Why Longer Polymer Chains Affect Viscosity and Strength

Two batches of molten polyethene are pushed through identical nozzles at the same temperature and pressure. One flows easily. The other moves much more slowly, even though both are made from the same monomer.

Suppose the slower-flowing batch contains polymer chains that are, on average, three times longer. Would you expect it to make a stronger solid as well?

Often, yes. Longer polymer chains tend to increase viscosity because they are more difficult to move past one another. They can also increase tensile strength because greater chain entanglement and more effective load transfer make the solid harder to pull apart.

But neither effect is guaranteed. Temperature, branching, chain alignment, crystallinity, and processing can change the result. The useful chemistry is understanding which property comes from molecular structure, which comes from processing, and what a measurement can actually prove.

01Why longer chains resist movement

Start with two bundles of cables.

One contains many short cables. The other contains fewer, much longer cables, with both bundles containing the same total amount of cable.

If you try to pull the cables through one another, the longer ones are more likely to loop around other cables. Moving one long cable may require several neighbouring cables to rearrange.

This is a useful first model for polymer chain entanglement.

In a polymer such as polyethene, repeating units are connected by strong covalent bonds. The repeating unit is:

\[ \ce{[-CH2-CH2-]_{n}} \]

Here, \(n\) represents the number of repeating units in a chain, called its degree of polymerisation.

The larger \(n\) becomes, the longer the molecule generally becomes.

What changes when the chains become longer?

Imagine comparing two samples of linear polyethene with the same branching pattern and chemical composition.

The longer-chain sample generally has:

• More opportunities for different chains to become physically entangled.
• More chain segments that can interact with neighbouring molecules.
• Fewer chain ends per unit mass of polymer.
• Greater difficulty rearranging its chains when the material is flowing or being stretched.

The covalent bonds within each chain have not become stronger. Instead, the overall movement of the molecules has become more restricted.

The cable analogy has limits. Polymer molecules are constantly changing shape through molecular motion, and their neighbouring segments interact through intermolecular forces. A physical entanglement is not necessarily a permanent knot or a chemical bond.

This matters when we heat the polymer.

A sufficiently heated, uncrosslinked thermoplastic can flow because its chains can eventually rearrange relative to one another. Longer chains generally take more time to escape the restrictions imposed by their neighbours.

That produces a higher melt viscosity.

Viscosity describes resistance to flow. A high-viscosity polymer melt is more difficult to move through a nozzle under specified conditions than a low-viscosity melt.

Why the same effect can increase strength

Now cool the polymer into a solid.

When the material is pulled, neighbouring chains must transfer forces between themselves. Chains may straighten, deform, slide, or eventually pull out from surrounding material.

Longer chains can resist this movement more effectively because they are more extensively entangled and may connect different regions of the solid.

For semicrystalline polymers such as many forms of polyethene, sufficiently long molecules can also connect ordered crystalline regions through less ordered material. These connecting chain segments help transfer forces.

This often improves tensile strength, resistance to fracture, or toughness, depending on the polymer and test conditions.

However, tensile strength, stiffness, and toughness are different properties:

A longer-chain polymer can be tougher without becoming proportionally stiffer. A material can also resist melting-flow movement while still being flexible when solid.

There is another limitation: increasing molecular mass eventually gives diminishing improvements in some mechanical properties. Once the material already contains extensive chain entanglement, making the molecules even longer may provide little additional tensile strength.

The UNSW School of Materials Science and Engineering’s explanation of molecular mass describes this levelling-off of strength as molecular mass increases.

02How experiments helped establish that polymers really contain long molecules

The connection between chain length and viscosity was part of a much older scientific problem.

In the early twentieth century, chemists disagreed about the structure of substances such as natural rubber.

One explanation treated rubber as an association of many relatively small molecules. Another proposed that rubber consisted of individual molecules containing very long chains of covalently bonded atoms.

These models could both help account for rubber’s unusual physical behaviour. The difficult part was finding an experiment that distinguished them.

The competing prediction

In 1920, Hermann Staudinger argued that materials such as rubber contained very large covalently bonded molecules, later called macromolecules.

An alternative explanation proposed that smaller rubber-related units were held together through associations involving their carbon-carbon double bonds.

This suggested a chemical test.

If the large structures depended on associations involving those double bonds, what should happen when the double bonds were removed?

