Polymers

Question 1

(a) Proof (10 marks)

Summation identity (prove one — 3 marks). Start from the geometric series $\sum_{x\ge0}p^x = (1-p)^{-1}$ for $|p|<1$. Differentiate w.r.t. $p$: $\frac{d}{dp}\sum_{x\ge0}p^x = \sum_{x\ge1}x p^{x-1} = \frac{d}{dp}(1-p)^{-1} = (1-p)^{-2}. \quad(1)$ (2 marks setup, 1 mark result.) Differentiating again, $\sum_{x\ge1}x^2 p^{x-2}\cdot$… more cleanly: differentiate $\sum x p^{x-1}=(1-p)^{-2}$ times $p$ then differentiate again gives $\sum x^2 p^{x-1}=(1+p)(1-p)^{-3}$. (Accept either identity proven.)

Number-average (3 marks). With $N_x \propto x_n = p^{x-1}(1-p)$ and $M_x = xM_0$: $\bar{X}_n = \frac{\sum_x x\,p^{x-1}(1-p)}{\sum_x p^{x-1}(1-p)} = \frac{(1-p)\,(1-p)^{-2}}{(1-p)\,(1-p)^{-1}} = \frac{(1-p)^{-1}}{1} = \frac{1}{1-p}.$ (Denominator $\sum p^{x-1}(1-p)=1$; numerator uses identity (1).)

Weight-average (3 marks). $\bar{X}_w = \frac{\sum_x x^2 p^{x-1}(1-p)}{\sum_x x\,p^{x-1}(1-p)} = \frac{(1-p)(1+p)(1-p)^{-3}}{(1-p)(1-p)^{-2}} = \frac{(1+p)(1-p)^{-2}}{(1-p)^{-1}} = \frac{1+p}{1-p}.$

PDI (1 mark). $\text{PDI} = \frac{\bar{M}_w}{\bar{M}_n} = \frac{\bar{X}_w}{\bar{X}_n} = \frac{(1+p)/(1-p)}{1/(1-p)} = 1+p. \qquad\blacksquare$

(b) Computation (5 marks)

$\bar{X}_n = 1/(1-0.990) = 1/0.010 = 100$. $\bar{M}_n = 100 \times 226 = 22600\ \text{g mol}^{-1}. \quad(2)$ $\bar{X}_w = (1+0.990)/(1-0.990) = 1.990/0.010 = 199$. $\bar{M}_w = 199 \times 226 = 44974\ \text{g mol}^{-1}. \quad(2)$ $\text{PDI} = 1 + 0.990 = 1.990. \quad(1)$

(c) Sensitivity (4 marks)

$\bar{X}_n = 100 \Rightarrow (1-p)=0.01$; $\bar{X}_n=500\Rightarrow(1-p)=0.002$. Change: $(1-p)$ must drop from $0.01$ to $0.002$, a factor of 5 decrease, i.e. $p$ from $0.990$ to $0.998$. (2) Comment: because $\bar{X}_n\propto 1/(1-p)$, high DP requires $p$ extremely close to 1. This demands exact stoichiometric balance of the two monomers, near-complete conversion, removal of the condensation by-product (water), and absence of monofunctional impurities that cap chains — all experimentally demanding. (2)

(d) Code (5 marks)

def flory(p, M0, xmax):
    x   = list(range(1, xmax+1))
    Nx  = [p**(xi-1)*(1-p) for xi in x]      # mole fractions (unnormalised OK)
    Mx  = [xi*M0 for xi in x]
    S0  = sum(Nx)                             # sum N_x
    S1  = sum(n*m for n, m in zip(Nx, Mx))    # sum N_x M_x
    S2  = sum(n*m*m for n, m in zip(Nx, Mx))  # sum N_x M_x^2
    Mn  = S1/S0
    Mw  = S2/S1
    return Mn, Mw, Mw/Mn

(3 marks for correct arrays & the three moment sums; 1 mark for correct $\bar M_n,\bar M_w$.) Numerical precaution (1 mark): choose xmax large enough that $p^{xmax}$ is negligible — the neglected tail scales as $\sim p^{xmax}$; near $p\to1$ (e.g. 0.99) one needs xmax of order several thousand (a few $\times\,1/(1-p)$) so truncation error in the moments is small. (Also: use float/double precision.)

Question 2

(a) Classification (10 marks — 2 each)

*Kevlar is technically a linear (thermoplastic-class) polymer though it decomposes before melting; award full marks for "thermoplastic/linear, not thermoset."

