The pressure exerted by `12 g` of an ideal gas at temperature `t^(@)C` in a vessel of volume `V` litre is `1 atm`. When the temperature is increased by `10^(@)C` at the same volume, the pressure increases by `10%`. Calculate the temperature `T` and volume `V`. (Molecular weight of the gas is `120`).

Explanation

`"Moles of gas"=(12)/(120)="0.1 mole"`
`p="1 atm, T"=t^(@)C=(273+t)K," Volume = V"`
According to ideal gas equation,
`pV=nRT`
`therefore" "1xxV=0.1xxRxx(273+t)" ….(i)"`
`"Now, p"=1+(10)/(100)="1.1 atm, T"=t+273+10=283+tK`
`"Volume"="V (same)"`
`therefore" "1.1xxV=0.1xxRxx(288+t)" ....(ii)"`
Dividing Eq. (ii) by (i), we have
`(1.1)/(1)=(283+t)/(273+t)`
`283+t=1.1(273+t)=308.3+1.1t`
`"or "-0.1t=17.3`
`"or "t=-173^(@)C`
Substituting the value of t in equation (i):
`1xxV=0.1xx0.082xx(273-173)`
`"or "V=0.1xx0.082xx100`
`=0.82L`