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

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

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