The enthalpy of vaporization of Substance X is 23.0 Round your answer to 2 significant digits. 0 atm kJ and its normal boiling point is 76. °C. Calculate the vapor pressure of X at -29. °C. mol 6
Enthalpy of vaporisation of substance X, ΔHvap = 23.0 kJ/mol
= 23000 J/mol
Normal boiling point of substance X, T1 = 76 °C
= (76+273.15) K
= 349.15 K
Temperature, T2 = -29 °C
= (-29+273.15) K
= 244.15 K
Since, normal boiling point is the temperature at which vapor pressure is equal to the the atmospheric pressure.Vapour pressure at boiling point, P1= 1 atm
Vapor pressure of X at -29°C, P2 =?
Step by step
Solved in 3 steps with 4 images
