A closed vessel with rigid walls contains 1mol of ${}_{92}^{238}U$ and 1 mole of air at 298K. 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:

Hint: The relation $PV=nRT$ is used for the given closed system, having constant temperature. So, the pressure is directly related to the number of moles of the gas present in the system after the reaction.

Complete answer:
It is given that initially 1mol of ${}_{92}^{238}U$ and 1mole of air is present in the closed system at temperature 298K. So, the volume remains constant throughout the system.
Also, during the reaction, the temperature is the same. Hence, the pressure is directly proportional to the number of moles present.
Now, the initial pressure of the system is directly proportional to one mole of air molecules present. That is, ${{\text{P}}_{\text{i}}}\propto \,\,1\,\text{mole}\,\,\text{of}\,\text{air present}$.
On complete decays of ${}_{92}^{238}U$, it takes place through the emission of radiation in the form of $\alpha -\,and\,\beta -$ particles, that is ${}_{2}^{4}He\,and\,{}_{-1}^{0}\beta $ as follows:
\[{}_{92}^{238}U\to {}_{82}^{206}Pb+8{}_{2}^{4}He+6{}_{-1}^{0}\beta\]
So, on decay of 1 mole of Uranium, it produces 1 mole of Lead along with 8 moles of alpha-particles and 6 moles of beta-particles.
The alpha-particles generated now contribute to the pressure over the system. So, we now have 8 moles of helium gas and 1 mole of air present, which form the final pressure. That is, ${{\text{P}}_{f}}\propto \,\,\text{1}\,\text{mole}\,\,\text{of}\,\text{air}\,\text{+}\,\text{8}\,\text{moles}\,\text{of}\,\text{He}\,\text{gas present}$.
Then, the ratio of the final pressure to the initial pressure, will be:
$\dfrac{{{\text{P}}_{f}}}{{{\text{P}}_{\text{i}}}}\propto \,\,\dfrac{\text{1}\,\text{mole}\,\,\text{of}\,\text{air}\,\text{+}\,\text{8}\,\text{moles}\,\text{of}\,\text{He}\,\text{gas present}}{\,\,1\,\text{mole}\,\,\text{of}\,\text{air present}}$
$\dfrac{{{\text{P}}_{f}}}{{{\text{P}}_{\text{i}}}}=\dfrac{9}{1}$
Therefore, the ratio of the final pressure to the initial pressure of the closed system at 298 K is 9:1.

Note:
The closed vessel with rigid walls, allows only the exchange of energy and not mass at constant volume and temperature. Also, note that the alpha particles are equivalent to the helium gas. So, it contributes to the pressure of the system.