10 g of `O_(2)` were introduced into an evacuated vessel of 5 litre capacity maintained at `27^(@)C`. Calculate the pressure of the gas in atmosphere in the container.

Explanation

Since the volume of a gas is equal to the volume of its container, therefore, V=5 litres
Further, molecular mass of `O_(2)=32.0`, therefore, the number of moles of `O_(2)` gas are :n`=(10)/(32)`moles
We are also given : `T=27^(@)C=27+273 K=300 K`
Taking R=0.0821 litres atm `K^(-1)mol^(-1)` and using the ideal gas equation, PV=nRT
we have `" "P=(nRT)/(V)=((10//32 mol)(0.0821 L atm K^(-1)mol^(-1))(300 K))/(5 L)=1.54" atm"`.