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)
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).
Form π/c. Dividing each pressure by its concentration:
c (g/cm³)
0.002
0.004
0.006
0.008
0.010
π/c (erg/g)
254 000
260 000
263 333
268 750
274 000
Least-squares line $\pi/c = \text{intercept}+\text{slope}\cdot c$:
$$\text{slope}=2.44\times10^{6}\ \text{erg}\cdot\text{cm}^3\text{g}^{-2},\qquad \text{intercept}=2.494\times10^{5}\ \text{erg}\cdot\text{g}^{-1}$$
Number-average molecular weight from the intercept, with $RT=(8.314\times10^{7})(296.15)=2.462\times10^{10}$ erg/mol:
$$\bar M_n=\frac{RT}{\text{intercept}}=\frac{2.462\times10^{10}}{2.494\times10^{5}}=\boxed{9.87\times10^{4}\ \text{g}\cdot\text{mol}^{-1}}$$
Second virial coefficient from the slope:
$$A_2=\frac{\text{slope}}{RT}=\frac{2.44\times10^{6}}{2.462\times10^{10}}=\boxed{9.9\times10^{-5}\ \text{cm}^3\cdot\text{mol}\cdot\text{g}^{-2}}$$
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).