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QJEEAdv15Ch1079: JEE Advanced 2015 Chemistry - Class 12 JEEAdv15Ch1079 Solution: A closed vessel with rigid walls contains 1 mol of and 1 mol of …

A closed vessel with rigid walls contains 1 mol of 92238U{}_{92}^{238}U and 1 mol of air at 298 K. Considering complete decay of 92238U{}_{92}^{238}U to 82206Pb{}_{82}^{206}Pb , the ratio of the final pressure to the initial pressure of the system at 298 K is

Step-by-Step Verified Solution

92U238→82Pb206+8 2He4(g)+6−1β0{}_{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 24He_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+11=9= {{8 + 1} \over 1} = 9

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