Question

The radioactive decay $B{i}_{83}^{211}\to T{l}_{81}^{207}$, takes place in a $100\text{L}$ closed vessel at ${27}^{\circ }C$. Starting with 2 moles of $B{i}_{83}^{211}$ ($t_{1/2}$ = 130 seconds), the pressure developed in the vessel after 520 seconds will be:

• A. $\text{1.875 atm}$
• B. $\text{0.2155 atm}$
• C. $\text{0.4618 atm}$
• D. $\text{4.618 atm}$

Answer 1

The correct option is C $\text{0.4618 atm}$

${}_{83}^{211}Bi\to {}_{81}^{207}Tl+{}_2^{4}He$
t=0 $2$
t=t $2-x$ $x$ $x$

$t_{\frac{1}{2}}$ = 130 seconds
$2\stackrel{130}{⟶}1\stackrel{130}{⟶}\frac{1}{2}\stackrel{130}{⟶}\frac{1}{4}\stackrel{130}{⟶}\frac{1}{8}$
$\frac{1}{8}$ is the remaining amount
$2-x$ = $\frac{1}{8}$
$x=\frac{15}{8}$
$PV=nRT$
$P\times 100$ = $\frac{15}{8}\times 0.0821\times 300$
$P=0.4618atm$