A vessel containing water is put in a dry sealed room of volume `76m^(3)`at a temperture of `15^(@)C`. The saturation vapour pressure of water at `15^(@)C`. is 15mm of mercury. How much water will evaporate before the water is in equilibrium with the vapour?

Explanation

Water will be in equilibrium with its vapour when the vapour gets saturated, In this case, the pressure of vapour =saturation vapour pressure =15mm of mercury
`(15xx10^(-3))(13600kg m^(-3))(9.8m s^(-2))`
`=2000Nm^(-2)`
Using gas law , `pV=(m)/(M)RT`
`m=(MpV)/(RT)`
`((18g mol^(-1))(2000N M^(-2))(76m^(3)))/((8.3JK^(-1)mol^(-1))(288K))`
`=1145g=1.14kg.`
Thus, `1.14kg` of water will evaporate.