Question 1 of 7: Viscosity-average molecular weight from dilute-solution viscometry
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-Chem-B8, Polymer Engineering — December 2014. Three hours, OPEN BOOK
(any non-communicating calculator). The paper is in four parts: Part A (Q1–2, 20 marks each),
Part B (Q3–4, 30 marks each), Part C (Q5–6, 30 marks each) and Part D (Q7, 20 marks).
A candidate answers ONE question from each of A, B, C and the single question in D — four questions
constitute a complete paper. For completeness this solution works all seven questions in full.
Reference texts: Rudin & Choi, The Elements of Polymer Science and Engineering,
3rd ed.; Sperling, Introduction to Physical Polymer Science, 4th ed.; Odian,
Principles of Polymerization, 4th ed.; Young & Lovell, Introduction to Polymers, 3rd ed.;
Painter & Coleman, Fundamentals of Polymer Science, 2nd ed.
Question 1: Viscosity-average molecular weight from dilute-solution viscometry (Part A — 20 marks)
Given. Efflux (flow) times in an Ostwald/Ubbelohde capillary viscometer for polystyrene (PS)
in toluene at 30 °C; the pure-solvent time is \(t_0 = 67.04\ \text{s}\). In a capillary viscometer
the relative viscosity is the ratio of efflux times (equal-density dilute-solution approximation),
\(\eta_r = \eta/\eta_0 = t/t_0\). Mark–Houwink constants \(K = 1.2\times10^{-4}\), \(\alpha = 0.71\).
Find. The viscosity-average molecular weight \(M_v\) from the intrinsic viscosity
\([\eta]\) via \([\eta] = K M_v^{\alpha}\).
Double extrapolation to infinite dilution: the Huggins line (\(\eta_{sp}/c\), rising) and the Kraemer line (\(\ln\eta_r/c\), falling) share a common intercept \([\eta]\approx1.28\ \text{dL/g}\).
Approach. Convert each efflux time to relative, specific, reduced and inherent
viscosities; extrapolate the reduced viscosity \(\eta_{sp}/c\) (Huggins) and inherent viscosity
\(\ln\eta_r/c\) (Kraemer) to \(c\to0\); their common intercept is the intrinsic viscosity
\([\eta]\); invert Mark–Houwink for \(M_v\).
Relative and specific viscosities. With \(\eta_r=t/t_0\) and
\(\eta_{sp}=\eta_r-1\), e.g. at \(c=0.402\): \(\eta_r=107.70/67.04=1.6065\),
\(\eta_{sp}=0.6065\). The full set of reduced viscosities \(\eta_{sp}/c\) is
\(1.509,\,1.595,\,1.648,\,1.750,\,1.987\ \text{dL/g}\) and inherent viscosities \(\ln\eta_r/c\) are
\(1.179,\,1.170,\,1.148,\,1.093,\,1.014\ \text{dL/g}\) (for \(c=0.402\ldots1.207\ \text{g/dL}\)).
Huggins extrapolation. A least-squares fit of \(\eta_{sp}/c = [\eta] + k_H[\eta]^2\,c\)
gives intercept \([\eta]_H = 1.293\ \text{dL/g}\) and slope \(0.576\), i.e. a Huggins constant
\(k_H = 0.576/[\eta]^2 = 0.34\) — the value expected for a flexible coil in a good solvent
(toluene is a good solvent for PS).
Kraemer extrapolation. Fitting \(\ln\eta_r/c = [\eta] - k_K[\eta]^2\,c\) gives
intercept \([\eta]_K = 1.272\ \text{dL/g}\). The two intercepts agree to within 2 %, confirming the
common value
$$[\eta] = \tfrac12\big([\eta]_H+[\eta]_K\big) = \boxed{1.28\ \text{dL/g}}$$
Invert Mark–Houwink. With \([\eta]=K M_v^{\alpha}\),
$$M_v = \left(\frac{[\eta]}{K}\right)^{1/\alpha}
= \left(\frac{1.28}{1.2\times10^{-4}}\right)^{1/0.71}
= \boxed{4.7\times10^{5}\ \text{g/mol}}$$
where \([\eta]\) and \(K\) are taken in the same units of dL/g (see the Verify note).
The tabulated constant
\(K=1.2\times10^{-4},\ \alpha=0.71\) is the standard literature Mark–Houwink pair for PS/toluene
when \([\eta]\) is expressed in dL/g; the problem's “cm³/g” label is a units slip.
Computing \([\eta]\) in dL/g (as done above) gives the physically reasonable \(M_v\approx4.7\times10^5\).
Taking the constant literally in cm³/g (\([\eta]=128\ \text{cm}^3/\text{g}\)) would give
\(M_v\approx3\times10^{8}\) — far above any real polystyrene — so the dL/g convention is intended.