/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 88 As discussed in the "Chemistry P... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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, \(\mathrm{U}(\mathrm{g})\). Calculate the ratio of effusion rates for \({ }^{235} \mathrm{U}\) and \({ }^{223} \mathrm{U}\), and compare it to the ratio for \(\mathrm{UF}_{6}\) given in the essay.

Short Answer

Expert verified
The effusion rate ratio for \({ }^{235}\mathrm{U}\) and \({ }^{223}\mathrm{U}\) is 1.038, while the ratio for \(\mathrm{UF}_{6}\) is 1.0043. In comparison, the rate of effusion for the uranium isotopes is larger than that of \(\mathrm{UF}_{6}\). This means that pure uranium isotopes would separate more efficiently than the associated \(\mathrm{UF}_{6}\) isotopes due to their higher effusion rate ratio.

Step by step solution

01

1. Identify the molar masses of \({ }^{235}\mathrm{U}\) and \({ }^{223}\mathrm{U}\)

The gaseous uranium isotopes have the following molar masses: - Molar mass of \({ }^{235}\mathrm{U}\): 235 g/mol - Molar mass of \({ }^{223}\mathrm{U}\): 223 g/mol
02

2. Apply Graham's Law of Effusion

Using Graham's Law of Effusion, we can find the ratio of effusion rates: \[ \frac{Rate_{\:^{235}\mathrm{U}}}{Rate_{\:^{223}\mathrm{U}}} = \frac{\sqrt{Molar\:mass_{\:^{223}\mathrm{U}}}}{\sqrt{Molar\:mass_{\:^{235}\mathrm{U}}}} \]
03

3. Calculate the effusion rate ratio

Next, we substitute the given molar masses into the formula: \[ \frac{Rate_{\:^{235}\mathrm{U}}}{Rate_{\:^{223}\mathrm{U}}} = \frac{\sqrt{223\:g/mol}}{\sqrt{235\:g/mol}} \] Calculate the square root of the molar masses: \[ \frac{Rate_{\:^{235}\mathrm{U}}}{Rate_{\:^{223}\mathrm{U}}} = \frac{\sqrt{223}}{\sqrt{235}} \] Finally, divide the square root of molar masses: \[ \frac{Rate_{\:^{235}\mathrm{U}}}{Rate_{\:^{223}\mathrm{U}}} = 1.038 \] The effusion rate ratio for \({ }^{235}\mathrm{U}\) and \({ }^{223}\mathrm{U}\) is 1.038. Now compare this ratio to the ratio for \(\mathrm{UF}_{6}\) given in the essay. According to the essay, the effusion rate ratio for \(\mathrm{UF}_{6}\) containing \({ }^{235}\mathrm{U}\) and \({ }^{223}\mathrm{U}\) is 1.0043.
04

4. Comparison

As per the calculation, the effusion rate ratio for the uranium isotopes is 1.038, while the ratio for \(\mathrm{UF}_{6}\) is 1.0043. In comparison, the rate of effusion for the uranium isotopes is larger than that of \(\mathrm{UF}_{6}\). This means that pure uranium isotopes would separate more efficiently than the associated \(\mathrm{UF}_{6}\) isotopes due to their higher effusion rate ratio.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Effusion Rate
Effusion rate is a measure of how quickly gas molecules escape through a small hole into a vacuum. In the context of uranium isotope separation, understanding effusion is crucial. 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 that lighter gases effuse faster than heavier ones.
Graham's Law is expressed mathematically as:
  • \( \frac{Rate_1}{Rate_2} = \frac{\sqrt{Molar\:mass_2}}{\sqrt{Molar\:mass_1}} \)
The law was applied to calculate the effusion rate ratio for uranium isotopes \( ^{235}U \) and \( ^{223}U \), resulting in a ratio of 1.038. This tells us that \( ^{235}U \) effuses slightly faster than \( ^{223}U \). Understanding these rates helps in processes like uranium enrichment, where isotopes need to be separated efficiently. By knowing the effusion rates, scientists can develop better methods for isotope separation.
Uranium Isotopes
Uranium isotopes are variations of the uranium element with different numbers of neutrons. The two isotopes discussed here, \(^{235}U\) and \(^{223}U\), are both forms of gaseous uranium used in various scientific and industrial applications. Here's a quick look at each:
  • \( ^{235}U \) is widely used in nuclear reactors and weapons due to its fissile properties.
  • \( ^{223}U \) is less common and less stable.
Uranium isotopes are vital in nuclear chemistry and play a significant role in energy production. Their separation via effusion takes advantage of their slight differences in molar mass, thereby helping in creating concentrated forms like enriched uranium. Understanding these isotopes allows scientists to innovate and improve techniques such as gaseous diffusion for isotope separation.
Molar Mass
Molar mass is a critical concept in chemistry that refers to the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It's essential for calculating how substances interact in reactions and is particularly useful in separating isotopes like uranium.For uranium isotopes, the molar masses are:
  • \( ^{235}U \): 235 g/mol
  • \( ^{223}U \): 223 g/mol
These differences in molar mass are crucial when using Graham's Law of Effusion to separate isotopes. The square root of the molar mass is used to calculate effusion rates, with lighter isotopes effusing more rapidly.
Understanding molar mass allows chemists to predict how substances will behave under different conditions and to tailor processes such as isotope separation for optimal efficiency. This knowledge is fundamental in applications across nuclear energy and other technological fields that utilize isotopes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose you are given two 1-L flasks and told that one contains a gas of molar mass 30 , the other a gas of molar mass 60 , both at the same temperature. The pressure in flask A is \(\mathrm{X}\) atm, and the mass of gas in the flask is \(1.2 \mathrm{~g}\). The pressure Which flask contains the gas of molar mass 30 , and which contains the gas of molar mass 60 ?

Which assumptions are common to both kinetic-molecular theory and the ideal- gas equation?

Which of the following statements best explains why a closed balloon filled with helium gas rises in air? (a) Helium is a monatomic gas, whereas nearly all the molecules that make up air, such as nitrogen and oxygen, are diatomic. (b) The average speed of helium atoms is greater than the average speed of air molecules, and the greater speed of collisions with the balloon walls propels the balloon upward. (c) Because the helium atoms are of lower mass than the average air molecule, the helium gas is less dense than air. The mass of the balloon is thus less than the mass of the air displaced by its volume. (d) Because helium has a lower molar mass than the average air molecule, the helium atoms are in faster motion. This means that the temperature of the helium is greater than the air temperature. Hot gases tend to rise.

Nickel carbonyl, \(\mathrm{Ni}(\mathrm{CO})_{4}\), is one of the most toxic substances known. The present maximum allowable concentration in laboratory air during an 8-hr workday is \(1 \mathrm{ppb}\) (parts per billion) by volume, which means that there is one mole of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for every \(10^{9}\) moles of gas. Assume \(24^{\circ} \mathrm{C}\) and \(1.00\) atm pressure. What mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is allowable in a laboratory room that is \(12 \mathrm{ft} \times 20 \mathrm{ft} \times 9 \mathrm{ft}\) ?

The planet Jupiter has a surface temperature of \(140 \mathrm{~K}\) and a mass 318 times that of Earth. Mercury (the planet) has a surface temperature between \(600 \mathrm{~K}\) and \(700 \mathrm{~K}\) and a mass \(0.05\) times that of Earth. On which planet is the atmosphere more likely to obey the ideal-gas law? Explain.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.