Question

A certain substance has a heat of vaporization of 64.13 kJ/mol. At what Kelvin temperature will the vapor pressure be 6.50 times higher than it was at 355 K?

Answer 1

we have to use the Clausius Clapeyron equation

ln(P₂/P₁) = (∆H/R)∙((1/T₁) - (1/T₂))

P₁,P₂ - the vapor pressures

lets P1 = 1, P2 = 6.5 atm

P2/P1 = 6.5/1 = 6.5

T1, T2 are absolute temperatures respectively

∆H - enthalpy of vaporization 64.13 kJ/mol we need to convert in to joules

= 64.13 x 10³Jolus / mol

R - universal gas const = 8.3145 J∙mol⁻¹∙K⁻¹

now put all these questions in above equation

ln (6.5) = (64.13 x 10³ / 8.31 ) x ∙((1/355) - (1/T₂))

1.8718 = 7.7172 x 10³ x (0.0028 - 1/T₂)

(0.0028 - 1/T₂) = 0.2425 x 10^-3

0.0028 + 0.2425 x 10^-3 = 1/T₂

1/T2 = 0.0005225

T2 = 1914 K