 Solution

ID: 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

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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