Problem 91 - Chemistry: The Central Science

Question

Naturally found uranium consists of \(99.274 \%^{238} \mathrm{U}\), \(0.720 \%{ }^{235} \mathrm{U}\), and \(0.006 \%{ }^{233} \mathrm{U}\). As we have \(\sec ,{ }^{235} \mathrm{U}\) is the isotope that can undergo a nuclear chain reaction. Most of the \({ }^{235} \mathrm{U}\) used in the first atomic bomb was obtained by gascous diffusion of uranium hexafluoride, \(\mathrm{UF}_{6}(g)\). (a) What is the mass of \(\mathrm{UF}_{6}\) in a \(30.0-\mathrm{L}\) vessel of \(\mathrm{UF}_{6}\) at a pressure of 695 torr at \(350 \mathrm{~K}\) ? (b) What is the mass of \({ }^{235} \mathrm{U}\) in the sample described in part (a)? (c) Now suppose that the \(\mathrm{UF}_{6}\) is diffused through a porous barricr and that the change in the ratio of \({ }^{23} \mathrm{U}\) and \({ }^{2 / 4} \mathrm{U}\) in the diffused gas can he deseribed by Equation 10.23. What

Step-by-Step Solution

Answer

(a) The mass of UF6 in a 30.0-L vessel is 316 g. (b) The mass of U-235 in the sample is 2.28 g. (c) We are unable to proceed further in solving the problem as Equation 10.23 is not provided.

Step 1: Applying Ideal Gas Law to find moles of UF6

First, let's find the number of moles of \( UF_6 \) in the 30.0L vessel. To do this, we will use the Ideal Gas Law: \( PV = n RT \) Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Here, we have \( P = 695\,\text{torr} \), \( V = 30.0\,\text{L} \), \( R = 62.36 \frac{\mathrm{L .torr}}{\mathrm{mol .K}} \), and \( T = 350\,\text{K} \). We need to find n: \( n = \frac{PV}{RT} \)

Step 2: Calculate moles of UF6

Now, we can find the number of moles of UF6 by plugging in the values: \( n = \frac{(695\,\text{torr})(30.0\,\text{L})}{(62.36 \frac{\mathrm{L .torr}}{\mathrm{mol .K}})(350\,\text{K})} \) \( n = 0.898\,\text{mol} \)

Step 3: Calculate mass of UF6

Next, we need to find the mass of UF6. To do this, we'll multiply the number of moles by the molar mass of UF6. The molar mass of UF6 is: \( M_{UF_6} = M_U + 6 \times M_F = 238\,\text{g/mol} + 6 \times 19\,\text{g/mol} = 352\,\text{g/mol} \) Now we can find the mass of UF6: \( \text{Mass}_{UF6} = n \times M_{UF_6} = 0.898\,\text{mol} \times 352\,\text{g/mol} = 316\,\text{g} \)

Step 4: Calculate mass of U-235

Now we'll find the mass of U-235 in the UF6 sample. Since the natural abundance of U-235 is 0.720%, we can find its mass as: \( \text{Mass}_{U-235} = 0.00720 \times 316\,\text{g} = 2.28\,\text{g} \)

Step 5: Calculate change in the ratio of U-233 and U-234

We'll proceed to part (c) and, using Equation 10.23, calculate the change in the ratio of U-233 and U-234 in the diffused gas. The equation is not provided in the question, but let's assume it has the variables \( X \) and \( Y \) representing the masses of U-233 and U-234, respectively, in the diffused gas: \( \frac{X}{Y} = \text{Equation 10.23} \) Unfortunately, as the equation 10.23 is not provided, we are unable to proceed further in solving the problem.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics that helps us understand the behavior of gases under various conditions. This law is expressed as \( PV = nRT \). Here:

• \( P \) represents pressure, usually measured in atmospheres (atm) or torr.
• \( V \) stands for volume, typically in liters (L).
• \( n \) is the number of moles of gas.
• \( R \) denotes the ideal gas constant, with a common value of \( 62.36 \frac{\mathrm{L .torr}}{\mathrm{mol .K}} \).
• \( T \) is the temperature in Kelvin (K).

To solve for the number of moles \( n \), rearrange the equation: \( n = \frac{PV}{RT} \). This formula allows you to determine the amount of gas in a container when you know the pressure, volume, and temperature, simplifying calculations in many chemical applications.

Isotopes

Isotopes are variants of a particular chemical element that share the same number of protons but have different numbers of neutrons in their nuclei. This difference in neutron count results in isotopes having distinct atomic masses.

• For example, uranium has several isotopes such as \(^{238}\mathrm{U}\), \(^{235}\mathrm{U}\), and \(^{233}\mathrm{U}\).
• While they all share the same number of protons (92 for uranium), the number of neutrons differs, affecting their stability and properties.

One noteworthy characteristic about isotopes is that not all are naturally abundant or stable. Some isotopes play significant roles in various nuclear processes:

• \(^{235}\mathrm{U}\) is crucial for nuclear reactions due to its potential to sustain chain reactions.
• The natural abundance of isotopes in an element influences its applications in both scientific and industrial areas.

Uranium Enrichment

Uranium enrichment refers to the process of increasing the concentration of \(^{235}\mathrm{U}\) in uranium compounds, typically for use in nuclear reactors or weapons. Natural uranium primarily consists of \(^{238}\mathrm{U}\), and only about 0.720% is \(^{235}\mathrm{U}\), which is not enough for most nuclear reactors.

• The primary method of enrichment involves using chemical and physical processes, such as gaseous diffusion or centrifugation, to separate the desired isotopes.
• In gaseous diffusion, uranium hexafluoride \(\mathrm{UF}_6 \) is converted into gas, and the lighter \(^{235}\mathrm{U}\) molecules are separated through porous barriers.

This enrichment process allows for the production of fuel-grade uranium, which is essential for maintaining controlled nuclear reactions in power plants or producing weapon-grade material.

Molar Mass

Molar mass is a crucial concept in chemistry that plays a key role in converting the number of moles of a substance to its mass, usually in grams.

• The molar mass of a compound is the mass of one mole of that compound, calculated by summing the atomic masses of all the atoms in its formula.
• For instance, uranium hexafluoride \(\mathrm{UF}_6 \) has a molar mass composed of the atomic mass of uranium (238 g/mol) plus six times the atomic mass of fluorine (6 \( \times \) 19 g/mol), totaling 352 g/mol.

Using molar mass allows chemists to easily relate the amount of substance in moles to its mass in grams, a necessary conversion for laboratory work, industrial applications, and problem-solving in nuclear chemistry and other fields.