Problem 86 - Chemistry The Central Science

Question

As discussed in the "Chemistry Put to Work" box in Section 10.8 , enriched uranium can be produced by effusion of gaseous \(\mathrm{UF}_{6}\) across a porous membrane. Suppose a process were developed to allow effusion of gaseous uranium atoms, U(g). Calculate the ratio of effusion rates for \(^{235} \mathrm{U}\) and \(^{238} \mathrm{U},\) and compare it to the ratio for \(\mathrm{UF}_{6}\) given in the essay.

Step-by-Step Solution

Answer

The ratio of effusion rates for gaseous uranium isotopes \(^{235}\mathrm{U}\) and \(^{238}\mathrm{U}\) is approximately 1.0063, which is slightly higher but still quite comparable to the given ratio of 1.0043 for \(\mathrm{UF_{6}}(g)\) containing these isotopes. This suggests that the effusion process would be more efficient for separating uranium isotopes in the gaseous form, although the difference is not significant.

Step 1: Recall Graham's law of effusion

Graham's law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it is expressed as: $$\frac{Rate_A}{Rate_B} = \sqrt{\frac{M_B}{M_A}}$$ where Rate_A and Rate_B are the effusion rates of gas A and B respectively, and M_A and M_B are their molar masses.

Step 2: Identify the given information

We are given the two isotopes of uranium: 1. \(^{235}\mathrm{U}\) 2. \(^{238}\mathrm{U}\) We need to find the ratio of effusion rates for these isotopes.

Step 3: Apply Graham's law of effusion

Applying the formula from Step 1: $$\frac{Rate_{^{235}\mathrm{U}}}{Rate_{^{238}\mathrm{U}}} = \sqrt{\frac{M_{^{238}\mathrm{U}}}{M_{^{235}\mathrm{U}}}$$

Step 4: Calculate the effusion rates ratio

Now we can plug in the molar masses of the two uranium isotopes: $$\frac{Rate_{^{235}\mathrm{U}}}{Rate_{^{238}\mathrm{U}}} = \sqrt{\frac{238 \mathrm{g/mol}}{235 \mathrm{g/mol}}}$$ Solving for the ratio: $$\frac{Rate_{^{235}\mathrm{U}}}{Rate_{^{238}\mathrm{U}}} = \sqrt{\frac{238}{235}} = 1.0063$$ The ratio of effusion rates for \(^{235}\mathrm{U}\) and \(^{238}\mathrm{U}\) is approximately 1.0063.

Step 5: Compare the ratio for uranium isotopes to 𝑈𝐹6 ratios

The ratio of effusion rates for \(\mathrm{UF_{6}}(g)\) containing \(^{235}\mathrm{U}\) and \(^{238}\mathrm{U}\) is given in the essay as 1.0043. The ratio for gaseous uranium isotopes, 1.0063, is slightly higher but still quite comparable to the ratio for \(\mathrm{UF_{6}}(g)\). This indicates that the effusion process would be more efficient for separating the uranium isotopes in the gaseous form, but the difference is not significant.

Key Concepts

Effusion Rates

Effusion is the process by which gas molecules escape through a small opening into a vacuum. Understanding effusion rates is essential when dealing with processes like uranium enrichment. According to Graham's law of effusion, the rate at which a gas effuses is inversely proportional to the square root of its molar mass. This means the lighter the gas, the faster it will effuse. Graham's law is mathematically expressed as:

• \( \frac{\text{Rate}_A}{\text{Rate}_B} = \sqrt{\frac{M_B}{M_A}} \)

Here, \( \text{Rate}_A \) and \( \text{Rate}_B \) are the effusion rates of gases A and B, and \( M_A \) and \( M_B \) are their respective molar masses. This equation helps us compare the rates of effusion between different gases, playing a crucial role in applications like isotopic separation. In the context of uranium isotopes, knowing effusion rates allows us to understand how efficiently an isotope can be separated based on its mass.

Uranium Isotopes

Uranium has several isotopes, but the most commonly discussed ones are \( ^{235}\mathrm{U} \) and \( ^{238}\mathrm{U} \). Isotopes are atoms with the same number of protons but different numbers of neutrons. This results in different atomic masses. In uranium's case, the mass of \( ^{235}\mathrm{U} \) is approximately 235 g/mol, while \( ^{238}\mathrm{U} \) is about 238 g/mol.The slight difference in mass between these isotopes makes it possible to separate them using techniques like effusion. Since the difference in mass is minimal, separation processes need to be incredibly precise. The ratio of effusion rates can give insights into how effectively a separation process might work. For example, \( ^{235}\mathrm{U} \), being slightly lighter, will effuse slightly faster than \( ^{238}\mathrm{U} \). Such information is valuable in nuclear applications where specific isotopes are required.

Molar Mass Calculation

Molar mass is a critical value in chemistry that represents the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Calculating molar mass is straightforward: simply sum up the atomic masses of all the atoms in a molecule.For isotopes like uranium, calculating the molar mass involves knowing the atomic mass for each isotope. For instance, \( ^{235}\mathrm{U} \) has a molar mass of about 235 g/mol, while \( ^{238}\mathrm{U} \) has around 238 g/mol. These values directly influence their effusion rates. When comparing isotopes, the slight mass difference results in variations in how quickly they effuse.Using the molar masses of these uranium isotopes in Graham's law allows us to compute the ratio of their effusion rates. This calculation can help determine the feasibility and efficiency of a separation process using effusion, aiding in developing techniques for isotopic enrichment.