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Stolb23 [73]
3 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]3 years ago
3 0
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
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