Boyle's law doesn't work here?

[Amith2006]

Sir,
I have a have a doubt in the following problem:
# A volume V of air saturated with water vapour exerts a pressure P. If the volume is made V/2 isothermally, what will be the final pressure?
I solved it using Boyle's law but that answer which is "2P" doesn't agree with the answer given in my book which is "less than 2P but greater than P". Could you please explain why?

 

[Andrew Mason]

Sir,
I have a have a doubt in the following problem:
# A volume V of air saturated with water vapour exerts a pressure P. If the volume is made V/2 isothermally, what will be the final pressure?
I solved it using Boyle's law but that answer which is "2P" doesn't agree with the answer given in my book which is "less than 2P but greater than P". Could you please explain why?

For a particular temperature, the air is saturated with water vapour at a unique pressure (or you could say, the pressure determines the saturation temperature-the boiling point).

So, as you start at saturation pressure (ie at boiling point) and then increase the pressure (decrease the volume) keeping the temperature the same, the air loses saturation (the air cannot hold as much water). This means that the number of moles of water molecules in the air decreases (turns to liquid). Since PV=nRT, PV is not constant because n decreases. So [itex]P_f = n_fRT/Vf < n_iRT/Vf = 2niRT/Vi = 2Pi = 2P[/itex]

AM