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The rate of \(effusion\)-that is, leakage of a gas through tiny cracks-is proportional to \(v_{rms}\) . If tiny cracks exist in the material that's used to seal the space between two glass panes, how many times greater is the rate of \(He\) leakage out of the space between the panes than the rate of \(Xe\) leakage at the same temperature? (a) 370 times; (b) 19 times; (c) 6 times; (d) no greater-the \(He\) leakage rate is the same as for \(Xe\).

Short Answer

Expert verified
(c) 6 times

Step by step solution

01

Understand the formula

The rate of effusion of a gas is proportional to its root-mean-square velocity, \(v_{rms}\). The formula for \(v_{rms}\) is given by \(v_{rms} = \sqrt{\frac{3kT}{m}}\), where \(k\) is the Boltzmann constant, \(T\) is the temperature, and \(m\) is the molar mass of the gas. Thus, the rate of effusion \(R\) is proportional to \(\frac{1}{\sqrt{m}}\), i.e. \(R \propto \frac{1}{\sqrt{m}}\).
02

Calculate the molar mass

We need to find the molar masses of Helium (\(He\)) and Xenon (\(Xe\)). Helium has a molar mass of approximately \(4\) g/mol, and Xenon has a molar mass of approximately \(131.3\) g/mol.
03

Calculate the effusion ratio

The ratio of the effusion rates of \(He\) and \(Xe\) can be found using the inverse square root of their molar masses: \(\frac{R_{He}}{R_{Xe}} = \frac{\sqrt{m_{Xe}}}{\sqrt{m_{He}}}\). Substituting the molar masses gives \(\frac{\sqrt{131.3}}{\sqrt{4}}\).
04

Solve the ratio

First, calculate the square roots: \(\sqrt{131.3} \approx 11.46\) and \(\sqrt{4}=2\). Thus the ratio \(\frac{11.46}{2} = 5.73\) which is approximately 6 times.
05

Compare with options

Since we calculated the ratio as approximately 6 times, compare it with the given options: (a) 370 times; (b) 19 times; (c) 6 times; (d) no greater-the \(He\) leakage rate is the same as for \(Xe\). The correct answer is option (c) as it matches our calculation.

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Key Concepts

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

Root-Mean-Square Velocity
The root-mean-square velocity, often abbreviated as \(v_{rms}\), is a vital concept in understanding the behavior of gas molecules. Essentially, it measures the average speed of gas molecules in a sample. The formula for \(v_{rms}\) is \[v_{rms} = \sqrt{\frac{3kT}{m}}\]where:
  • \(k\) is the Boltzmann constant, a constant in physics that relates the average kinetic energy of particles in a gas with the temperature of the gas.
  • \(T\) is the temperature in Kelvin.
  • \(m\) is the molar mass of the gas, which is crucial because it significantly influences the \(v_{rms}\).

At a given temperature, gases with smaller molar masses will have higher \(v_{rms}\) compared to the gases with larger molar masses. This directly affects properties such as diffusion and effusion, where gases with higher \(v_{rms}\) tend to disperse or leak quicker.
Molar Mass
Molar mass is a fundamental property of chemical substances. It's the mass of one mole of a substance, usually expressed in grams per mole (g/mol). For gases, molar mass is critical in determining their effusion rates and behavior in various conditions.

For example, Helium (\(He\)) has a molar mass of about \(4\, g/mol\), which is quite low compared to many other gases. Meanwhile, Xenon (\(Xe\)) has a much higher molar mass of approximately \(131.3\, g/mol\).

This difference in molar mass is key to explaining why Helium can effuse (leak through small openings) much faster than Xenon. The formula for effusion, \(R \propto \frac{1}{\sqrt{m}}\), shows that as molar mass decreases, the rate of effusion increases.
Gas Properties
Gases exhibit several distinct properties that differentiate them from liquids and solids. Understanding these properties can help clarify why rates of processes like diffusion and effusion vary among different gases.

Here are some important gas properties:
  • Gases have no fixed shape or volume, allowing them to expand and fill a container.
  • They are highly compressible, meaning their volume can change significantly with pressure or temperature alterations.
  • Gas particles are in constant, random motion, contributing to properties like pressure and temperature.
  • Collisions between gas molecules are perfectly elastic, meaning no kinetic energy is lost.

In the context of gases like Helium and Xenon, such properties explain how easily they can escape through small leaks. Their behavior is strongly influenced by factors such as temperature, which ties back into the root-mean-square velocity and molar masses. Understanding these gas properties provides a framework for predicting and explaining real-world scenarios in which gases interact.

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Most popular questions from this chapter

A welder using a tank of volume 0.0750 m\(^3\) fills it with oxygen (molar mass 32.0 g/mol) at a gauge pressure of 3.00 \(\times\) 10\({^5}\) Pa and temperature of 37.0\(^\circ\)C. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 22.0\(^\circ\)C, the gauge pressure of the oxygen in the tank is 1.80 \(\times\) 10\({^5}\) Pa. Find (a) the initial mass of oxygen and (b) the mass of oxygen that has leaked out.

What is one reason the noble gases are \(preferable\) to air (which is mostly nitrogen and oxygen) as an insulating material? (a) Noble gases are monatomic, so no rotational modes contribute to their molar heat capacity; (b) noble gases are monatomic, so they have lower molecular masses than do nitrogen and oxygen; (c) molecular radii in noble gases are much larger than those of gases that consist of diatomic molecules; (d) because noble gases are monatomic, they have many more degrees of freedom than do diatomic molecules, and their molar heat capacity is reduced by the number of degrees of freedom.

How many moles are in a 1.00-kg bottle of water? How many molecules? The molar mass of water is 18.0 g/mol.

Calculate the mean free path of air molecules at 3.50 \(\times\) 10\({^-}{^1}{^3}\) atm and 300 K. (This pressure is readily attainable in the laboratory; see Exercise 18.23.) As in Example 18.8, model the air molecules as spheres of radius 2.0 \(\times\) 10\({^-}{^1}{^0}\) m.

The \(vapor\) \(pressure\) is the pressure of the vapor phase of a substance when it is in equilibrium with the solid or liquid phase of the substance. The \(relative\) \(humidity\) is the partial pressure of water vapor in the air divided by the vapor pressure of water at that same temperature, expressed as a percentage. The air is saturated when the humidity is 100%. (a) The vapor pressure of water at 20.0\(^\circ\)C is 2.34 \(\times\) 103 Pa. If the air temperature is 20.0\(^\circ\)C and the relative humidity is 60%, what is the partial pressure of water vapor in the atmosphere (that is, the pressure due to water vapor alone)? (b) Under the conditions of part (a), what is the mass of water in 1.00 m\(^3\) of air? (The molar mass of water is 18.0 g/mol. Assume that water vapor can be treated as an ideal gas.)

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