Question

At 25 oC, the vapor pressure of water is 23.8 torr and the heat of vaporization is 43.9 kJ/mol. Calculate the vapor pressure of water at 50. oC

Answer 1

Answer: The vapor pressure of water at
`50^0C` is 93.8 torr

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
`25^oC` = 23.8 torr

`P_2` = final pressure at
`50^oC` = ?

`\Delta H_(vap)` = enthalpy of vaporisation = 43.9 kJ/mol = 43900 J/mol

R = gas constant = 8.314 J/mole.K

`T_1` = initial temperature =
`25^oC=273+25=298K`

`T_2` = final temperature =
`50^oC=273+50=323K`

Now put all the given values in this formula, we get

`\log ((P_2)/(23.8)=(43900)/(2.303* 8.314J/mole.K)[(1)/(298K)-(1)/(323K)]`

`(P_2)/(23.8)=antilog(0.5955)`

`P_2=93.8torr`

Therefore, the vapor pressure of water at
`50^0C` is 93.8 torr