Animated Solution for Chemistry - States of Matter: A closed vessel with rigid walls contains 1 mol of ${}_{92}^{238}\text{U}$ and 1 mol of air at 298 K. Considering complete decay of ${}_{92}^{238}\text{U}$ to ${}_{82}^{206}\text{Pb}$, the ratio of the final pressure to the initial pressure of the system at 298 K is -

Imagine you are looking at a sealed, rigid steel container. Inside this container, we have two distinct substances: $1\text{ mol}$ of solid Uranium-238 (${}_{92}^{238}\text{U}$) and $1\text{ mol}$ of air. The temperature is kept perfectly constant at $298\text{ K}$.

Before any reaction happens, we need to ask ourselves: What is actually causing the pressure inside this vessel? Pressure in a closed container is exerted by gas molecules colliding with the walls. Since Uranium is a solid, it sits quietly at the bottom and contributes absolutely nothing to the gas pressure. Therefore, the initial pressure ($P_i$) is entirely due to the $1\text{ mol}$ of air.

Now, the Uranium-238 undergoes a complete radioactive decay to become Lead-206 (${}_{82}^{206}\text{Pb}$). This isn't just a simple phase change; it's a nuclear transformation. During this decay, alpha ($\alpha$) particles and beta ($\beta$) particles are emitted. We need to write down the balanced nuclear equation to see exactly what is produced.

To find the number of alpha particles ($x$), we balance the mass numbers (the superscripts). The total mass number on the left must equal the total mass number on the right.

So, the decay of $1\text{ mol}$ of Uranium-238 produces $8\text{ moles}$ of alpha particles. We could also balance the atomic numbers to find $y$ (which turns out to be 6), but beta particles are just high-energy electrons. They don't act as an independent gas that contributes to pressure in this context.

What happens to those $8\text{ moles}$ of alpha particles? An alpha particle is simply a Helium nucleus (${}_2^4{\text{He}}^{2+}$). Inside the closed vessel, these nuclei will quickly capture electrons (like the emitted beta particles) and become neutral, stable Helium gas atoms.

So, in our final state, the solid Uranium is gone, replaced by solid Lead (which still doesn't contribute to pressure). But now, alongside our original $1\text{ mol}$ of air, we have $8\text{ moles}$ of newly formed Helium gas! The total number of gaseous moles in the final state ($n_f$) is:

We are asked for the ratio of the final pressure to the initial pressure. Let's bring in the Ideal Gas Law:

We know the vessel has rigid walls, which means the volume ($V$) is strictly constant. The problem also states the temperature ($T$) remains at $298\text{ K}$. Since $R$, $T$, and $V$ are all constant, the pressure is directly proportional to the number of moles of gas.

Therefore, the ratio of the pressures is simply the ratio of the moles:

Substituting our values:

The pressure inside the vessel has increased exactly nine times due to the generation of Helium gas from the radioactive decay. The final answer is 9.