Step-by-Step Solution

1. Analyze the initial state of the system. The closed vessel has rigid walls, which means its volume (V) is constant. The temperature (T) is also kept constant at 298 K. Initially, the vessel contains:

   • 1 mol of ${}_{92}^{238}U$, which is a solid at 298 K and does not contribute to the pressure.
   • 1 mol of air, which is a gas.

   Therefore, the initial number of moles of gas ($n_i$) in the vessel is solely from the air. $n_i = n_{air} = 1 \text{ mol}$ According to the Ideal Gas Law, the initial pressure is $P_i = \frac{n_i R T}{V}$.

2. Determine the products of the nuclear decay. The problem states that ${}_{92}^{238}U$ undergoes complete decay to ${}_{82}^{206}Pb$. This decay involves the emission of alpha (${}_{2}^{4}He$) and beta (${}_{-1}^{0}e$) particles. Let's write the balanced nuclear reaction: ${}_{92}^{238}U \rightarrow {}_{82}^{206}Pb + x({}_{2}^{4}He) + y({}_{-1}^{0}e)$ where x is the number of alpha particles and y is the number of beta particles.

   • Balance the mass number (superscript): $238 = 206 + 4x + 0y$ $32 = 4x$ $x = 8$ So, 8 alpha particles are emitted.

   • Balance the atomic number (subscript): $92 = 82 + 2x - y$ Substitute the value of x = 8: $92 = 82 + 2(8) - y$ $92 = 82 + 16 - y$ $92 = 98 - y$ $y = 6$ So, 6 beta particles are emitted.

   The complete balanced nuclear reaction is: ${}_{92}^{238}U \rightarrow {}_{82}^{206}Pb + 8{}_{2}^{4}He + 6{}_{-1}^{0}e$

3. Analyze the final state of the system. The decay of 1 mol of ${}_{92}^{238}U$ produces:

   • 1 mol of ${}_{82}^{206}Pb$, which is a solid at 298 K and does not contribute to the pressure.
   • 8 mol of alpha particles (${}_{2}^{4}He$). Alpha particles are helium nuclei ($He^{2+}$). In the vessel, they will capture electrons (including the beta particles emitted) to form stable, neutral Helium (He) atoms. Helium is a gas at 298 K.
   • 6 mol of beta particles (${}_{-1}^{0}e$), which are electrons and will be captured by the alpha particles.

   The total number of gaseous moles in the final state ($n_f$) is the sum of the moles of air and the moles of Helium gas produced. $n_f = n_{air} + n_{He}$ $n_f = 1 \text{ mol} + 8 \text{ mol} = 9 \text{ mol}$ The final pressure is $P_f = \frac{n_f R T}{V}$.

4. Calculate the ratio of final pressure to initial pressure. Since the volume (V) and temperature (T) are constant, the pressure of the gas is directly proportional to the number of moles of the gas ($P \propto n$). $\frac{P_f}{P_i} = \frac{\frac{n_f R T}{V}}{\frac{n_i R T}{V}} = \frac{n_f}{n_i}$ Substituting the values of $n_i$ and $n_f$: $\frac{P_f}{P_i} = \frac{9 \text{ mol}}{1 \text{ mol}} = 9$ Thus, the ratio of the final pressure to the initial pressure is 9.