Question

Propane has a normal boiling point of -42.04 °C and a heat of vaporization of 24.54 kJ/mole. What is the vapor pressure of propane at 25.0 °C?

Answer 1

Answer : The vapor pressure of propane at $25.0^oC$ is 17.73 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 propane at $25.0^oC$ = ?

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

$T_1$ = temperature of propane = $25.0^oC=273+25.0=298.0K$

$T_2$ = normal boiling point of propane = $-42.04^oC=230.96K$

$\Delta H_{vap}$ = heat of vaporization = 24.54 kJ/mole = 24540 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{1atm}{P_1})=\frac{24540J/mole}{8.314J/K.mole}\times (\frac{1}{298.0K}-\frac{1}{230.96K})$

$P_1=17.73atm$

Hence, the vapor pressure of propane at $25.0^oC$ is 17.73 atm.