Question

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 1

Correct Answer - 7
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"`