Question

Knowing that the enthalpy of vaporization for water is 40.7 kJ/mol and its normal boiling point is 100 ^oC, calculate Vapor pressure of water at 58°C.

456 torr

14.8 torr

144 torr

5.3 torr

759 torr

Answer 1

According to Clausius-Clayperon equation,
ln (P'/P) = (-L/R) x [(1/T') - (1/T)]
Where
P' = vapour pressure at 58 oC = ?
P = vapour pressure at 100oC = 760.0 torr
R = gas constant = 8.314 J/(mol-K)
T = initial temperature = 100oC = 100+273 = 373 K
T' = final temperature = 58oC oC = 58+273 = 331 K
L = heat of vaporization of this substance = 40.7kJ*1000 J/ kJ
Plug the values we get
ln (P'/P) = (-L/R) x [(1/T') - (1/T)] = -1.66

P'/P= e^ -1.66= 0.19

P'= 0.19*760 torr= 144 torr