Question

A substance has a vapor pressure of 0.138 atm at 371 K. What is the normal boiling point of the substance in kelvin? Normal boiling is when vapor pressure = 1 atm or 760 mmHg. ΔHvap = 12.3 kJ/mol

Answer 1

Answer: Normal boiling point of the substance is 248 K

Step-by-step explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

`ln((P_2)/(P_1))=(\Delta H_(vap))/(R)((1)/(T_1)-(1)/(T_2))`

where,

`P_1`= initial pressure at normal boiling point= 1 atm (standard atmospheric pressure

`P_2` = final pressure at 371 K= 0.138 atm

= enthalpy of vaporisation = 12.3 kJ/mol = 12300 J/mol

R = gas constant = 8.314 J/mole.K

`T_1`= normal boiling point = ?

`T_2` = boiling point at pressure of 0.138 atm = 371 K

Now put all the given values in this formula, we get

`\log ((1atm)/(0.138atm))=(12300)/(2.303* 8.314J/mole.K)[(1)/(T_1)-(1)/(371)]`

`0.860=(12300)/(2.303* 8.314J/mole.K)[(1)/(T_1K)-(1)/(371K)]`

`T_1=248K`

Thus the normal boiling point of the substance in kelvin is 248