To an evacuated vessel with movable piston under external pressure of 1 atm 0.1 mole of He and 1.0 mole of an 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

7

Explanation

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"`