.500\times10^{-3}\times6.02\times10^{23}=\boxed{9.03\times10^{20}\ \text{chains}}.$$