JEE Advance - Chemistry (2015 - Paper 2 Offline - No. 2)
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 $${}_{82}^{206}Pb$$, the ratio of the final pressure to the initial pressure of the system at 298 K is
Answer
9
Explanation
$${}_{92}{U^{238}} \to {}_{82}P{b^{206}} + 8\,{}_2H{e^4}(g) + 6{}_{ - 1}{\beta ^0}$$
To calculate pressure, only gaseous products need to be considered.
Initially, only 1 mol of air is present and finally, after complete decay, 8 moles of $$_2^4He$$ gas are produced and 1 mol of air will also remain in the mixture.
Ratio of the final pressure to the initial pressure $$ = {{8 + 1} \over 1} = 9$$
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