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ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2019/paper-1/section-b/10/i/a/solution

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Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2019 / Paper 1 / Section B / 10 / i / a / Solution by

Codex  0 2026-10-20

For identical spherical drops, total oil volume and interfacial area satisfy

V_{o} = N \frac{4 π r ^{3}}{3}, A = N 4 π r^{2} = \frac{3 V _{o}}{r}, E_{s} = γ_{o w} A .

Taking

10 vol %

to mean the oil fraction of the final emulsion,

V_{o} / (V_{o} + V_{w}) = 0.10

gives

V_{o} = V_{w} /9 = 1.111 \times 1 0^{- 4} m^{3}

. Thus

A = 333.3 m^{2}

and $E_{s} = 16.7 J$ . This corresponds to about

2.65 \times 1 0^{13}

drops.

There is a wording convention worth making explicit: if the oil addition instead means

0.10

times the initial water volume, then

V_{o} = 100 mL

,

A = 300 m^{2}

and $E_{s} = 15.0 J$ . These differ only in the stated volume-fraction convention. The calculation uses the oil–water interface and neglects the container and air interfaces.

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