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Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2017 / Paper 1 / Section B / 10 / iv / a / Solution 2026-10-20

Use the normal boiling point as the reference,

T_{B} = 373.15 K

and

P_{B} = 101325 Pa

. At

T = 313.15 K

,

P_{sat} ≃ 101325 exp [\frac{41000}{8.314} (\frac{1}{373.15} - \frac{1}{313.15})] = 8.05 \times 1 0^{3} Pa .

This is approximately 0.0795 atmosphere. It uses the supplied constant enthalpy of vaporization approximation across a 60 K interval; it should not be presented as an exact tabulated vapor pressure. Absolute temperature, rather than Celsius values, belongs in the exponential.

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