A closed vessel of fixed volume is filled with 3.2g `O_(2)` at a pressure of P atm and at a temperature of TK. The container is then heated to a temperature of (T+30)K. to maintain the pressure of P atm inside the container at (T+30)K, a certain amount of gas is removed from the container. the gas removed is found to have a volume of 246 mL at 1 atm and `27^(@)C`. calculate T.

Explanation

Suppose, number of moles of `O_(2)` removed`=n_(1)mol`
`therefore n_(1)=(PV)/(RT)=(1xx246xx10^(-3))/(0.0821xx300)=9.98xx10^(-3)mol`
Number of moles of oxygen remained in the container
`(n_(2))=((3.2)/(32)-9.98xx10^(-3))mol=0.09mol`
Initial state: `PV=nRT=(3.2)/(32)RT=0.1RT`
Final state: `PV=n_(2)RT=0.09xxR(T+30)`
`therefore 0.1 RT=0.09xxR(T+30):` hence, T=270 K