Question

An unknown liquid has a vapor pressure of 88 mmhg at 45°c and 39 mmhg at 25°c. What is its heat of vaporization?

Answer 1

The relation between vapour pressure , enthalpy of vapourisation and temperature is

$ln(\frac{P1}{P2} ) = \frac{deltaH}{R} (\frac{1}{T2} - \frac{1}{T1})$

ln (88/ 39) = DeltaH / 8.314 (1 / 318 - 1 / 298)

0.814 = DeltaH / 8.314 (2.11 X 10^-4 )

DeltaH = -32.07 kJ