Ratio of Final to Initial Pressure Following Complete Decay of Uranium to Lead

To find the ratio of the final pressure to the initial pressure of the system, we analyze the states of the system before and after the radioactive decay.

1. Initial State:

The closed rigid container contains:

• $1\text{ mol}$ of solid $^{238}_{92}\text{U}$
• $1\text{ mol}$ of air (gas)

Since solid uranium has a negligible vapor pressure and volume compared to the gas, the initial pressure inside the container is exerted solely by the air. Assuming ideal gas behavior: $P_i = \frac{n_i R T}{V}$ where:

• $n_i = 1\text{ mol}$ (initial moles of gaseous particles)
• $V$ is the constant volume of the rigid container
• $T = 298\text{ K}$

2. Radioactive Decay:

The nuclear decay of $^{238}_{92}\text{U}$ to stable $^{206}_{82}\text{Pb}$ involves the emission of $\alpha$ particles ($^{4}_{2}\text{He}^{2+}$) and $\beta^-$ particles ($_{-1}^{\ \ 0}\text{e}$): $^{238}_{92}\text{U} \longrightarrow\ ^{206}_{82}\text{Pb} + x \, ^{4}_{2}\text{He} + y \, _{-1}^{\ \ 0}\text{e}$

Balancing the mass numbers (nucleons): $238 = 206 + 4x \implies 4x = 32 \implies x = 8$

Balancing the atomic numbers (charge): $92 = 82 + 2x - y \implies 92 = 82 + 2(8) - y \implies y = 6$

Each emitted $\alpha$ particle eventually captures two electrons to form a neutral, monoatomic helium gas atom ($^{4}_{2}\text{He}$).

Therefore, the complete decay of $1\text{ mol}$ of $^{238}_{92}\text{U}$ produces:

• $1\text{ mol}$ of $^{206}_{82}\text{Pb}$ (solid)
• $8\text{ mol}$ of $\text{He}$ (gas)

3. Final State:

The final gaseous mixture at $298\text{ K}$ consists of the original air and the produced helium gas: $n_f = n_{\text{air}} + n_{\text{He}} = 1\text{ mol} + 8\text{ mol} = 9\text{ mol}$

The final pressure inside the rigid container is: $P_f = \frac{n_f R T}{V}$

4. Ratio of Final to Initial Pressure:

Since the volume $V$ and temperature $T$ remain constant: $\frac{P_f}{P_i} = \frac{n_f}{n_i} = \frac{9\text{ mol}}{1\text{ mol}} = 9$