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
Answer: 9
Solution
The decay equation is:
²³⁸₉₂U → ²⁰⁶₈₂Pb + 8 ⁴₂He + 6 ⁰₋₁e
Initial gaseous moles = 1 mol (air)
From 1 mol of U, 8 mol of He gas is produced.
Final gaseous moles = 1 + 8 = 9 mol
At constant volume and temperature:
Pf/Pi = nf/ni = 9/1 = 9.
²³⁸₉₂U → ²⁰⁶₈₂Pb + 8 ⁴₂He + 6 ⁰₋₁e
Initial gaseous moles = 1 mol (air)
From 1 mol of U, 8 mol of He gas is produced.
Final gaseous moles = 1 + 8 = 9 mol
At constant volume and temperature:
Pf/Pi = nf/ni = 9/1 = 9.
JEE Advanced · 2015 · 24 May 2015 · Paper 2 · Integer answer