Uranium has two naturally occurring isotopes. 238 U ^{238}\text{U} has a natural abundance of 99.3 % 99.3\% and 235 U ^{235}\text{U} has an abundance of 0.7 % 0.7\% . It is the rarer 235 U ^{235}\text{U} that is needed for nuclear reactors. The isotopes are separated by forming uranium hexafluoride, UF 6 \text{UF}_6 , which is a gas, then allowing it to diffuse through a series of porous membranes. 235 U F 6 ^{235}UF_6 has a slightly larger rms speed than 238 U F 6 ^{238}UF_6 and diffuses slightly faster. Many repetitions of this procedure gradually separate the two isotopes. What is the ratio of the rms speed of 235 U F 6 ^{235}UF_6 to that of 238 U F 6 ^{238}UF_6 ?

Hey everyone. So this problem has to do with the root mean squared velocity of ideal gasses. There is a lot of background information in this problem. But essentially, we are asked to calculate the ratio of the root mean squared velocities of these two different gasses hydrogen and deuterium deuterium. They tell us that the hydrogen has a mass of 1.008 U. And the deuterium has a mass of 2.014 U. Our multiple choice answers are a 1.414 B, 1.732 D 1.9984 C 3.992. And so the key to solving this problem is we're calling our equation for root mean squared velocity is equal to the square root of three case of B which is our Boldman constant divided by T or multiple, sorry, multiplied by T all divided by M. Now, the problem does tell us that we're, that this experiment is kept at a fixed temperature. So the temperature is a constant. So if we take our VR MS for hydrogen divided by RV R MS for deter for deuterium, we have three, the square root of three K BT divided by M sub H for hydrogen divided by the square root of three K BT divided by M sub D four deter. And so everything in on the top cancels because they are constants. And so we're left with the square root of one divided by the mass of the hydrogen divided by one divided by the mass of the deter. And so when we plug that into our calculators, the ratio we get one divided by, sorry, the mass of the hydrogen was given as 1. U and divided by the square of one divided by 2.014 U for the mass of the deteriorate. Given in the problem, we plug that in and we get 1.414. And so it's a pretty straightforward problem as long as we can recall uh this initial equation. And so that aligns with answer choice A so A is the correct one for this problem. That's all we have for this one. We'll see you in the next video.

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21. Kinetic Theory of Ideal Gases

Root-Mean-Square Velocity of Gases

Physics

21. Kinetic Theory of Ideal Gases

Root-Mean-Square Velocity of Gases

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Problem 53

Textbook Question

Uranium has two naturally occurring isotopes. $to the 238th power cap U$ has a natural abundance of $99.3 percent sign$ and $to the 235th power cap U$ has an abundance of $0.7 percent sign$ . It is the rarer $to the 235th power cap U$ that is needed for nuclear reactors. The isotopes are separated by forming uranium hexafluoride, $UF sub 6$ , which is a gas, then allowing it to diffuse through a series of porous membranes. $to the 235th power U F sub 6$ has a slightly larger rms speed than $to the 238th power U F sub 6$ and diffuses slightly faster. Many repetitions of this procedure gradually separate the two isotopes. What is the ratio of the rms speed of $to the 235th power U F sub 6$ to that of $to the 238th power U F sub 6$ ?

Verified step by step guidance

Step 1: Recall the formula for the root mean square (rms) speed of a gas molecule, which is given by $ v_{rms} = \sqrt{\frac{3RT}{M}} $, where $ R $ is the gas constant, $ T $ is the temperature, and $ M $ is the molar mass of the gas.

Step 2: Note that the temperature $ T $ and the gas constant $ R $ are the same for both isotopes, so the ratio of their rms speeds depends only on the molar masses of $ ^{235}UF_6 $ and $ ^{238}UF_6 $.

Step 3: Calculate the molar mass of $ ^{235}UF_6 $ and $ ^{238}UF_6 $. The molar mass of $ UF_6 $ is the sum of the molar mass of uranium and six fluorine atoms. For $ ^{235}UF_6 $, the molar mass is $ 235 + 6 \times 19 $, and for $ ^{238}UF_6 $, the molar mass is $ 238 + 6 \times 19 $.

Step 4: Write the ratio of the rms speeds as $ \frac{v_{rms, ^{235}UF_6}}{v_{rms, ^{238}UF_6}} = \sqrt{\frac{M_{^{238}UF_6}}{M_{^{235}UF_6}}} $, where $ M_{^{235}UF_6} $ and $ M_{^{238}UF_6} $ are the molar masses calculated in Step 3.

Step 5: Substitute the molar masses into the ratio formula and simplify. This will give the numerical ratio of the rms speeds of $ ^{235}UF_6 $ to $ ^{238}UF_6 $.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Root Mean Square Speed

The root mean square (rms) speed is a measure of the average speed of particles in a gas, calculated as the square root of the average of the squares of the speeds of the particles. It is directly related to the temperature and molar mass of the gas, providing insight into the kinetic energy of the particles. For two different gases, the rms speed can be compared using the formula: v_rms = sqrt(3RT/M), where R is the gas constant, T is the temperature, and M is the molar mass.

Diffusion

Diffusion is the process by which particles spread from areas of high concentration to areas of low concentration, driven by the random motion of particles. In the context of gases, lighter molecules diffuse faster than heavier ones due to their higher average speeds. This principle is crucial in the separation of isotopes, as the different masses of uranium hexafluoride isotopes lead to different diffusion rates through porous membranes.

Isotope Separation

Isotope separation is the process of concentrating specific isotopes of an element, which can have different physical properties, such as mass. In the case of uranium, the separation of ²³⁵U from ²³⁸U is essential for nuclear applications. Techniques like gas diffusion exploit the slight differences in rms speeds of the isotopes to achieve separation, allowing for the enrichment of the desired isotope for use in nuclear reactors.

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