Answer to Question #129515 in General Chemistry for MEGA

Question

Propane (C3H8) has a normal boiling point of -42.0°C and a heat of Vaporization of 19.04Kj/mol. What is the Vapour Pressure of propane at 25.0°C?

Answer 1

T₁ = -42.0 + 273.15 = 231.15 K

T₂ = 25 + 273.15 = 298.15

∆H_vap = 19.04 kJ/mol = 19040 J/mol

At normal boiling point, the vapor pressure is equal to 1 atm.

p₁ = 1 atm

Clausius Clapeyron equation:

ln(p₂/p₁) = (∆H_vap/R)×(1/T₁ – 1/T₂)

R — the gas constant (8.3145 J/mol K).

ln(p₂/1) = (19040/8.3145)×(1/231.15 – 1/298.15)

ln(p₂/1) = 2289.97×(0.00432 – 0.00335)

ln(p₂/1) = 2289.97×0.00097

ln(p₂/1) = 2.22

p₂ = e^2.22 = 9.207 atm

Answer: 9.207 atm