Question

A certain substance has a heat of vaporization of 39.90 kJ/mol. At what Kelvin temperature will the vapor pressure be 5.50 times higher than it was at 325 K?

Answer 1

Given:

Delta Hvap = 39.90 kJ/mol = 39900 J /mol

Vapor pressure of substance is 5.50 time higher than that of the vapor pressure at 325 K

Solution:

We use Classius clapeyron equation to get temperature at which the pressure is 5.50 times.

Equation:

Ln (p2/p1) = (-Delta H/R) x ( 1/T2-1/T1)

Here p2 is the vapor pressure at Temperature T2, P1 is vapor pressure at temperature T1.

Lets P2/p1 = 5.50

Delta H = heat of vaporization in J /mol , R = 8.314 J / (K mol ) , T is in Kelvin.

Let T1 = 325 K.

Lets plug all the values to get T2.

Ln (5.50) = (- 39900 / 8.314 ) x ( 1/T2 - 1/ 325 )

T2 = 367.42 K

The temperature of the substance will be = 367.42 K