Question

Propane has a normal boiling point of -42.0 ∘C and a heat of vaporization (ΔHvap) of 19.04 kJ/mol. What is the vapor pressure of propane at 40.0 ∘C

Answer 1

By definition, the vapor pressure of a substance at its normal boiling point is 1 atm.

Clausius-Clapeyron Equation is

ln (P₂ / P₁) = (ΔH / R) (1/T₁ - 1/T₂)

P1 = 1 atm

T1 = -42.0 ∘C = -42 + 273 = 231 K

P2 = ?

T2 = 40.0 ∘C = 40 + 273 = 313 K

ΔHvap = 19.04 kJ/mol = 19040 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) = (19040 / 8.314) (1/231 - 1/313)

P2 = 13.42 atm

Therefore, vapor pressure of propane at 40.0 ∘C = 13.42 atm