Question

A certain substance has a heat of vaporization of 63.25 kJ/mol. At what Kelvin temperature will the vapor pressure be 5.00 times higher than it was at 363 K?

Answer 1

Here we will use the Clasius- Clapeyron equation--

ln ( P₁ / P₂) = H_vap / R ( 1 /T₂ - 1 /T₁)

where

P₁ is the vapor pressure measured at T1;
P₂ is the vapor pressure measured at P2;
H_vap is the heat of vaporization;
R is the gas constant = 8.314 J/molK

Since the pressure measured at this new temperature will be 5.00 times higher than P₁ , so P₂ = 5P₁

Now,

ln ( P₁ / 5P₁) = 63250 Jmol^-1 / 8.314 Jmol^-1K^-1 ( 1 /T₂ -1/363 K)

=> ln ( 1/5) = 7607.6 x 1/T₂ -7607.6 x 1 / 363

=> -1.6094 = (7607.6/T₂ ) - 20.9575

=> -1.6094 + 20.9575 = 7607.6/T₂

=> 19.3481 = 7607.6/T₂

=> T₂ =  7607.6 / 19.3481

=> T₂ = 393.19 K = 393 K