Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 306.0 Torr at 45.0 °C. Calculate the value of ΔH°vap for this liquid

Answer 1

Sol :-

According to clauses-clapyron equation , we have

log P₂/P₁ = ΔH_vap. / 2.303 R . [T₂ -T₁ / T₁T₂] .............(1)

P₁ = 92.0 torr Vapor pressure at temperature (T₂ = 23⁰C) 300 K

P₂ = 306.0 torr Vapor pressure at temperature (T₂ = 45⁰C) 318 K

R = Gas constant = 8.314 J K^-1 mol^-1

From equation (1), we have

log 306.0 / 92.0 = ΔH_vap. / 2.303 x 8.314 J K^-1 mol^-1 . [318 - 300 / (300). (318 )] K

log 3.326 = ΔH_vap. / (19.147 J/mol) [1.887 x 10^-4]

0.522 x 19.147 J/mol = 1.887 x 10^-4 .ΔH_vap.

ΔH_vap. = 52966.3 J/mol

ΔH_vap. = 52.97 KJ/mol