Question

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?

Answer 1

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.