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Decay Products of Uranium Minerals Radon and helium are both by-products of the radioactive decay of uranium minerals. A fresh sample of carnotite, \(\mathrm{K}_{2}\left(\mathrm{UO}_{2}\right)_{2}\left(\mathrm{VO}_{4}\right)_{2}\) \(\cdot 3 \mathrm{H}_{2} \mathrm{O},\) is put on display in a museum. Calculate the relative rates of diffusion of helium and radon under fixed conditions of pressure and temperature. Which gas diffuses more rapidly through the display case?

Short Answer

Expert verified
Short Answer: Under fixed conditions of pressure and temperature, helium diffuses more rapidly compared to radon in the display case, as the ratio of their diffusion rates is greater than 1 (\(Rate_{He} / Rate_{Rn} = \sqrt{55.5}\)).

Step by step solution

01

Recall Graham's Law of Diffusion formula

Graham's law of diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it is written as: \(Rate_{1} / Rate_{2} = \sqrt{M_{2} / M_{1}}\) Where \(Rate_{1}\) and \(Rate_{2}\) are the diffusion rates of the gases, and \(M_{1}\) and \(M_{2}\) are their molar masses respectively.
02

Find the Molar Masses of Helium and Radon

We need to find the molar masses of helium and radon in order to apply Graham's law. For helium: \(\mathrm{He}\), molecular weight = 4 g/mol. For radon: \(\mathrm{Rn}\), molecular weight = 222 g/mol.
03

Calculate the Relative Rates of Diffusion

Now, we will use Graham's law to find the ratio of their diffusion rates. \(Rate_{He} / Rate_{Rn} = \sqrt{M_{Rn} / M_{He}}\) \(Rate_{He} / Rate_{Rn} = \sqrt{(222\ g/mol) / (4\ g/mol)}\) \(Rate_{He} / Rate_{Rn} = \sqrt{55.5}\)
04

Determine which gas diffuses more rapidly

Since the ratio \(Rate_{He} / Rate_{Rn}\) is greater than 1, it means that helium diffuses more rapidly than radon through the display case. Therefore, under the fixed conditions of pressure and temperature, helium diffuses more rapidly compared to radon in the display case.

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Key Concepts

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

Rates of Diffusion
The rates of diffusion are an essential concept in understanding how gases spread through different mediums. You can visualize diffusion like the way a drop of ink spreads in water. The rate at which this spreading occurs depends on several factors, one of them being Graham's Law of Diffusion.
Graham's Law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. This relationship allows us to compare how fast two gases will diffuse under the same conditions. By knowing the molar masses of these gases, we use the formula:
  • \( Rate_{1} / Rate_{2} = \sqrt{M_{2} / M_{1}} \)
Where \( Rate_{1} \) and \( Rate_{2} \) are the diffusion rates and \( M_{1} \) and \( M_{2} \) are the respective molar masses. A higher rate indicates faster diffusion, aiding in predicting how quickly a gas will penetrate through materials, which is crucial for chemical reactions and industrial applications.
Molar Mass
Molar mass is the weight of one mole of a substance, measured in grams per mole (g/mol). It's like a heavier element is made up of heavier 'building blocks,' which affects how it moves. In the context of diffusion, gases with lower molar mass diffuse more quickly because lighter particles move faster.
Helium and radon in the given problem have very different molar masses. Helium has a molar mass of 4 g/mol, making it very light. On the other hand, radon has a molar mass of 222 g/mol. This significant difference directly impacts their rates of diffusion.
Using the concept of molar mass, students can determine which gas will diffuse faster by applying Graham’s Law. Simply, the lighter the gas (lower the molar mass), the faster it will diffuse through a medium.
Uranium Decay Products
Understanding uranium decay products brings us into the world of radioactivity and isotopes. Uranium, a heavy metal, undergoes radioactive decay, producing several by-products including radon and helium.
Radon is a noble gas, heavier than many other gases due to its high molar mass (222 g/mol). It is a significant by-product because it is radioactive, posing health risks if not properly managed. Helium, in this context, is lighter and non-radioactive. It's produced when alpha particles (which are helium nuclei) are emitted during radioactive decay.
The diffusion properties of these decay products are significant for safe storage and management in environments like museums. Because helium diffuses more quickly than radon, understanding these properties ensures effective strategies to monitor and manage these gases, protecting both artifacts and people from potential harm.

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Most popular questions from this chapter

A flask of ammonia is connected to a flask of an unknown acid HX by a \(1.00 \mathrm{m}\) glass tube. As the two gases diffuse down the tube, a white ring of \(\mathrm{NH}_{4} \mathrm{X}\) forms \(68.5 \mathrm{cm}\) from the ammonia flask. Identify element X.

Which noble gas is expected to deviate the most from ideal behavior in a graph of PV/RT versus \(P ?\)

Denitrification in the Environment In some aquatic ecosystems, nitrate \(\left(\mathrm{NO}_{3}^{-}\right)\) is converted to nitrite \(\left(\mathrm{NO}_{2}^{-}\right),\) which then decomposes to nitrogen and water. As an example of this second reaction, consider the decomposition of ammonium nitrite: $$\mathrm{NH}_{4} \mathrm{NO}_{2}(a q) \rightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ What is the change in pressure in a sealed 10.0 L. vessel due to the formation of \(\mathrm{N}_{2}\) gas when the ammonium nitrite in \(1.00 \mathrm{L}\) of \(1.0 \mathrm{M} \mathrm{NH}_{4} \mathrm{NO}_{2}\) decomposes at \(25^{\circ} \mathrm{C} ?\)

Why do real gases behave non-ideally at very low temperatures and very high pressures?

The reaction between potassium super-oxide and carbon dioxide is used to produce \(0.200 \mathrm{L}\) of \(\mathrm{O}_{2},\) which is collected over water at \(25.0^{\circ} \mathrm{C} .\) The atmospheric pressure is 750.0 torr. The vapor pressure of water at \(25.0^{\circ} \mathrm{C}\) is 24.0 torr. How many moles of \(\mathrm{O}_{2}\) have becn collected? $$4 \mathrm{KO}_{2}(s)+2 \mathrm{CO}_{2}(g) \rightarrow 2 \mathrm{K}_{2} \mathrm{CO}_{3}(s)+3 \mathrm{O}_{2}(g)$$

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