Ch.6 - Gases

Tro - Chemistry: A Molecular Approach 6th Edition

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Tro 6th Edition

Ch.6 - Gases

Problem 96

Chapter 6, Problem 96

A sample of CO2 effuses from a container in 55 seconds. How long will it take the same amount of gaseous Xe to effuse from the same container under identical conditions?

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Identify the problem as one involving effusion, which can be solved using Graham's Law of Effusion.

Recall Graham's Law of Effusion: \( \frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}} = \sqrt{\frac{M_2}{M_1}} \), where \( M_1 \) and \( M_2 \) are the molar masses of the gases.

Set up the equation using the given information: \( \frac{\text{Rate of effusion of CO}_2}{\text{Rate of effusion of Xe}} = \sqrt{\frac{M_{\text{Xe}}}{M_{\text{CO}_2}}} \).

Recognize that the rate of effusion is inversely proportional to the time taken for effusion, so \( \frac{t_{\text{Xe}}}{t_{\text{CO}_2}} = \sqrt{\frac{M_{\text{Xe}}}{M_{\text{CO}_2}}} \).

Substitute the known values: \( t_{\text{CO}_2} = 55 \) seconds, \( M_{\text{CO}_2} = 44.01 \) g/mol, and \( M_{\text{Xe}} = 131.29 \) g/mol, and solve for \( t_{\text{Xe}} \).

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Key Concepts

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

Graham's Law of Effusion

Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This means that lighter gases effuse faster than heavier gases. In this scenario, the effusion rates of CO2 and Xe can be compared using their molar masses to determine the time it will take for Xe to effuse.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). For this question, the molar masses of CO2 (approximately 44 g/mol) and Xe (approximately 131 g/mol) are crucial for applying Graham's Law. The difference in molar mass directly affects the effusion rates of the gases.

Effusion

Effusion is the process by which gas molecules escape from a container through a small opening into a vacuum or lower pressure area. The rate of effusion is influenced by factors such as temperature and the size of the gas molecules. Understanding effusion is essential for solving the problem, as it involves comparing the escape rates of CO2 and Xe.

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Textbook Question

Calculate the root mean square velocity and kinetic energy of CO, CO₂, and SO₃ at 298 K. Which gas has the greatest velocity? The greatest kinetic energy? The greatest effusion rate?

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Textbook Question

Use the van der Waals equation and the ideal gas equation to calculate the volume of 1.000 mol of neon at a pressure of 500.0 atm and a temperature of 355.0 K. Explain why the two values are different. (Hint: One way to solve the van der Waals equation for V is to use successive approximations. Use the ideal gas law to get a preliminary estimate for V.)

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Textbook Question

The graph shows the distribution of molecular velocities for the same molecule at two different temperatures (T₁ and T₂). Which temperature is greater? Explain.

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Textbook Question

A sample of argon effuses from a container in 112 seconds. The same amount of an unknown noble gas requires 79.6 seconds. Identify the second gas.

Textbook Question

Calculate the ratio of effusion rates for Ar and Kr.

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