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 seen, \({ }^{235} \mathrm{U}\) 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 gaseous diffusion of uranium hexafluoride, \(\mathrm{UF}_{6}(g) .(\mathbf{a})\) What is the mass of \(\mathrm{UF}_{6}\) in a 30.0 -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 UF \(_{6}\) is diffused through a porous barrier and that the change in the ratio of \({ }^{238} \mathrm{U}\) and \({ }^{235} \mathrm{U}\) in the diffused gas can be described by Equation \(10.23 .\) What is the mass of \({ }^{235} \mathrm{U}\) in a sample of

Short answer

(a) The mass of UF6 in the 30.0-L vessel is 35.2 g. (b) The mass of \({ }^{235}\mathrm{U}\) in the sample is 0.173 g. (c) We cannot solve part (c) without knowing Equation 10.23.

Step-by-step solution

Use the Ideal Gas Law to find the moles of UF6 in the gas

To calculate the moles of UF6 in the gas, we will use the Ideal Gas Law equation: \(PV = nRT\) Where P is the pressure in atm, V is the volume in liters, n is the moles of UF6, R is the gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin. First, we need to convert the pressure from torr to atm: 695 torr * (1 atm/760 torr) = 0.914 atm Now, plug in the values and solve for n: \( (0.914 \mathrm{atm})(30.0 \mathrm{L}) = n(0.0821\frac{\mathrm{ L\cdot atm}}{\mathrm{mol\cdot K}})(350 \mathrm{K})\)

Calculate the moles of UF6

We can now solve for n and find the number of moles: \(n = \frac{(0.914 \mathrm{atm})(30.0 \mathrm{L})}{(0.0821\frac{\mathrm{ L\cdot atm}}{\mathrm{mol\cdot K}})(350 \mathrm{K})} = 0.100 \mathrm{mol}\)

Convert moles of UF6 to mass

To convert the moles of UF6 to mass, we need to find the molar mass of UF6, which is: 1 U: 238 g/mol 6 F: 6 * 19 g/mol = 114 g/mol Total: 238 + 114 = 352 g/mol Now we can multiply the moles of UF6 by its molar mass to find the mass: (0.100 mol)(352 g/mol) = 35.2 g

Calculate the mass of \({ }^{235}\mathrm{U}\) in the sample

First, calculate the total molar mass of uranium in the sample by multiplying the mass of UF6 by the mass fraction of uranium: (35.2 g) * (238 g/mol) / (352 g/mol) = 24.0 g Next, find the mass of \({ }^{235}\mathrm{U}\) in the sample, knowing that natural uranium contains 0.720% of \({ }^{235}\mathrm{U}\): (24.0 g) * (0.00720) = 0.173 g

Calculate the mass of \({ }^{235}\mathrm{U}\) in the sample of diffused gas

We are not given a direct method for finding the mass of \({ }^{235}\mathrm{U}\) after diffusion has occurred. We are however told the change in the ratio can be described by Equation 10.23. Without knowing this equation, we cannot solve part (c).

Key concepts

Uranium Isotopes

Uranium isotopes are naturally occurring variants of the element uranium, distinguished by the number of neutrons within their nuclear core. The most common isotopes are

• \(^{238}\mathrm{U}\): This is the most abundant isotope of uranium, making up approximately 99.274% of natural uranium. It has 146 neutrons.
• \(^{235}\mathrm{U}\): This isotope constitutes about 0.720% of uranium and is notable for being fissile, meaning it can sustain a chain reaction when bombarded by neutrons, making it crucial for nuclear reactors and weapons.
• \(^{233}\mathrm{U}\): Although present in negligible amounts of 0.006%, it's used in certain nuclear applications such as thorium fuel cycles.

Understanding these isotopes is key, especially since the nuclear reactions of \(^{235}\mathrm{U}\) provide a powerful energy source. The unique properties and reactivities of these isotopes highlight the complexity and the potential of nuclear chemistry. Each isotope has its own set of properties, uses, and implications for both energy generation and weaponry.

Molar Mass Calculation

Molar mass is a fundamental concept in chemistry, representing the mass of one mole of a given substance. It is calculated by summing the atomic masses of each element in a compound, adjusted for the number of each type of atom present.
For example, for uranium hexafluoride (\( \mathrm{UF}_6 \)), we calculate its molar mass as follows:

• Determine the atomic mass of uranium (U), which is approximately 238 g/mol.
• Each fluorine atom (F) has an atomic mass of roughly 19 g/mol. As there are six fluorine atoms in \( \mathrm{UF}_6 \), their combined mass is \(6 \times 19 = 114 \text{ g/mol} \).
• Adding these, the molar mass of \( \mathrm{UF}_6 \) becomes \(238 + 114 = 352 \text{ g/mol} \).

Knowing the molar mass allows us to convert between the moles and grams, which is crucial in stoichiometric calculations as seen in the context of uranium isotope separation and use in nuclear processes.

Gaseous Diffusion

Gaseous diffusion is a method used to separate isotopes of uranium, particularly to increase the concentration of \(^{235}\mathrm{U}\) in uranium hexafluoride (\(\mathrm{UF}_6\)).
This separation technique exploits the small differences in the molecular weights of the isotopes:

• \(\mathrm{UF}_6\) containing \(^{235}\mathrm{U}\) is slightly lighter than that with \(^{238}\mathrm{U}\).
• By passing the gas through a barrier with tiny pores, the lighter \(^{235}\mathrm{U}F_6\) molecules diffuse more readily than their heavier \(^{238}\mathrm{U}F_6\) counterparts.

This process gradually collects a sample with a higher proportion of \(^{235}\mathrm{U}\). While effective, it's energy-intensive, requiring several diffusion stages to achieve significant enrichment levels.
Understanding gaseous diffusion is fundamentally important for nuclear fuel preparation, influencing everything from power generation to international policies on nuclear material handling and non-proliferation.

Stoichiometry in Chemistry

Stoichiometry in chemistry involves the calculation of reactants and products in chemical reactions. It is essential for predicting the results of reactions and for scaling them for various applications.
Key aspects of stoichiometry include:

• Molar Ratios: These are derived from the balanced chemical equation and indicate the proportion of reactants and products involved.
• Mole to Mass Conversions: By using molar masses, chemists convert between the mass of a substance and the number of moles.
• Limiting Reactant: This component determines the maximum extent of the reaction, as it's completely consumed. Understanding this allows for optimization of reactant quantities to maximize yield and efficiency.

In the context of uranium processing for nuclear reactions, stoichiometry helps determine the precise amounts of \(^{235}\mathrm{U}\) required and the byproducts expected in reactions, ensuring both safety and efficiency in the handling of nuclear materials.