The smaller units should no longer be held together in the same way. The material should lose characteristics associated with the original large structures.

But if rubber contained long covalently bonded backbones, removing the double bonds would not necessarily break those backbones into small molecules.

The hydrogenation experiment

Staudinger investigated hydrogenation of rubber.

Hydrogenation adds hydrogen across carbon-carbon double bonds, converting them into single bonds under suitable reaction conditions.

The important sequence was:

1. Begin with rubber containing carbon-carbon double bonds.
2. React it with hydrogen under conditions that hydrogenate those bonds.
3. Examine whether the resulting material retains polymer-like behaviour.

The hydrogenated rubber remained broadly similar to the original material in its polymer-like characteristics.

That observation challenged the proposed model in which associations involving the double bonds were mainly responsible for the apparent large molecular structures.

It supported the explanation that the material contained long covalently bonded molecules.

The experiment did not prove that every polymer had the same structure or that hydrogenation left every property unchanged.

A chemical reaction can alter several material properties. What mattered was that the original explanation did not adequately predict the persistence of rubber-like behaviour.

Why viscosity measurements provided further evidence

During the late 1920s, Staudinger also investigated the viscosity of polymer solutions.

A solution-viscosity experiment can use a narrow glass capillary connected to a larger reservoir. A measured amount of solution moves through the capillary under controlled conditions, and the time taken for the liquid level to pass between two marks is recorded.

Longer flow times can indicate greater resistance to flow, although quantitative interpretation also depends on the instrument and liquid properties.

The important idea was to compare carefully prepared solutions and examine how their flow behaviour related to the proposed molecular structures.

These investigations contributed evidence that polymers could contain very large molecules rather than simply behaving as collections of small associated particles.

There is a distinction worth keeping: a dilute polymer solution is not the same as a molten polymer.

A long chain can increase the viscosity of a dilute solution even when different polymer chains are too far apart to become extensively entangled. Its large molecular dimensions affect how the surrounding liquid moves.

In a concentrated polymer melt, interactions and entanglements between chains become especially important.

The historical solution-viscosity experiments therefore supported the macromolecular model, but they were not direct measurements of modern melt entanglement.

The American Chemical Society’s historical account of Staudinger’s work documents the development of this evidence. Staudinger received the 1953 Nobel Prize in Chemistry for his contributions to macromolecular chemistry.

03A controlled experiment: Does chain length change melt flow?

Suppose we want to test the relationship directly using two modern samples of polyethene.

The first idea might be to heat both samples in separate containers and compare how quickly they drip through openings.

But this design has a problem.

If one sample is hotter, contains more branching, or passes through a wider opening, the observed flow difference might not be caused by chain length.

Even identical heating times do not guarantee identical melt temperatures.

We need a revised experiment that controls these conditions.

The apparatus: A heated capillary rheometer

A capillary rheometer measures how a material flows through a narrow passage under controlled conditions.

Imagine the apparatus from top to bottom.

At the top, a piston applies pressure to molten polymer inside a heated barrel.

A temperature sensor monitors the material’s temperature. At the bottom of the barrel, the melt enters a narrow capillary nozzle.

Polymer emerges from the nozzle and is collected or measured over time.

For our comparison, we hold constant:

• Polymer chemistry and branching structure.
• Melt temperature.
• Applied pressure.
• Capillary length and diameter.
• Additives and relevant sample preparation conditions.

We deliberately change average polymer chain length.

We then compare the volume extruded per minute.

At the same temperature, pressure, and capillary geometry, a lower flow rate indicates greater resistance to flow in that apparatus.

Real polymer melts can be shear-thinning: their apparent viscosity changes with the rate of deformation. Therefore, flow-rate ratios are not automatically exact viscosity ratios.

Illustrative experimental data

The following measurements are invented for teaching. They are not published measurements of commercial polyethene.

Assume two batches of linear polyethene have relatively narrow chain-length distributions.

Both are tested at \(190^\circ\mathrm{C}\), using the same capillary and applied pressure. Their solid tensile specimens are later prepared using the same intended cooling conditions and tested at \(23^\circ\mathrm{C}\).

The number-average molecular mass is an average calculated by counting polymer molecules. It does not mean every chain has exactly that mass.

