Answer to Question #91988 in Chemistry for Shali

Question

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

Answer 1

According to Clausius Clapeyron equation:

ln(P₂/P₁) = (-∆H_vap)/R × (1/T₂ - 1/T₁),

where T₁ - boiling point (-42°C = 315.15 K), T₂ - temperature (25°C = 298.15 K), ∆H_vap - heat of vaporizayion (19.04 Kj/mol), R - gas constant (8.314 J/mol K), P₁ - atmospheric pressure (1 atm = 101.325 kPa), P₂ - vapor pressure (?).

From here:

ln(P₂) - ln(P₁) = (-∆H_vap)/R × (1/T₂ - 1/T₁)

ln(P₂) = (-∆H_vap)/R × (1/T₂ - 1/T₁) + ln(P₁)

P₂ = e^{[(-∆Hvap)/R × (1/T2 - 1/T1) + ln(P1)]}

P₂ = 939 kPa

Answer: 939 kPa