Mystery liquid X has a vapor pressure of 238 torr at 33.0 celsius and 918 torr at 83.2 celsius . What is the normal boiling point of mystery liquid X in Kelvin?

Answer 1

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Use the Clausius Clapeyron equation to solve for ∆Hvap and then use that value and the CC equation again to solve for T2 setting P2 = 760 torr (normal boiling point pressure). Just change the 200 torr to 238 torr and the 819 torr to 918 torr and repeat the calculations.

ln (P1/P2) = ∆H/R (1/T2 - 1/T1)

P1 = 200 torr

T1 = 33 + 273 = 306K

P2 = 819 torr

T2 = 80.5 + 273 = 354K

R = 8.314 J/Kmol

∆H = ?

ln(200/819) = ∆H/8.314 (1/354 - 1/306)

-1.41 = ∆H/8.314 (-0.000448)

∆H = 26,167 J/mol = 26.2 kJ/mol

ln (P1/P2) = ∆H/R (1/T2 - 1/T1)

ln (200/760) = 26,167/8.314 (1/T2 - 1/306)

-1.34 = 3147 (1/T2 - 0.00327)

-1.34 = 3147/T2 - 10.3

T2 = 352K = 79ºC = normal boiling point