Question

Naturally found uranium consists of 99.274\(\%^{238} \mathrm{U}\) \(0.720 \%^{233} \mathrm{U},\) 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 \(^{255}\) U used in the first atomic bomb was obtained by gaseous diffusion of uranium hexafluoride, UF \(_{6}(g) .\) (a) What is the mass of UF \(_{6}\) in a 30.0 -L vessel of 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 barrier and that the change in the ratio of 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 the diffused gas analogous to that in part (a)? (d) After one more cycle of gaseous diffusion, what is the percentage of \(^{235} \mathrm{UF}_{6}\) in the sample?

Short answer

Short answer: (a) The mass of UF6 in a 30.0 L vessel of UF6 at a pressure of 695 torr at 350 K is approximately 336.87 g. (b) The mass of 235U in the sample described in part (a) is approximately 2.426 g. (c, d) Without the provided Equation 10.23, we cannot calculate the mass of 235U in a sample of the diffused gas and the percentage of 235UF6 after one more cycle of gaseous diffusion.

Step-by-step solution

(Step 1: Calculate the mass of UF6 in the vessel)

We are given the volume, pressure, and temperature of the UF6 gas in the vessel. Using the ideal gas law, we can calculate the number of moles of UF6, and then convert it to mass using the molar mass of UF6. The ideal gas law is given by PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. First, we should convert pressure from torr to atm, as the ideal gas constant R=0.0821 L⋅atm/mol⋅K : P(atm) = 695 torr * (1 atm / 760 torr) ≈ 0.915 atm Now, we can calculate the number of moles: n = PV / RT ≈ (0.915 atm * 30.0 L) / (0.0821 L⋅atm/mol⋅K * 350 K) ≈ 0.956 mol Finally, we can calculate the mass using the molar mass of UF6: Molar mass of UF6 ≈ (1*238) + (6*19) = 238 + 114 = 352 g/mol Mass of UF6 = n * molar mass = 0.956 mol * 352 g/mol ≈ 336.87 g

(Step 2: Calculate the mass of 235U in the initial sample)

Now, we will use the mass fraction of 235U in naturally found uranium to calculate the mass of 235U in the initial sample. Mass fraction of 235U = 0.720% Mass of 235U in the initial sample = (Mass of UF6) * (Mass fraction of 235U / Molar Mass of UF6) = 336.87 g * (0.0072) ≈ 2.426 g

(Step 3: Calculate the mass of 235U in the diffused gas sample)

Now, we will use Equation 10.23 (which is not provided but is assumed to describe the change in the ratio of 238U and 235U in the diffused gas) to find the mass of 235U in the diffused gas sample. Assuming the mass ratios of 238U and 235U in the diffused gas are given by: Mass_ratio_diffused = Mass_ratio_initial * Equation_10.23 We will determine the mass of 235U in the diffused gas: Mass(235U)_diffused = Mass(235U)_initial * Equation_10.23 Unfortunately, as Equation 10.23 is not provided, we cannot further solve for the mass of 235U in the diffused gas sample.

(Step 4: Calculate the percentage of 235UF6 after one more cycle of gaseous diffusion)

Similar to step 3, this part requires Equation 10.23 to determine the percentage of 235UF6 after one more cycle of gaseous diffusion. Since Equation 10.23 is not provided, we cannot complete this step either.

Key concepts

Uranium Isotopes

Uranium isotopes play a crucial role in nuclear chemistry, particularly in the realm of nuclear energy and weaponry. Uranium naturally occurs as several isotopes, with Uranium-238 ((U^{238}U)) and Uranium-235 ((U^{235}U)) being the most abundant.

(U^{238}U) is more stable and accounts for the majority of natural uranium. However, it is (U^{235}U) that is of significant interest due to its ability to sustain a nuclear chain reaction, which is fundamental in both nuclear reactors and atomic bombs. While natural uranium contains about 99.3% of (U^{238}U), it’s the 0.7% of (U^{235}U) that requires enrichment for use in nuclear technology.

Enrichment is a process by which the proportion of (U^{235}U) is increased, a topic we explore further when discussing gaseous diffusion. To obtain a clearer understanding of uranium uses and why certain isotopes are preferred over others, students should recognize the distinction in nuclear properties between the isotopes. (U^{235}U)'s ability to easily absorb neutrons and fission makes it especially sought after in nuclear applications.

Ideal Gas Law

A fundamental concept in understanding gas behavior in nuclear chemistry is the ideal gas law, which is expressed as (PV = nRT). Here, (P) stands for the pressure of the gas, (V) is the volume it occupies, (n) represents the number of moles of gas, (R) is the universal gas constant, and (T) refers to the absolute temperature.

This law allows us to calculate the amount of a gaseous substance under certain conditions, as demonstrated in the textbook exercise for uranium hexafluoride ((UF_6)). By manipulating the ideal gas law, one can find the number of moles of (UF_6) in a container at a known pressure and temperature, and subsequently determine its mass. The ideal gas law assumes no interactions between gas particles and that they occupy negligible space, which simplifies calculations but can introduce errors when applied to real gases at high pressures or low temperatures. Understanding the ideal gas law is crucial for calculating gas densities, which are necessary for separating isotopes using gaseous diffusion.

Gaseous Diffusion

Gaseous diffusion is an intricate process used in the enrichment of uranium isotopes, specifically separating (U^{235}U) from (U^{238}U). This method takes advantage of the slight mass difference between the isotopes. When uranium hexafluoride gas is passed through a porous barrier or membrane, lighter (^{235}UF_6) molecules tend to pass through the barrier slightly faster than the heavier (^{238}UF_6) molecules.

This results in a gradual increase in the concentration of (U^{235}U) on one side of the barrier. Over multiple iterative cycles, the process yields uranium hexafluoride gas with a higher percentage of (U^{235}U), which is desired for nuclear reactors and weapons. The exercise from the textbook illustrates this process, showing how the mass of (U^{235}U) changes before and after gaseous diffusion. Understanding gaseous diffusion requires not only a grasp of molecular behavior but also an understanding of how equilibrium and rates of diffusion can affect the enrichment outcome. To improve comprehension, students should study examples of gaseous diffusion and explore how varying conditions can influence the rate of isotope separation.