To an evacuated vessel with movable piston under external pressure of 1 atm, 0.1 mol of He and 1.0 mol of unknown compound (vapour pressure 0.68 atm. At 0^(@)C) are introduced. Considering the ideal gas behaviour the total volume (in litre) of the gases at 0^(@)C is close to

To an evacuated vessel with movable piston under external pressure of

1.	To an evacuated vessel with movable piston under external pressure of 1 atm, 0.1 mol of He and 1.0 mol of unknown compound (vapour pressure 0.68 atm. At 0^(@)C) are introduced. Considering the ideal gas behaviour the total volume (in litre) of the gases at 0^(@)C is close to
Answer» Solution :Total volume of the gases in the VESSEL will be equal to the volume of the vessel total pressure of the gases will be equal to the external pressure of 1 atm. SUPPOSE volume of the vessel=V L Considering ideal GAS behaviour, for vapour of the compound (solid or liquid ) with 0.68 atm pressure at `0^(@)C`, PV=nRT `:.` n (no of moles of vapour) `=(PV)/(RT)=(0.68xxV)/(0.0821xx273)` `:.` Total no. of moles (He+vapour) `=0.1+(0.68V)/(0.0821xx273)` Total pressure =1 atm Applying PV=nRT for the mixture, `1xxV=(0.1+(0.68V)/(0.0821xx273))xx0.0821xx273` or `""V=0.1xx0.0821xx273+0.68" V"` or `0.32" V"=2.24" or " V= 7" L"` Alternatively, total pressure =1 atm Pressure of the vapour=0.68 atm `:.` Pressure of Helium =1-0.68=0.32 atm For ideal gas behaviour, PV=nRT `:. 0.32xxV=0.1xx0.0821xx273" or "V=7" L"`