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  • The radioactive decay <img alt="_{Z}^{A}\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}" src="https://entrancecorner.oncodecogs.com/gif.latex?_%7BZ%7D%5E%7BA%7D%5Ctextrm%7BX%7D%5Crightarrow%20_%7BZ-2%

The radioactive decay _{Z}^{A}\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y} takes place in 200 \mathrm{~L} closed vessel at \mathrm{27^{\circ} \mathrm{C}}. Starting with 3 moles of \mathrm{X\left(t_{1 / 2}=150\right)}, the pressure developed in the vessel after 10 minutes, will be:

Option: 1

0.352 \mathrm{~atm}


Option: 2

0.462 \mathrm{~atm}


Option: 3

0.5 \mathrm{~atm}


Option: 4

3.52 \mathrm{~atm}


Answers (1)

best_answer

\quad _{Z}^{A}\textrm{X}\rightarrow{ }_{2-2}^{A-4} Y+{ }_{2}^{4} \mathrm{He}

\mathrm{Total \: time =n \times t_{1 / 2} \Rightarrow 10 \times 60=n \times 150 \Rightarrow n=4}

Moles of Substance left after \mathrm{n} halves \mathrm{=\frac{n_{\text {initial }}}{2^{n}}=\frac{3}{2^{4}}=\frac{3}{16}}

\mathrm{\text{Moles of the produced} =3-\frac{3}{16}=\frac{45}{16}}
\mathrm{Pressure \: P=\frac{45}{16} \times \frac{1}{12} \times \frac{300}{200}=\frac{45}{128}=0.352 \mathrm{~atm}}

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