Answer: 0.05470atm
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 = 1 atm (standard atmospheric pressure
[tex]P_2[/tex] = final pressure at [tex]15^oC[/tex] = ?
[tex]\Delta H_{vap}[/tex] = enthalpy of vaporisation = 38.56 kJ/mol = 38560 J/mol
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex]= initial temperature = [tex]78.4^oC=273+78.4=351.4K[/tex]
[tex]T_2[/tex] = final temperature =[tex]15^oC=273+15=288K[/tex]
Now put all the given values in this formula, we get
[tex]\log (\frac{P_2}{1atm})=\frac{38560}{2.303\times 8.314J/mole.K}[\frac{1}{351.4K}-\frac{1}{288K}][/tex]
[tex]\log (\frac{P_2}{1atm})=-1.262[/tex]
[tex]P_2=0.05470atm[/tex]
Thus the vapor pressure of ethanol at [tex]15^0C[/tex] is 0.05470atm