The first result supports the chain-length explanation: Sample L has a greater average molecular mass and extrudes less material in the same time.

The tensile results also suggest that Sample L is stronger under these testing conditions.

But we should calculate the actual stresses before comparing strength.

04Why longer chains do not always give the stronger product

The data above were deliberately controlled.

In real manufacturing, two polymer products may differ in several ways besides molecular mass.

Think about a thin polymer strap.

One strap is made from shorter chains but has been stretched during manufacture so that many chain segments become aligned along its length.

Another contains longer chains but has relatively little molecular orientation.

Which will resist a force applied along the strap more effectively?

You cannot decide from chain length alone.

The shorter-chain strap may be stronger along its length because processing has oriented its molecules in a useful direction.

This gives us three levels of explanation.

Molecular structure influences what processing can achieve. Processing influences the final arrangement of the material. The arrangement then affects its measured bulk properties.

These levels are connected, but they are not interchangeable.

The contrasting case: Longer chains, lower measured strength

Imagine two samples made from the same type of polyethene.

Sample A has shorter chains but has been drawn into an oriented film.

Sample B has longer chains but has been moulded into a film with much less chain alignment.

Suppose the illustrative tensile measurements are:

Sample A has greater tensile strength along the film, even though its chains are shorter.

Does this mean chain length has no effect on strength?

No. It means the experiment changed two important factors: average chain length and molecular orientation.

The result cannot isolate the effect of chain length.

It also shows why a single tensile measurement can be misleading. Sample A is stronger in one direction but weaker in another.

If you want to test chain length alone, you should compare samples with similar orientation, crystallinity, processing histories, and testing conditions.

If you want to test orientation, you should instead keep the polymer molecular structure as similar as possible while varying the drawing conditions.

What about temperature?

Temperature creates another common trap.

Suppose a longer-chain polymer at \(210^\circ\mathrm{C}\) flows more readily than a shorter-chain polymer at \(180^\circ\mathrm{C}\).

That does not overturn the usual chain-length trend.

Heating increases molecular mobility and generally reduces the viscosity of a polymer melt over a suitable processing range.

A hotter long-chain melt can therefore flow faster than a cooler short-chain melt.

There is also an important difference between temporary softening and permanent chemical change.

Heating a thermoplastic enough to melt it does not necessarily shorten its chains. However, sufficiently severe processing conditions can cause polymer degradation, including chain scission, which breaks covalent bonds.

If chain scission occurs, the molecular mass distribution changes. A lower viscosity measured after the material has returned to its original testing temperature could then have a different explanation from ordinary thermal softening.

The correct comparison always asks what changed during the experiment and whether that change was reversible.

05Questions and solutions

All numerical data in the following questions are illustrative, not commercial polymer specifications. Unless otherwise stated, assume the polymers are linear thermoplastics with the same chemical repeating unit and no covalent crosslinks.

Question 1

Two polyethene samples have number-average molecular masses of 42,000 g mol\(^{-1}\) and 70,000 g mol\(^{-1}\).

The repeating unit has a molar mass of 28.0 g mol\(^{-1}\).

a) Calculate the number-average degree of polymerisation of each sample.

b) Predict which sample would usually have the greater melt viscosity at the same temperature and under comparable flow conditions.

c) Explain why the difference does not mean that the carbon-carbon covalent bonds in one sample are stronger.

Solution 1

The degrees of polymerisation are 1,500 and 2,500 respectively. The second sample would generally have the greater melt viscosity because its longer chains are more restricted by neighbouring chains.

Part a

Using:

\[ n=\frac{M_n}{M_0} \]

For the first sample:

\[ n_1=\frac{42\,000}{28.0}=1500 \]

For the second:

\[ n_2=\frac{70\,000}{28.0}=2500 \]

Both calculations divide molar masses expressed in g mol\(^{-1}\), so the degree of polymerisation is dimensionless.

Part b

The second sample has longer chains on average.

These chains generally experience more entanglement and require greater rearrangement to flow past their neighbours.

Therefore, the second sample is expected to have greater melt viscosity when the other relevant conditions are comparable.

This is a qualitative prediction, not a calculation of an exact viscosity.

Part c

The individual carbon-carbon covalent bonds remain chemically similar in both samples.

Increasing chain length adds more repeating units and changes how entire molecules interact and move. It does not inherently increase the strength of each backbone bond.

