Clausius - Clapeyron Equation

Question

At 34.0 °C, the vapor pressure of isooctane is 10.0 kPa, and

at 98.8 °C, its vapor pressure is 100.0 kPa. Use this

information to calculate the enthalpy of vaporization for

isooctane in kJ/mol.

Answer 1

ln (P2/P1) = -∆Hvap/R(1/T2 - 1/T1) .. Clausius-Claypeyron equation

P1 = 10.0 kPa

P2 = 100.0 kPa

T1 = 34.0ºC + 273 = 307K

T2 = 98.8ºC + 273 = 372K

R = 8.314 J/Kmol = 0.008314 kJ/Kmol

ln (100/10) = -∆Hvap/0.008314 (1/372 - 1/307)

2.303 = -∆Hvap/0.008314 (0.002688 - 0.003257)

2.303 = -∆Hvap/0.008314 (-0.000569)

∆Hvap = 33.7 kJ/mole

(be sure to check all of the math)