Question

What is the heat of vaporization of a liquid that has a vapor pressure of 641 torr at 85.2
oC and a boiling point of 95.6 oC at 1.00 atm?

Answer 1

$ln(P_2/P_1)=-\Delta H_v(1/T_2-1/T_1)/R$

$P_1 =641 torr=641atm/760=0.8434atm$

$P_2=1atm$

$T_1=85.2^oC=358.35K$

$T_2=95.6^oC=368.75K$

$\Delta H_v=R*ln(P_1/P_2)/(1/T_2-1/T_1)=$

$=8.31J/(mol*K)*ln(0.8434)/(1/368.75K-1/358.35K)=$

$=17991.5J/mol=17.991kJ/mol$

$\Delta H_v=17.991kJ/mol$