The radioactive decay $_{83} {B i}^{211} ⟶_{81} {T i}^{207}$ , takes place in $100 L$ closed vessel at $27^{o} C$ . Starting with $2$ moles of $_{83} {B i}^{211} (t_{1 / 2} = 130 s e c)$ , the pressure development in the vessel after $520$ sec will be:

1.875

atm

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0.2155

atm

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0.4618

atm

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4.618

atm

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Solution

The correct option is C $0.4618$ atm
After 520 sec (4 half life) moles of product forms is 1.87 mole.
So pressure developed is,
$P = 1.87 * 0.082 * 300 / 100 = 0.46 a t m$

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Similar questions

Q. The radioactive decay

B i_{83}^{211} \to T l_{81}^{207}

, takes place in a

100 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:

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