Question

The enthalpy of vaporization for water is given by ∆Hvap = 44.0 kJ/mol. Given the normal boiling point of water as 100.0 Celsius, estimate the vapor pressure of water at 80.0 Celcius.

Answer 1

By definition, the vapor pressure of a substance at its normal boiling point is 760 mmHg or 760 torr or

1 atm.

Clausius-Clapeyron Equation is

ln (P₂ / P₁) = (ΔH / R) (1/T₁ - 1/T₂)

P1 = 1 atm

T1 = 100 ∘C = 100 + 273 = 373 K

P2 = ?

T2 = 80 ∘C = 80 + 273 = 353 K

ΔHvap = 44 kJ/mol = 44000 J/mol

R= 8.314 J/K/mol

Substitute all the values in ln (P₂ / P₁) = (ΔH / R) (1/T₁ - 1/T₂)

ln (P₂ / 1) = (44000 / 8.314) (1/373 - 1/353)

P2 = 0.447 atm

Therefore, vapor pressure of water at 80 ∘C = 0.447 atm