Question

The following temperature and pressure data were collected for an unknown liquid Table: Temperature and Pressure of an Unknown Liquid Temperature (oC) Pressure (mmHg) 5 10 15 20 25 30 35 40 45 50 24.158 32.951 44.417 59.320 78.771 103.12 134.08 172.70 221.78 281.24 353.71 443.98 551.40 681.15 60 65 70 Using the relationship In Р--дн graph the data and determine the AHvap for the liquid. (Enter your answer to three significant figures.) 440.6 X kJ/mol

Answer 1

According to Clausius-Clayperon equation,
ln (P'/P) = (-L/R) x [(1/T') - (1/T)]
Where
P' = vapour pressure at 10 oC = 24.158 mm Hg
P = vapour pressure at 5 oC = 32.951 mm Hg
R = gas constant = 8.314*10^-3 kJ/(mol-K)
T' = initial temperature = 10oC=10+273=283K
T = final temperature = 5oC=5+273=278 K
L = heat of vaporization of this substance = ?
Plug the values we get
ln (P'/P) = (-L/R) x [(1/T') - (1/T)]
L = - [ln (P'/P) xR] / [ [(1/T') - (1/T)]]
= -40.6 kJ/mol
Therefore the heat of vaporization of this substance is -40.6 kJ/mol