Calculate the boiling point of a liquid

Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 ∘C and 354.0 Torr at 45.0 ∘C.

calculate the value of this liquid - kJ/mol

Calculate the normal boiling point of this liquid.- degress C

Answer 1

Clausius Clapeyron equation:

ln(P2/P1) = - ∆H_vap/R (1/T2 - 1/T1) and we must solve for ∆H_vap

P1 = 92.0 torr

P2 = 354 torr

T1 = 23+273 = 296K

T2 = 45+273 = 318K

R = 8.314 J/kmol

ln(354/92) = -∆Hvap/8.314 (1/318 - 1/296)

1.3475 = -∆Hvap/8.314 (0.003145 - 0.003378)

1.3475 = -∆Hvap/8.314 (-0.000233)

∆H_vap = (1.3475)(8.314)/0.000233 = 48,082 J = 48.1 kJ

Normal boiling will be when pressure is 760 torr. Use the C-C equation again and solve for T2

P1 = 92.0 torr

P2 = 760 torr

T1 = 296K

T2 = ?

R = 8.314 J/Kmol

∆Hvap = 48,082 J

ln(P2/P1) = - ∆Hvap/R (1/T2 - 1/T1)

ln(760/92) = -(48082/8.314) (1/T2 - 1/296)

2.112 = -5783 (1/T2 - 0.003378)

T2 = 332K = 59ºC