The tempting but incorrect explanation is that stronger plastic must contain stronger individual covalent bonds.

The relevant change here is molecular length and collective chain movement.

Question 2

An engineer compares two linear polyethene grades:

• Grade A: number-average molecular mass of 28,000 g mol\(^{-1}\).
• Grade B: number-average molecular mass of 84,000 g mol\(^{-1}\).

The first experiment uses identical capillaries but different operating conditions.

The engineer concludes that longer polymer chains decrease viscosity because Grade B flows faster.

A second experiment tests both samples at \(190^\circ\mathrm{C}\) and 0.60 MPa using the same capillary geometry.

The measured flow rates are:

• Grade A: 2.0 cm³ min\(^{-1}\).
• Grade B: 0.50 cm³ min\(^{-1}\).

A component requires 6.0 cm³ of molten polymer to pass through this apparatus in no more than 5.0 minutes. The second experiment’s operating conditions must be used.

a) Explain why the first experiment does not establish the effect of chain length.

b) Calculate the time required for each grade to deliver 6.0 cm³ in the second experiment.

c) Identify which grade meets the production requirement.

d) Explain what the second experiment supports and what it still cannot prove.

Solution 2

Grade A meets the 5.0-minute production requirement; Grade B does not. The controlled experiment supports greater melt-flow resistance for the longer-chain Grade B, while the original comparison was confounded by temperature and pressure.

Part a

The first experiment changed three relevant conditions:

• Average molecular chain length.
• Temperature.
• Applied pressure.

Increasing temperature generally lowers melt viscosity by increasing molecular mobility.

Increasing pressure driving the melt through the capillary can also raise the measured flow rate.

Therefore, Grade B’s faster flow does not establish that longer chains have lower viscosity.

The observation could be explained by the different operating conditions.

Part b

At a constant measured volumetric flow rate:

\[ t=\frac{V}{Q} \]

Here, \(t\) is the required time in minutes, \(V\) is the volume in cm³, and \(Q\) is volumetric flow rate in cm³ min\(^{-1}\).

For Grade A:

\[ \begin{aligned} t_A&=\frac{6.0}{2.0}\\ &=3.0\ \mathrm{min} \end{aligned} \]

For Grade B:

\[ \begin{aligned} t_B&=\frac{6.0}{0.50}\\ &=12\ \mathrm{min} \end{aligned} \]

Part c

The required time is no more than 5.0 minutes.

Grade A takes 3.0 minutes, so it meets the requirement.

Grade B takes 12 minutes, so it does not.

This decision concerns throughput under the specified conditions. It does not determine which grade makes the better finished component.

Part d

When temperature, pressure, and capillary geometry are the same, the longer-chain Grade B has a lower measured flow rate.

Grade A delivers four times the volume per minute:

\[ \frac{2.0}{0.50}=4.0 \]

That supports the prediction that longer chains can produce greater resistance to melt flow.

However, even the controlled experiment does not prove that chain length is the only molecular difference between the grades.

Branching and chain-length distribution must also be considered.

In addition, the measured flow-rate ratio is not automatically the ratio of true viscosities because polymer melts may be shear-thinning.

Question 3

A manufacturer must choose a polymer film for two different applications.

One application is a narrow strap that is pulled along its length. It requires a tensile strength of at least 36 MPa in the pulling direction.

The second is a film section that will be loaded across its width. It requires a tensile strength of at least 25 MPa in that direction.

Two samples of the same polymer chemistry are available.

• Sample U has shorter chains and has been strongly drawn in one direction.
• Sample V has longer chains but has not been deliberately drawn.

Their measured tensile strengths are:

All tensile tests use the same temperature and strain-rate conditions.

a) Choose the suitable sample for each application without changing the processing.

b) Can one of these unchanged samples meet both requirements?

c) A student argues that Sample U proves shorter chains always make stronger polymers. Evaluate this claim.

d) Propose a controlled experiment to investigate whether increasing chain length improves tensile strength independently of molecular orientation.

Solution 3

Sample U meets the strap requirement, Sample V meets the across-width requirement, and neither unchanged sample satisfies both. The data do not isolate the effect of chain length because molecular orientation also differs.

Part a

For the strap, strength must be at least 36 MPa along the film.

