Answer to Question #231522 in General Chemistry for Savannah

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 30.0 ∘C?Express the pressure in torrs to three significant figures.

Answer 1

T1 = -42.0 + 273.15 = 231.15 K

T2 = 25 + 273.15 = 298.15

∆Hvap = 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) = (∆Hvap/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 = e2.22 = 9.207 atm

Answer: 9.207 atm