Question

Ethanol has a heat of vaporization of 38.56kj/mol and a normal boiling point of 78.4 ∘c.

Answer 1

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 $14.0^oC$ is $5.174\times 10^{-2}atm$

Explanation :

The Clausius- Clapeyron equation is :

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

where,

$P_1$ = vapor pressure of ethanol at $14.0^oC$ = ?

$P_2$ = vapor pressure of ethanol at normal boiling point = 1 atm

$T_1$ = temperature of ethanol = $14.0^oC=273+14.0=287K$

$T_2$ = normal boiling point of ethanol = $78.4^oC=273+78.4=351.4K$

$\Delta H_{vap}$ = 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:

$\ln (\frac{1atm}{P_1})=\frac{38560J/mole}{8.314J/K.mole}\times (\frac{1}{287K}-\frac{1}{351.4K})$

$P_1=5.174\times 10^{-2}atm$

Hence, the vapor pressure of ethanol at $14.0^oC$ is $5.174\times 10^{-2}atm$