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23-Chem-B8 Polymer Engineering · May 2016

Question 3 of 6: Membrane Osmometry — M̄ n and Second Virial Coefficient

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: Open-book, 3 hours; six numbered problems of equal value (20 marks each), of which five constitute a complete paper (only the first five in the answer book are marked). All six problems are solved below so the set is complete for study.

Reference texts: Odian, Principles of Polymerization (4th ed., Wiley) — chain- and step-growth kinetics, emulsion polymerization, molecular-weight distributions; Rudin & Choi, The Elements of Polymer Science and Engineering (3rd ed., Academic Press) — dilute-solution properties, osmometry, viscoelasticity; Sperling, Introduction to Physical Polymer Science (4th ed., Wiley) — linear viscoelasticity, terminal-zone moduli; Bird, Armstrong & Hassager, Dynamics of Polymeric Liquids, Vol. 1 — non-Newtonian tube flow; supporting polyolefin process detail from Young & Lovell, Introduction to Polymers (3rd ed.).

Question 3: Membrane Osmometry — M̄n and Second Virial Coefficient (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Find. M̄n (from the c → 0 intercept) and the second virial coefficient A2 (from the slope) of the osmotic virial expansion.

Approach. The van 't Hoff–virial form $\pi/c = RT/\bar M_n + RT A_2\,c + \dots$ is linear in c, so a least-squares fit of π/c against c yields M̄n from the intercept and A2 from the slope. Work in CGS units to match the data (π in dyne/cm², c in g/cm³, R = 8.314×107 erg·mol-1K-1).

  1. Form π/c. Dividing each pressure by its concentration:
    c (g/cm³)0.0020.0040.0060.0080.010
    π/c (erg/g)254 000260 000263 333268 750274 000
  2. Least-squares line $\pi/c = \text{intercept}+\text{slope}\cdot c$: $$\text{slope}=2.44\times10^{6}\ \text{erg}\cdot\text{cm}^3\text{g}^{-2},\qquad \text{intercept}=2.494\times10^{5}\ \text{erg}\cdot\text{g}^{-1}$$
  3. Number-average molecular weight from the intercept, with $RT=(8.314\times10^{7})(296.15)=2.462\times10^{10}$ erg/mol: $$\bar M_n=\frac{RT}{\text{intercept}}=\frac{2.462\times10^{10}}{2.494\times10^{5}}=\boxed{9.87\times10^{4}\ \text{g}\cdot\text{mol}^{-1}}$$
  4. Second virial coefficient from the slope: $$A_2=\frac{\text{slope}}{RT}=\frac{2.44\times10^{6}}{2.462\times10^{10}}=\boxed{9.9\times10^{-5}\ \text{cm}^3\cdot\text{mol}\cdot\text{g}^{-2}}$$
concentration c (g/cm3)pi/c (erg/g)Osmotic pressure: pi/c vs c (Zimm-type plot)250k256k261k266k271k277k0.0000.0020.0040.0070.0090.011intercept = RT/Mn
Fig. 3 — osmometry plot of π/c versus c. The intercept (c → 0) gives RT/M̄n; the positive slope RT·A2 shows the solvent is thermodynamically good (A2 > 0).
QuantityValue
Intercept, RT/M̄n2.494×105 erg/g
Slope, RT·A22.44×106 erg·cm³/g²
Number-average MW, M̄n≈ 9.87×104 g/mol
Second virial coefficient, A2≈ 9.9×10-5 cm³·mol/g²