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

Step-by-step explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

`ln((P_2)/(P_1))=(\Delta H_(vap))/(R)((1)/(T_1)-(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 ((P_2)/(1atm))=(38560)/(2.303* 8.314J/mole.K)[(1)/(351.4K)-(1)/(288K)]`

`\log ((P_2)/(1atm))=-1.262`

`P_2=0.05470atm`

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