Respuesta :
This is an incomplete question, here is a complete question.
Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 °C. What is the vapor pressure of ethanol at 14 °C?
Answer : The vapor pressure of ethanol at [tex]14.0^oC[/tex] is [tex]5.174\times 10^{-2}atm[/tex]
Explanation :
The Clausius- Clapeyron equation is :
[tex]\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})[/tex]
where,
[tex]P_1[/tex] = vapor pressure of ethanol at [tex]14.0^oC[/tex] = ?
[tex]P_2[/tex] = vapor pressure of ethanol at normal boiling point = 1 atm
[tex]T_1[/tex] = temperature of ethanol = [tex]14.0^oC=273+14.0=287K[/tex]
[tex]T_2[/tex] = normal boiling point of ethanol = [tex]78.4^oC=273+78.4=351.4K[/tex]
[tex]\Delta H_{vap}[/tex] = heat of vaporization = 38.56 kJ/mole = 38560 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
[tex]\ln (\frac{1atm}{P_1})=\frac{38560J/mole}{8.314J/K.mole}\times (\frac{1}{287K}-\frac{1}{351.4K})[/tex]
[tex]P_1=5.174\times 10^{-2}atm[/tex]
Hence, the vapor pressure of ethanol at [tex]14.0^oC[/tex] is [tex]5.174\times 10^{-2}atm[/tex]
