A closed vessel with rigid walls contains 1mole of 92238 U and 1 mole of air at 298 K- Considering complete decay of 92238 U to 206-82 P b- the ratio of the final pressure to the initial pressure of the system at 298 K is

^238_92U → ^206_82Pb + 8 ^4_2He(g) + 6 ^∘_ 1β n(gas) [Initial] = 1 (air) n(gas) [Final] = 8(He) + 1(air) = 9 ⇒ At constant temperature and volume p ∝ n So, p_ ...

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Standard XII

Chemistry

Law of Radioactivity

A closed vess...

Question

A closed vessel with rigid walls contains 1mole of $_{92}^{238} U$ and 1 mole of air at 298 K. Considering complete decay of $_{92}^{238} U$ to $_{82}^{206} P b$ , the ratio of the final pressure to the initial pressure of the system at 298 K is ___

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Solution

$_{92}^{238} U \to_{82}^{206} P b + 8_{2}^{4} H e (g) + 6_{- 1}^{\circ} β$

n(gas) [Initial] = 1 (air)

n(gas) [Final] = 8(He) + 1(air) = 9

$\Rightarrow$ At constant temperature and volume

$p \propto n$

So, $\frac{p_{f}}{t_{i}} = \frac{n_{f}}{n_{i}} = \frac{9}{1} = 9$

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A closed vessel with rigid walls contains 1mole of 23892U and 1 mole of air at 298 K. Considering complete decay of 23892U to 20682Pb, the ratio of the final pressure to the initial pressure of the system at 298 K is ___

23892U → 20682Pb+8 42He(g)+6 ∘−1β n(gas) [Initial] = 1 (air) n(gas) [Final] = 8(He) + 1(air) = 9 ⇒ At constant temperature and volume p∝n So, pfti=nfni=91=9

A closed vessel with rigid walls contains 1mole of $_{92}^{238} U$ and 1 mole of air at 298 K. Considering complete decay of $_{92}^{238} U$ to $_{82}^{206} P b$ , the ratio of the final pressure to the initial pressure of the system at 298 K is ___

$_{92}^{238} U \to_{82}^{206} P b + 8_{2}^{4} H e (g) + 6_{- 1}^{\circ} β$

n(gas) [Initial] = 1 (air)

n(gas) [Final] = 8(He) + 1(air) = 9

$\Rightarrow$ At constant temperature and volume

$p \propto n$

So, $\frac{p_{f}}{t_{i}} = \frac{n_{f}}{n_{i}} = \frac{9}{1} = 9$