Sample U provides 42 MPa, which exceeds the requirement by:

\[ 42-36=6\ \mathrm{MPa} \]

Sample V provides only 32 MPa.

Therefore, Sample U is suitable for the strap.

For the across-width application, strength must be at least 25 MPa in that direction.

Sample V provides 31 MPa and exceeds the requirement by:

\[ 31-25=6\ \mathrm{MPa} \]

Sample U provides only 18 MPa.

Therefore, Sample V is suitable for the across-width application.

Part b

Neither unchanged sample meets both requirements.

Sample U fails the across-width requirement, while Sample V fails the longitudinal strap requirement.

Choosing the sample with the highest strength measurement overall would be a mistake because the direction of loading matters.

Part c

The claim is not supported.

Sample U differs from Sample V in both average chain length and processing history.

Drawing can align polymer chain segments and ordered regions along one direction, making the film better able to transfer and resist tensile forces in that direction.

The sharp difference between U’s longitudinal strength of 42 MPa and transverse strength of 18 MPa is consistent with orientation-dependent behaviour.

Its high longitudinal strength therefore does not show that shorter chains intrinsically produce stronger polymers.

Part d

Prepare samples with different number-average molecular masses but the same polymer chemistry, branching pattern, and additive content.

Use controlled processing conditions intended to produce similar crystallinity and orientation. Verify these characteristics as far as practical rather than assuming identical processing guarantees identical microstructure.

Prepare specimens with equal original cross-sectional areas and test them at the same temperature and strain rate.

Measure tensile strength and compare it with molecular mass.

If tensile strength rises with chain length while the other relevant factors remain comparable, that supports the proposed relationship more convincingly.

The important change is removing the processing difference that made the original comparison ambiguous.

Question 4

A manufacturer investigates whether prolonged heating has changed the molecular structure of a linear polyethene sample.

The initial number-average molecular mass is 84,000 g mol\(^{-1}\).

In a short heating experiment, the sample gives these melt flow rates through the same capillary at the same pressure:

In a separate prolonged-heating experiment, a fresh portion of the same polymer is held at \(210^\circ\mathrm{C}\) for substantially longer.

When returned to \(190^\circ\mathrm{C}\), it now flows at 1.5 cm³ min\(^{-1}\).

Independent molecular-mass measurements show that its number-average molecular mass has decreased to 42,000 g mol\(^{-1}\).

Assume no polymer mass is lost, and that chain scission is the only chemical reaction affecting the polymer backbones in the prolonged-heating experiment.

a) Explain the behaviour in the short experiment.

b) Explain why the prolonged-heating result is different.

c) Calculate the ratio of the final number of polymer molecules to the initial number.

d) Does your calculation prove that every original chain was cut exactly once? Explain.

Solution 4

The short heating cycle shows reversible temperature-dependent flow, whereas the prolonged-heating experiment provides evidence of permanent chain shortening. Under the stated assumptions, the number of polymer molecules has doubled, but this does not prove that every chain broke exactly once.

Part a

In the short experiment, the polymer flows faster when heated from \(190^\circ\mathrm{C}\) to \(210^\circ\mathrm{C}\).

The flow rate rises from 0.60 to 1.8 cm³ min\(^{-1}\).

Heating increases molecular mobility, allowing chains to rearrange more readily.

When the polymer returns to \(190^\circ\mathrm{C}\), the original flow rate of 0.60 cm³ min\(^{-1}\) is recovered.

This is consistent with reversible thermal softening without a significant permanent change in the molecular structure.

A higher flow rate at the higher temperature alone would not prove that covalent bonds had broken.

Part b

After prolonged heating, the sample returns to the original temperature but does not return to its original flow behaviour.

Its flow rate is now 1.5 cm³ min\(^{-1}\), rather than 0.60 cm³ min\(^{-1}\).

The independent molecular-mass measurement also shows that average chain mass has halved.

Under the stated assumption that chain scission is the only backbone reaction, the evidence indicates that polymer molecules have been broken into shorter chains.

These shorter chains can rearrange more readily, reducing resistance to flow.

The essential evidence is not simply that the polymer flowed faster while hot. It is the persistent change at the original testing temperature together with the lower measured molecular mass.

Part c

Let \(m\) represent the total polymer mass in grams, \(N\) the number of molecules, and \(N_A\) Avogadro’s constant in mol\(^{-1}\).

