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Engineering LibreTexts

16.13: Questions

  • Page ID
    36608
  • Quick questions

    You should be able to answer these questions without too much difficulty after studying this TLP. If not, then you should go through it again!

    Below are pictures of 3 polymers, one each of atactic, syndiotactic and isotactic. Which option has the correct sequence?



    a atactic, syndiotactic, isotactic
    b atactic, isotactic, syndiotactic
    c syndiotactic, atactic, isotactic
    d syndiotactic, isotactic, atactic
    e isotactic, syndiotactic, atactic
    f isotactic, atactic, syndiotactic
    Answer
    D

    Which of the following polymers would be likely to have the highest crystallinity:

    a An atactic polymer with many long side branches
    b An atactic polymer with few short side branches
    c An isotactic polymer with few short side branches
    d An isotactic polymer with many long side branches
    Answer

    C

    Deeper questions

    The following questions require some thought and reaching the answer may require you to think beyond the contents of this TLP.

    Kuhn length calculations

    a) A polyethylene chain is formed from 3000 ethene monomers. Given that the length of a single carbon-carbon bond is 0.154 nm, calculate the expected end-to-end distance using the random walk model, assuming that each bond is freely jointed.

    b) Given that the Kuhn length of polyethylene is equal to 3.5 C-C bond lengths, calculate an improved estimate of the expected end-to-end distance of the polymer chain.

    Answer

    a) If the polymer chain is made up of 3000 ethene monomers, the number of C-C bonds along the backbone is 6000 (technically 5998 since the monomers at the start and end of the chain only contribute one C-C bond to the polymer), so the expected end-to-end distance is 11.9 nm (⟨ Rn2 = ln)

    b) If the same polymer chain is divided up into segments, each with a length of 3.5 C-C bonds, the number of segments decreases from 6000 to 1714. The new estimate of the end-to-end distance is 22.3 nm (⟨ Rn2 = ln).

    A polymer sample was found to have the following molecular weight distribution:

    Number of molecules
    Molecular weight / g/mol
    200
    5,000
    300
    10,000
    400
    20,000
    100
    40,000

    Calculate the number average and weight average molecular weights, and the polydispersity index.

    Answer

    The number average molecular weight is calculated by multiplying the numbers of molecules by the molecular weights, adding them together, and dividing by the total number of molecules

    \[\overline{M_N} = \frac{\sum^{\infty}_{i=1}N_iM_i}{\sum^{\infty}_{i=1}N_i}\].
    This gives 16,000 g/mol.

    The weight average molecular weight is calculated by multiplying the numbers of molecules by the molecular weights squared, adding them together, and dividing by the sum of the numbers of molecules multiplied by the molecular weights

    \[\overline{M_W} = \frac{\sum^{\infty}_{i=1}N_iM_i^2}{\sum^{\infty}_{i=1}N_iM_i}\].
    This gives 22,200 g/mol (to 3 significant figures).

    The polydispersity index is given by the weight average molecular weight divided by the number average molecular weight. This gives 1.39.

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