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 ((P_2)/(P_1))=(\Delta H_(vap))/(R)* ((1)/(T_1)-(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 ((1atm)/(P_1))=(24540J/mole)/(8.314J/K.mole)* ((1)/(298.0K)-(1)/(230.96K))`

`P_1=17.73atm`

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