The radioactive decay ${ }_{83}{\mathrm{Bi}}^{211}{⟶}_{81}{\mathrm{Tl}}^{207},$ takes place in 100 L closed vessel at 27°C. Starting with 2 moles of ${ }_{83}{\mathrm{Bi}}^{211}\left(t_{1/2}=130\sec \right)$,the pressure development in the vessel after 520 sec will be:

${ }_{83}{\mathrm{Bi}}^{211}{⟶}_{81}{\mathrm{Tl}}^{207}{+}_2{\mathrm{H}}^4$

Total time = n × half-life moles of substance left after n halves 

$=\frac{\text{ initial moles }}{2^n}=\frac{2}{2^4}=0.125$

mole of He produced = 2 - 0.125 = 1.875

Pressure developed due to He

$=\frac{1.875\times 0.0821\times 300}{100}=0.4618 \mathrm{atm}$