Answer :
Answer: The vapor pressure of water at 298 K is 3.565kPa.
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 298 K = ?
[tex]P_2[/tex] = final pressure at 373 K = 101.3 kPa
[tex]\Delta H_{vap}[/tex] = enthalpy of vaporisation = 41.1 kJ/mol = 41100 J/mol
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = 298 K
[tex]T_2[/tex] = final temperature = 373 K
Now put all the given values in this formula, we get
[tex]\log (\frac{101.3}{P_1})=\frac{41100}{2.303\times 8.314J/mole.K}[\frac{1}{298K}-\frac{1}{373K}][/tex]
[tex]\frac{101.3}{P_1}=antilog(1.448)[/tex]
[tex]P_1=3.565kPa[/tex]
Therefore, the vapor pressure of water at 298 K is 3.565kPa.