For a fixed polymer mass:

\[ N=\frac{m}{M_n}N_A \]

Here, \(M_n\) is measured in g mol\(^{-1}\). The factor \(m/M_n\) gives the amount of polymer molecules in moles.

Because no mass is lost:

\[ \begin{aligned} \frac{N_f}{N_i} &=\frac{M_i}{M_f}\\ &=\frac{84\,000}{42\,000}\\ &=2 \end{aligned} \]

The subscripts \(i\) and \(f\) mean initial and final.

The total number of polymer molecules has doubled.

Part d

No. The calculation proves an overall change in molecule count, not the exact history of every molecule.

For example, some original chains could remain unbroken while others experience several scission events.

Each successful scission of one linear polymer molecule into two separate molecules increases the total molecule count by one.

If the initial number of molecules was \(N_i\), and the final number is \(2N_i\), then the total number of scission events must be \(N_i\) under the stated assumptions.

That is an average of one scission event per original molecule, not proof of one event on each molecule.

The distinction matters because average molecular mass does not reveal the complete distribution of chain lengths.

Question 5

A polymer scientist prepares two idealised batches of linear polyethene.

Both contain the same total number of repeating units per representative group of ten molecules.

The chain populations are:

• Batch P: ten molecules, each containing 1,000 repeating units.
• Batch Q: nine molecules containing 100 repeating units each, and one molecule containing 9,100 repeating units.

Assume these proportions are repeated throughout large bulk samples. Both batches have identical repeating-unit chemistry, no branching, no additives, and no covalent crosslinks.

Under identical capillary testing conditions, the following illustrative results are obtained:

The solid specimens were manufactured using the same nominal processing settings and tested under identical conditions.

A technician makes three claims:

1. Both batches must have the same melt-flow behaviour because they have the same number-average degree of polymerisation.
2. Batch Q’s lower tensile strength proves that long polymer chains weaken a material.
3. Removing the longest 10% of molecules from Batch Q will reduce its viscosity while retaining 90% of its original polymer mass.

a) Calculate the number-average degree of polymerisation for each batch.

b) Calculate the percentage of Batch Q’s polymer mass contained in its longest 10% of molecules.

c) Evaluate the first two claims using the molecular populations and measured properties.

d) Evaluate the third claim quantitatively. Determine the number-average degree of polymerisation of the material remaining after the proposed removal.

e) Explain why the experiment illustrates a limitation of using a single average chain length to predict both viscosity and tensile strength.

Solution 5

Both batches have the same number-average degree of polymerisation of 1,000, but their chain-length distributions are very different. In Batch Q, the longest 10% of molecules contain 91% of the polymer mass, so removing them would retain only 9%, not 90%.

The observed viscosity and strength differences are compatible with the given molecular structures. They do not establish a universal rule that longer chains weaken polymers.

Part a

The number-average degree of polymerisation is:

\[ n_n=\frac{\text{total repeat units}}{\text{total molecules}} \]

For Batch P, the total number of repeating units is:

\[ 10(1000)=10\,000 \]

There are ten molecules, so:

\[ n_{n,P}=\frac{10\,000}{10}=1000 \]

For Batch Q, the total number of repeating units is:

\[ 9(100)+9100=10\,000 \]

There are also ten molecules, so:

\[ n_{n,Q}=\frac{10\,000}{10}=1000 \]

The two batches have identical number-average degrees of polymerisation.

This is the tempting stopping point, but it hides a major difference.

Batch P contains only molecules with 1,000 repeating units.

Batch Q contains nine short molecules and one extremely long molecule.

The average does not tell us how the repeating units are distributed among molecules.

Part b

All repeating units have the same chemical composition and molar mass.

Ignoring the negligible contribution of end groups, polymer mass is therefore proportional to the total number of repeating units.

The longest chain in Batch Q contains 9,100 out of 10,000 repeating units.

Its mass fraction is:

\[ \begin{aligned} w_{\mathrm{long}} &=\frac{9100}{10\,000}\\ &=0.91 \end{aligned} \]

Here, \(w_{\mathrm{long}}\) is the fraction of polymer mass in the longest chains.

The corresponding percentage is:

\[ 0.91\times100=91\% \]

Only one in ten molecules is extremely long, but that molecule contains 91% of the polymer’s repeating units.

