Liquid ammonia boils at - 33.4ºC and has a heat of vaporization of 23.5 kJ/mol. What is its vapor pressure at - 50.0ºC? (R = 8.314 J/K•mol)

Question

Chemistry

Omar Zavala

19 December, 15:12

Liquid ammonia boils at - 33.4ºC and has a heat of vaporization of 23.5 kJ/mol. What is its vapor pressure at - 50.0ºC? (R = 8.314 J/K•mol)

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Jesus Hicks · Answer 1

We use the formula expressed as:

ln (P1 / P2) = - (ΔH / R) (1/T1 - 1/T2)

were P is the vapor pressures, T is the temperature, ΔH is the heat of vaporization and R is the gas constant.

ln (P1 / P2) = - (ΔH / R) (1/T1 - 1/T2)

ln (P1 / 760) = - (23.5 / 8.314) (1/223.15 - 1/239.75)

P1 = 759.33 mmHg