Question

If 1.00 mL of water is placed in a 5.00 L closed flask at 26°C, what volume of water would remain as liquid after equilibrium is allowed to establish between the liquid water and the water vapor in the flask ? (The vapor pressure of water at 26°C is 25.2 torr, and the density of water at 26°C is 1.00 g/mL.)

Answer 1

at Equilibrium, partital pressrue of water vapor= vapor pressrue of liquid = 25.2 Torr= 25.2 mm Hg= 25.2/760 atm

=0.033 atm. Partial pressure is the presssure exerted it the gas alone occupies the entire volume, Temperature= 26 deg.c= 26+273= 299 K, V= Volume of flask , R = gas constant =0.0821 L.atm/mole.K

n= moles of water vapor =PV/RT =0.033*5/(0.0821* 299)=0.006722 moles

mass of water vapor =moles of water vapor* molar mass =0.006722*18 gm =0.121 gm

mass of liquid water = volume of liquid* density of liquid= 1*1= 1 gm

mass of liquid water remaining after equilibrium = 1-0.121 gm = 0.879 gm

volume of liquid remaining= 0.879 gm/1gm/ml=0.879 ml