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

`PV=nRT`
`PV=nRT=(w)/(m)RT`
`1xxV=(12)/(120xxRxx(273+t)` (`i`)
`1.1xxV=(12)/(120)xxRxx(273+t+0)` (`ii`)
Dividing equation (`ii`) by equation (`i`), we get
`1.1=((273+t+10))/((273+t))`
or `t=-173^(@)C`
Putting this value in equation (`i`), we get
`1xxV=(12)/(120xx0.0821xx(273-173)`
or `V=0.821 L`