Question

The radioactive decay, ${}_{83}B{i}^{211}\to {}_{81}T{l}^{207},$ takes place in 10 L closed container at $0^{\circ }C.$ Half - life of ${}_{83}B{i}^{211}$ is 2.15 min. Starting with 1 mole of ${}_{83}B{i}^{211}$, the pressure developed in the container after 4.30 min. will be

• A. 16.8 atm
• B. 22.4 atm
• C. 2.24 atm
• D. 1.68 atm

Answer 1

The correct option is D

1.68 atm

${}_{83}B{i}^{211}\to {}_{81}T{l}^{207}{+}_{2}H{e}^4$
Fraction left $=(\frac{1}{2}{)}^n=(\frac{1}{2}{)}^{\frac{4.30}{2.15}}=\frac{1}{4}$
Moles of He produced = moles of Bi decayed
$=1-\frac{1}{4}=0.75$
Pressure developed due to He
$=\frac{22.4\times 0.75}{10}=1.68atm$