(b) HDPE vs LDPE (6 marks)

• Ziegler–Natta: monomer inserts into a metal–carbon bond at a well-defined catalytic site; insertion is stereo- and regio-controlled, growth stays at the metal centre, so chains grow linearly with few branches → close packing → high crystallinity → high density and modulus. (3)
• Free-radical (LDPE): the propagating radical is reactive and undergoes intramolecular chain transfer ("backbiting"), in which the radical abstracts a H six atoms back (favourable 6-membered transition state) to form a short (butyl) branch; intermolecular transfer gives long branches. Branches prevent tight chain packing → lower crystallinity, lower density. (3)

(c) Kevlar & Nomex (6 marks)

Kevlar repeat unit (poly-para-phenylene terephthalamide): $-\!\left[\,\mathrm{NH}\!-\!\underset{para}{\mathrm{C_6H_4}}\!-\!\mathrm{NH}\!-\!\mathrm{CO}\!-\!\underset{para}{\mathrm{C_6H_4}}\!-\!\mathrm{CO}\,\right]_n\!- \quad(2)$ Physical explanation (2): the para linkage gives rod-like, extended, rigid chains that align nearly parallel; dense inter-chain hydrogen bonds between amide N–H and C=O of neighbouring chains form sheets, and $\pi$-stacking of rings between sheets → highly ordered, crystalline fibre with very high specific tensile strength (strong yet light) → body armour, parachute cords. Nomex (2): meta substitution kinks the chain, preventing the fully extended alignment, so it is less stiff/strong; however the aromatic amide backbone is thermally very stable and the material is highly flame-resistant, so Nomex is used for fire-protective clothing rather than for stiffness.

Question 3

(a) Composite moduli (8 marks)

Longitudinal (2): $E_{\parallel} = 0.60(230) + 0.40(3.5) = 138 + 1.4 = 139.4\ \text{GPa}.$ Transverse (3): $\frac1{E_\perp} = \frac{0.60}{230} + \frac{0.40}{3.5} = 0.0026087 + 0.114286 = 0.116894\ \text{GPa}^{-1},$ $E_\perp = 8.555\ \text{GPa} \approx 8.56\ \text{GPa}.$ Anisotropy (1): $\frac{E_\parallel}{E_\perp} = \frac{139.4}{8.555} \approx 16.3.$ Design consequence (2): the panel is $\sim16\times$ stiffer along the fibres than across them; parts must be laid up with plies at multiple orientations (e.g. $0/\pm45/90$) so loads in all in-plane directions are carried — a unidirectional laminate is dangerously weak transversely.

(b) Biodegradability (6 marks)

• PHBV (poly-3-hydroxybutyrate-co-3-hydroxyvalerate): contains hydrolysable ester linkages in the backbone; microorganisms secrete depolymerase enzymes that cleave these esters, so it is biodegradable. (2)
• Nylon-2-nylon-6 (a poly-amide of glycine and amino-caproic acid): its regularly spaced amide bonds derived in part from an $\alpha$-amino acid (glycine) are enzymatically/hydrolytically cleavable, making it biodegradable unlike ordinary nylon-6,6. (2)
• Epoxy thermosets: cure into a 3-D covalently cross-linked network; there are no linear chains to slip or dissolve and no readily hydrolysable regular repeat linkages exposed, so they cannot melt, reprocess, or be attacked by enzymes → essentially non-recyclable and non-biodegradable. (2)

[
  {"claim":"Nylon-6,6 Mn at p=0.99 with M0=226 equals 22600","code":"p=Rational(99,100); M0=226; Xn=1/(1-p); Mn=Xn*M0; result = (Mn==22600)"},
  {"claim":"Nylon-6,6 Mw at p=0.99 equals 44974 and PDI=1.99","code":"p=Rational(99,100); M0=226; Xw=(1+p)/(1-p); Mw=Xw*M0; PDI=1+p; result = (Mw==44974 and PDI==Rational(199,100))"},
  {"claim":"Sum identity sum x p^(x-1) = 1/(1-p)^2","code":"x=symbols('x'); p=symbols('p'); S=summation(x*p**(x-1),(x,1,oo)); result = simplify(S-1/(1-p)**2)==0"},
  {"claim":"CFRP E_parallel=139.4 GPa and E_perp approx 8.555 GPa","code":"Vf=Rational(6,10); Ef=230; Em=Rational(35,10); Epar=Vf*Ef+(1-Vf)*Em; Eperp=1/(Vf/Ef+(1-Vf)/Em); result = (Epar==Rational(1394,10) and abs(float(Eperp)-8.5546)<0.01)"}
]