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

Question

Stolb23 [73]

4 years ago

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)

Chemistry

1 answer:

blondinia [14] · Answer 1 · 2021-12-10T14:25:15+03:00

We use the formula expressed as:

<span>ln (P1 / P2) = - (ΔH / R) (1/T1 - 1/T2)
</span>
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 / <span>8.314</span>) (1/223.15 - 1/239.75)
P1 = 759.33 mmHg