Answer: Normal boiling point of the substance is 248 K
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:
[tex]ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})[/tex]
where,
[tex]P_1[/tex]= initial pressure at normal boiling point= 1 atm (standard atmospheric pressure
[tex]P_2[/tex] = 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
[tex]T_1[/tex]= normal boiling point = ?
[tex]T_2[/tex] = boiling point at pressure of 0.138 atm = 371 K
Now put all the given values in this formula, we get
[tex]\log (\frac{1atm}{0.138atm})=\frac{12300}{2.303\times 8.314J/mole.K}[\frac{1}{T_1}-\frac{1}{371}][/tex]
[tex]0.860=\frac{12300}{2.303\times 8.314J/mole.K}[\frac{1}{T_1K}-\frac{1}{371K}][/tex]
[tex]T_1=248K[/tex]
Thus the normal boiling point of the substance in kelvin is 248
