Question

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 °C and 262.0 Torr at 45.0 °C. Calculate the value of ΔH°vap for this liquid.

Calculate the normal boiling point of this liquid.

Answer 1

for calculating Hvap :
P1 = 92 torr
T1 = 23.0 oC = 296.15 K
P2 = 262 torr
T2 = 45 oC = 318.15 K

use:
ln (P2/P1) = (HVap/R)*(1/T1 - 1/T2)
ln(262/92) = (HVap/8.314)*(1/296.15 - 1/318.15)
1.047 = (HVap/8.314)*(2.335*10^-4)
Hvap = 37280 J/mol
= 37.28 KJ/mol
Answer: 37.28 KJ/mol

To calculate boiling point,
pressure should be 760 torr
P1 = 92 torr
T1 = 23.0 oC = 296.15 K
P2= 760 torr
T2 = boiling point = ?
Hvap = 37280 J/mol

use:
ln (P2/P1) = (HVap/R)*(1/T1 - 1/T2)
ln(760/92) = (37280/8.314)*(1/296.15 - 1/T2)
(1/296.15 - 1/T2)= 4.709*10^-4
T2 = 344.14 K
=71.0 oC
Answer: Boiling point is 71.0 oC