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(D) The nuclear decay reaction for ${ }_{92}^{238} U$ to ${ }_{82}^{206} Pb$ is given by: ${ }_{92}^{238} U \rightarrow { }_{82}^{206} Pb + 8 { }_{2}^

Vedclass · Accepted Answer

$9$

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(D) The nuclear decay reaction for ${ }_{92}^{238} U$ to ${ }_{82}^{206} Pb$ is given by: ${ }_{92}^{238} U \rightarrow { }_{82}^{206} Pb + 8 { }_{2}^{4} He + 6 { }_{-1}^{0} e$.Initially, the system contains $1 \ mol$ of air (gaseous) and $1 \ mol$ of solid ${ }_{92}^{238} U$.Since the vessel is rigid and the temperature is constant at $298 \ K$, the pressure is directly proportional to the number of moles of gas present $(P \propto n_{gas})$.Initial moles of gas $(n_i)$ = $1 \ mol$ (air).After complete decay, the solid ${ }_{92}^{238} U$ is converted into $1 \ mol$ of solid ${ }_{82}^{206} Pb$ and $8 \ mol$ of gaseous ${ }_{2}^{4} He$ ($\alpha$-particles).The final moles of gas $(n_f)$ = $1 \ mol$ (air) + $8 \ mol$ (He) = $9 \ mol$.The ratio of the final pressure to the initial pressure is given by $P_f / P_i = n_f / n_i = 9 / 1 = 9$.

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

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