A certain liquid has a vapor pressure of 92.0 Torr at 23.0 ∘C and 257.0 Torr at 45.0 ∘C.

Question

1. Calculate the value of Δ𝐻∘vapΔ⁢Hvap∘ for this liquid.
2. Calculate the normal boiling point of this liquid.

Answer 1

Clausius Clapeyron equation

ln (P2/P1) = -∆Hvap / R (1/T2 - 1/T1)

P1 = 92.0 torr

T1 = 23.0C + 273 = 296K

P2 = 257.0 torr

T2 = 45C + 273 = 318K

R = 8.314 J/Kmol

∆Hvap = ?

ln (257/92) = -∆Hvap / 8.314 (1/318 - 1/296)

1.027 = -∆Hvap / 8.314 (0.003145 - 0.003378)

1.027 = -∆Hvap / 8.314 (-0.000233)

1.027 = 0.000233 ∆Hvap / 8.314

∆Hvap = 36,646 J/mol = 36.6 kJ/mol

To find the normal boiling point of this liquid, we use the ∆Hvap calculated in the previous part and solve for temperature at 760 torr pressure.

ln (P2/P1) = -∆Hvap / R (1/T2 - 1/T1)

P1 = 92.0 torr

T1 = 296K

P2 = 760 torr

T2 = ?

∆Hvap = 36.6 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol (note to change units to agree with those of ∆H)

ln (760/92) = -36.6 / 0.008314 (1/T2 - 1/296)

2.11 = -4402 (1/T2 - 0.003378)

2.11 = -4402/T2 + 14.87

T2 = 345K - 273 = 72ºC