The radioactive decay  83Bi211⟶81Tl207, takes place in 100 L closed vessel at 27°C. Starting with 2 moles of  83Bi211t1/2=130sec,the pressure development in the vessel after 520 sec will be:

83Bi211⟶81Tl207+2H4Total time = n × half-life moles of substance left after n halves = initial moles 2n=224=0.125mole of He produced = 2 - 0.125 = 1.875Pressure developed due to He=1.875×0.0821×300100=0.4618 atm