Question

The normal boiling point of ethanol is 78.4 oC and its heat of vaporization is 38.56 kJ/mol. What is the vapor pressure of ethanol at 15 oC?

Answer 1

Answer: 0.05470atm

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

$ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = initial pressure at  = 1 atm (standard atmospheric pressure

$P_2$ = final pressure at $15^oC$ = ?

$\Delta H_{vap}$ = enthalpy of vaporisation = 38.56 kJ/mol = 38560 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$= initial temperature = $78.4^oC=273+78.4=351.4K$

$T_2$ = final temperature =$15^oC=273+15=288K$

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

$\log (\frac{P_2}{1atm})=\frac{38560}{2.303\times 8.314J/mole.K}[\frac{1}{351.4K}-\frac{1}{288K}]$

$\log (\frac{P_2}{1atm})=-1.262$

$P_2=0.05470atm$

Thus the vapor pressure of ethanol at $15^0C$ is 0.05470atm