Answer: The vapor pressure of water at [tex]50^0C[/tex] is 93.8 torr
Explanation:
The vapor pressure is determined by Clausius Clapeyron equation:
[tex]ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})[/tex]
where,
[tex]P_1[/tex] = initial pressure at [tex]25^oC[/tex] = 23.8 torr
[tex]P_2[/tex] = final pressure at [tex]50^oC[/tex] = ?
[tex]\Delta H_{vap}[/tex] = enthalpy of vaporisation = 43.9 kJ/mol = 43900 J/mol
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = [tex]25^oC=273+25=298K[/tex]
[tex]T_2[/tex] = final temperature = [tex]50^oC=273+50=323K[/tex]
Now put all the given values in this formula, we get
[tex]\log (\frac{P_2}{23.8}=\frac{43900}{2.303\times 8.314J/mole.K}[\frac{1}{298K}-\frac{1}{323K}][/tex]
[tex]\frac{P_2}{23.8}=antilog(0.5955)[/tex]
[tex]P_2=93.8torr[/tex]
Therefore, the vapor pressure of water at [tex]50^0C[/tex] is 93.8 torr
