Question 3 of 7: Melt Index and Molecular Weight; PTFE Degree of Polymerization
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — December 2017. 3 hours,
closed-book examination (approved Casio or Sharp calculator only). Notes on the paper state that
any five questions constitute a complete paper and only the first five questions appearing in the
answer book are marked, with all questions of equal value. All seven questions are solved below
for completeness.
Reference texts: Callister & Rethwisch, Materials Science and
Engineering: An Introduction, 9th ed. (crystal structure and defects, diffusion, mechanical
behaviour and tensile testing, polymers, phase diagrams and the lever rule, precipitation
hardening, corrosion, casting and solidification).
Question 3: Melt Index and Molecular Weight; PTFE Degree of Polymerization (20 marks)
Given. Degree of polymerization $\overline{DP}=8000$ for PTFE, whose repeat
unit is –CF$_2$–CF$_2$–; sample mass $=1200$ g; atomic masses $M_C=12.0$,
$M_F=19.0$ g/mol (page-1 table).
Find. (a)(i)–(ii) Qualitative relationships between melt index, molecular
weight, and the choice of averaging method. (b)(i) Chain molecular weight. (b)(ii) Number of
chains in 1200 g.
Approach
Melt index is essentially an inverse measure of melt viscosity: it measures how much polymer
flows through a standard die under a standard load in a fixed time. Melt viscosity rises steeply
with chain length (empirically $\eta\propto \overline{M}_w^{3.4}$ above the entanglement
threshold, via reptation of entangled chains), so molecular weight and melt index move in
opposite directions. Part (b) is a direct repeat-unit bookkeeping problem: multiply the repeat-unit
molecular weight by the degree of polymerization to get the chain molecular weight, then divide the
sample mass by the chain molecular weight and multiply by Avogadro's number to get the number of
chains.
(a)(i) Melt index vs. molecular weight. The melt index decreases
as molecular weight increases. Longer chains entangle more and resist flow more strongly (higher
melt viscosity), so under the same standard load and temperature, less material is extruded in the
fixed 10-minute period — melt index and molecular weight are inversely related.
(a)(ii) Why $\overline{M}_w$ matters more than $\overline{M}_n$ for melt index.
Melt flow is controlled by chain entanglement and reptation, both of which are dominated by the
largest, most entangled chains in the distribution — these longest chains create the
most drag and are the slowest to disentangle and flow. $\overline{M}_w=\sum w_iM_i$ weights each
chain by its own mass, so long chains contribute disproportionately more to $\overline{M}_w$ than to
$\overline{M}_n=\sum N_iM_i/\sum N_i$ (a simple number average, which weights every chain, long or
short, equally by count). Because $\overline{M}_w$ reflects the population of large, viscosity-
controlling chains far better than $\overline{M}_n$ does, it correlates much more directly with melt
viscosity (and hence melt index) than the number average does.
(b)(i) Chain molecular weight. The PTFE repeat unit –CF$_2$–CF$_2$–
has formula C$_2$F$_4$:
$$M_{repeat}=2(12.0)+4(19.0)=24.0+76.0=100.0\ \text{g/mol}$$
$$M_{chain}=\overline{DP}\times M_{repeat}=8000(100.0)$$
$$\boxed{M_{chain}=8.00\times10^5\ \text{g/mol}}$$
(b)(ii) Number of chains in 1200 g. Since every chain has the same
molecular weight (monodisperse, by the problem statement), the number of moles of chains is simply
mass$/M_{chain}$:
$$N_{chains}=\frac{m}{M_{chain}}N_A=\frac{1200}{8.00\times10^5}(6.02\times10^{23})$$
$$\boxed{N_{chains}\approx9.03\times10^{20}\ \text{chains}}$$
