Domanda del libro di testo
Which postulate of the kinetic molecular theory breaks down under conditions of high pressure? Explain.
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Ch.6 - Gases
Tro6th EditionChemistry: A Molecular ApproachISBN: 9780137832217
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TOC.tableOfContents
- Ch.1 - Matter, Measurement & Problem Solving118
- Ch.2 - Atoms & Elements106
- Ch.3 - Molecules and Compounds131
- Ch.4 - Chemical Reactions and Chemical Quantities56
- Ch.5 - Introduction to Solutions and Aqueous Solutions82
- Ch.6 - Gases114
- Ch.7 - Thermochemistry96
- Ch.8 - The Quantum-Mechanical Model of the Atom60
- Ch.9 - Periodic Properties of the Elements90
- Ch.10 - Chemical Bonding I: The Lewis Model89
- Ch.11 - Chemical Bonding II: Molecular Shapes, VSEPR & MO Theory85
- Ch.12 - Liquids, Solids & Intermolecular Forces51
- Ch.13 - Solids & Modern Materials47
- Ch.14 - Solutions90
- Ch.15 - Chemical Kinetics115
- Ch.16 - Chemical Equilibrium68
- Ch.17 - Acids and Bases136
- Ch.18 - Aqueous Ionic Equilibrium153
- Ch.19 - Free Energy & Thermodynamics93
- Ch.20 - Electrochemistry106
- Ch.21 - Radioactivity & Nuclear Chemistry74
- Ch.22 - Organic Chemistry140
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Capitolo 6, Problema 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?
Guida verificata passo dopo passo1
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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Concetti chiave
Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.
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.
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Percorso guidato
2:03
Graham's Law of Effusion
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.
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Percorso guidato
02:11Molar Mass Concept
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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Percorso guidato
01:31Effusion Rate Example
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We separate U-235 from U-238 by fluorinating a sample of uranium to form UF6 (which is a gas) and then taking advantage of the different rates of effusion and diffusion for compounds containing the two isotopes. Calculate the ratio of effusion rates for 238UF6 and 235UF6. The atomic mass of U-235 is 235.054 amu and that of U-238 is 238.051 amu.
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