That is the central surprise.

The proportion of molecules and the proportion of polymer mass are not the same.

Part c

The first claim is incorrect.

The number-average degree of polymerisation is identical, but the complete chain-length distributions differ.

In Batch Q, most of the polymer mass lies in very long molecules. These molecules can produce extensive intermolecular entanglement and strongly influence melt flow.

The supplied measurements show that Batch Q has the lower flow rate:

\[ \frac{1.2}{0.40}=3.0 \]

Batch P extrudes three times the volume per minute under the same test conditions.

The observations are compatible with Batch Q having greater melt-flow resistance because of its long-chain-rich mass distribution.

The exact viscosity difference cannot be predicted from the number-average chain length alone.

The second claim is also unjustified.

Batch Q has the lower measured tensile strength:

\[ 34-29=5\ \mathrm{MPa} \]

However, Batch Q is not simply Batch P with every molecule made longer.

Its molecular population contains both very short and extremely long chains.

The short chains may influence molecular mobility, load transfer, and the material’s final structure. Differences in crystallinity, chain arrangement, or defects may also arise despite identical nominal processing conditions.

The measured result establishes that Batch Q is weaker under these particular conditions.

It does not establish that increasing the length of otherwise comparable polymer chains reduces strength.

A controlled comparison of narrowly distributed chain lengths would be needed to test that claim.

Part d

Removing the longest 10% of molecules means removing the one 9,100-unit chain from each representative group of ten molecules.

Part b showed that this chain contains 91% of the mass.

Therefore, the fraction of mass remaining is:

\[ \begin{aligned} w_{\mathrm{remaining}} &=1-0.91\\ &=0.09 \end{aligned} \]

Only 9% of the polymer mass remains.

The remaining material consists entirely of chains containing 100 repeating units.

Its number-average degree of polymerisation is therefore:

\[ n_{n,\mathrm{remaining}} =\frac{9(100)}{9} =100 \]

Removing the very long chains is expected to reduce the melt’s resistance to flow under comparable conditions because the remaining molecules are much shorter.

But the technician’s mass-balance claim is wrong.

The procedure removes most of the polymer material, not a small fraction.

It could also sacrifice mechanical properties associated with having sufficiently long chains. The exact final tensile strength cannot be calculated from these data.

Part e

Number-average chain length describes one aspect of molecular structure, not the whole structure.

Two polymers can have the same average chain length but very different proportions of short, medium, and long molecules.

Those distributions can influence melt viscosity, how neighbouring chains interact, and the way the solid transfers stress.

The same numerical average can therefore hide important molecular differences.

There is a second limitation: even knowing the complete molecular mass distribution would not uniquely determine tensile strength without information about processing and final microstructure.

The correct reasoning has two stages.

First, identify the molecular population, including its chain-length distribution, rather than relying on one average.

Second, establish how that population was processed and what bulk properties were actually measured.

The final result is not a contradiction. It shows that equal averages do not necessarily mean equivalent molecular materials, and greater melt viscosity does not guarantee greater tensile strength.

06The next useful decision: Change the chains or change the processing?

Suppose a polymer is too viscous to mould easily, but its finished product needs high tensile strength.

Reducing average chain length may improve melt flow, yet could also reduce resistance to chain sliding or pull-out in the solid.

Increasing processing temperature may improve flow without deliberately changing chain length, provided the polymer remains chemically stable under those conditions.

Changing how a film or fibre is drawn may improve strength in a particular direction without requiring the same change to its melt viscosity.

These are different interventions, and they solve different problems.

To understand the processing route in more detail, see How Chain Alignment Changes Polymer Fibre Tensile Strength. If a heat-treated polymer behaves permanently differently after returning to its original temperature, How to Distinguish Polymer Degradation from Melting explains the next diagnostic step.

For NSW HSC Chemistry students in 2026, the applicable course is the Chemistry Stage 6 Syllabus (2017). The Chemistry 11-12 Syllabus (2025) begins implementation in 2028, with its first HSC examination in 2029. The experimental comparisons here develop structure-property reasoning and scientific investigation skills; they are not presented as prescribed named practicals.

The important decision is whether the problem belongs to the molecules, the way they were processed, or the way the finished material is being tested. Only then can you decide whether making the polymer chains longer will improve the property that actually matters.