Question

A liquid has a vapor pressure of 150 torr at 60ºC. The molar heat of vaporization for this substance is

         40.8 kJ/mol. Determine the normal boiling point of this liquid.

                {ln(P₁/P₂ ) = ∆H_vap/R(T₁-T₂/T₁T₂) = ∆H_vap/R(1/T₂ - 1/T₁); ln P = 2.303 log P}

Answer 1

P = 150 torr, T = 60°C = 60+273 = 333; dHvap = 40.8 kJ/mol

a)

Find normal BP --> P = 760 torr, T = X

so

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

ln(760/150) = 40800/8.314 * (1/(60+273) - 1/T2))

-1/T2 = 8.314*ln(760/150) /40800 - 1/(60+273) = -0.002672

T2 = -(-0.002672)^-1 = 374.25149 K

T2 = 374.25149-273 = 101.2 C