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

The normal boiling point of propane is -42.0 ^oC i.e. vapor pressure of propane at -42.0 ^oC (273 – 42.0 = 231 K) is 760 mm of Hg.

Vapor pressure at 40.0 ^oC (273 + 40.0 = 313 K) can be calculated by using Clausius-Clapeyron equation. The expression to be used can be given as follows:

ln (P1/P2) = delta Hvap / R (1/T2 – 1/T1)

Where, P1 = 760 mm of Hg and T1 = 231 K

T2 = 313 K and delta Hvap = 19.04 kJ/mol = 19040 J/mol

Thus,

ln (760 / P2) = (19040 J/mol) / 8.31 J/K (1/313 – 1/231) = 2291 (-0.001134 )

ln (760 / P2) = -2.5985

760 / P2 = 0.07438

P2 = 760 / 0.07438

P2 = 10217 mm of Hg

Thus, at 40^oC vapor pressure of propane will be 10217 mm of Hg