Question

Carbon disulfide has a molar enthalpy of vaporization of +29.2 kJmol^-1. At 268 K it has a vapor pressure of 100mmHg. Calculate the normal temperature of carbon disulfide.

Answer 1

n(p₁/p₂) = (∆Hv/R)(1/T₂ - 1/T₁)
where
p₁,p₂ are the vapor pressure at T₁ and at T₂ respectively, ∆Hv heat of vaporization, R universal gas constant.

For this problem let
p₂= 100mmHg at T₂=268K
The pressure at normal boiling point equals standard atmospheric pressure. i.e.
p₁ =760mmHg at T₁=Tb

Solve the equation above for T₁:
ln(p₁/p₂) = (∆Hv/R)∙(1/T₂ - 1/T₁)
<=>
(1/T₂ - 1/T₁) = (R/∆Hv)∙ln(p₂/p₁)
<=>
Tb = T₂ = 1/[ 1/T₂ - (R/∆Hv)∙ln(p₂/p₁) ]
= 1/[ 1/268K - (8.314472J/molK / 29200J/mol)∙ln(760/100) ]
= 317K