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

During the reaction, ${ }_{92}^{238}\mathrm{U}{\to }_{82}^{206}\mathrm{Pb}$ happens by a series of alpha and beta emissions. 

The number of alpha particles emitted by ${}_{92}{}^{238}\mathrm{U}$ is:

$\frac{238-206}{4}=8$

This means that 1 mole of  ${}_{92}{}^{238}\mathrm{U}$ 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.