How to Find Heat of Vaporization

Heat (or enthalpy) of vaporization is the energy needed to transform one mole of substance into gas. The heat of vaporization can be found using the Clausius-Clapeyron equation if the vapor pressure over the liquid phase is known at several temperatures: ln(P1 / P2 ) = Hvap/R x (1/T2 - 1/T1). "P1"is the vapor pressure at the temperature "T1" measured in Kelvin(K), and "P2"is the vapor pressure at temperature "T2." "Hvap" is the molar enthalpy (heat) of vaporization. "R" is the molar gas constant that is equals to 8.3145 J/mol x K. As an example, we'll find the heat of vaporization if the vapor pressure is 0.66 atm at 346 K and 0.55 atm at 343 K.

Instructions

1

Divide the pressure at the first point by the pressure at the second point. In our example, it is P1/P2 = 0.66 atm / 0.55 atm = 1.2.

2

Calculate the natural logarithm of the value from Step 1 using the "ln" function on your calculator. In the example, ln1.2 = 0.18232.

3

Divide 1 by the first temperature. In the example, 1/346 = 0.00289.

4

Divide 1 by the second temperature. In the example, 1/343 = 0.00291.

5

Subtract the value from Step 4 from the one in Step 3. In our example, the difference is 0.00291 -- 0.00289 = 0.00002.

6

Divide the value from Step 1 by the difference from Step 5, and then multiply the quotient by the constant "R" to find the heat of vaporization. In our example, heat of vaporization (Hvap) = (0.18232 / 0.00002) x 8.3145 = 75795 J/mole = 75.795 kJ/mole.