Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 279.0 Torr at 45.0 °C.

Calculate the value of ΔH°vap for this liquid.
Calculate the normal boiling point of this liquid.

Answer 1

1)
P1 = 92.0 torr
P2 = 279.0 torr
T1 = 23.0 oC= (23.0 + 273) K = 296 K
T2 = 45.0 oC = (45.0 + 273) K = 318 K

ln (P2/P1) = (ΔHo/R) *(1/T1 - 1/T2)
ln (279/92) = (ΔHo/8.314) * (1/296 - 1/318)
1.109 = (ΔHo/8.314) * (2.337*10^-4)
ΔHo = 39468 J/mol
= 39.47 KJ/mol
Answer: 39.47 KJ/mol

------------------------------------------------
2)
at boiling point P = 760 torr
P1 = 92.0 torr
P2 = 760 torr
T1 = 23.0 oC= (23.0 + 273) K = 296 K
T2 = ?

ln (P2/P1) = (ΔHo/R) *(1/T1 - 1/T2)
ln (760/92) = (39468 /8.314) * (1/296 - 1/T2)
(1/296 - 1/T2) = 4.448*10^-4
T2 = 341 K
= (341 - 273) oC
= 68 oC
Answer: 68 oC