A closed vessel with rigid walls contains 1 mol of 23892U and 1 mol 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-

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Standard XII

Chemistry

Avogadro's Law & Gay Lussac's Law

A closed vess...

Question

A closed vessel with rigid walls contains 1 mol of $_{92}^{238} U$ and 1 mol 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

$_{92}^{238} U$ decays to $_{82}^{206} P b$ by a series of alpha and beta emissions.
The number of alpha particles emitted by 1 $_{92}^{238} U$ atom is $\frac{238 - 206}{4} = 8$ .
Hence 1 mole of $_{92}^{238} U$ will produce 8 moles of He.
Initially, only air was present.
Hence, an initial number of moles of gaseous substances $= 1$ mole.
The final number of moles of gaseous substances $= 1$ mole air $+ 8$ mole He $= 9$ moles.
The pressure of the gas is directly proportional to the number of moles of gas.
Hence, the ratio of the final pressure to the initial pressure of the system at 298 K is $9 : 1$ .

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A closed vessel with rigid walls contains 1 mol of 23892U and 1 mol 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:

A closed vessel with rigid walls contains 1 mol of $_{92}^{238} U$ and 1 mol 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: