Question

Entalpy of vaporization of water is 41.1k/mol. if the vapor pressure of water at 373k is 101.3 kpa, what is the vapor pressure of water at 298k?

Answer 1

Answer: The vapor pressure of water at 298 K is 3.565kPa.

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

$ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = initial pressure at 298 K = ?

$P_2$ = final pressure at 373 K = 101.3 kPa

$\Delta H_{vap}$ = enthalpy of vaporisation = 41.1 kJ/mol = 41100 J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 298 K

$T_2$ = final temperature = 373 K

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

$\log (\frac{101.3}{P_1})=\frac{41100}{2.303\times 8.314J/mole.K}[\frac{1}{298K}-\frac{1}{373K}]$

$\frac{101.3}{P_1}=antilog(1.448)$

$P_1=3.565kPa$

Therefore, the vapor pressure of water at 298 K is 3.565kPa.