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Calculate the rms speed of an oxygen gas molecule, O2, at 21.0 °C. Express your answer numerically in meters per second. • View Available Hint(s)

Calculate the rms speed of an oxygen gas molecule, O2, at 21.0 °C. Express your answer numerically in meters per second. • View Available Hint(s)

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Learning Goal: To calculate the rms speed and relative rates of effusion of gas molecules. Part A In a given sample of gas, the particles move at varying speeds. The root mean square speed (rms speed) of particles in a gas sample, u, is given by the formula Calculate the rms speed of an oxygen gas molecule, O2, at 21.0 °C. Express your answer numerically in meters per second. 3RT V M where T is the Kelvin temperature, M is the molar mass in kg/mol, and R= 8.314 J/(mol - K) is the gas constant. Effusion is the gas molecules through a tiny hole into a vacuum. The rate of effusion of a gas is directly related to the rms speed of the gas molecules, so it's inversely proportional to the square root of its mass. The rms speed is related to kinetic energy, rather than average speed, and is the speed of a molecule possessing a kinetic energy identical to the average kinetic energy of the sample. Given its relationship to the mass of the molecule, you can conclude that the lighter the molecules of the gas, the more rapidly it effuses. Mathematically, this can be • View Available Hint(s) ? escape of m/s Submit Previous Answers expressed as effusion rate ox - X Incorrect; Try Again; 5 attempts remaining The relative rate of effusion can be expressed in terms of molecular masses ma and mp as
Transcribed Image Text:Learning Goal: To calculate the rms speed and relative rates of effusion of gas molecules. Part A In a given sample of gas, the particles move at varying speeds. The root mean square speed (rms speed) of particles in a gas sample, u, is given by the formula Calculate the rms speed of an oxygen gas molecule, O2, at 21.0 °C. Express your answer numerically in meters per second. 3RT V M where T is the Kelvin temperature, M is the molar mass in kg/mol, and R= 8.314 J/(mol - K) is the gas constant. Effusion is the gas molecules through a tiny hole into a vacuum. The rate of effusion of a gas is directly related to the rms speed of the gas molecules, so it's inversely proportional to the square root of its mass. The rms speed is related to kinetic energy, rather than average speed, and is the speed of a molecule possessing a kinetic energy identical to the average kinetic energy of the sample. Given its relationship to the mass of the molecule, you can conclude that the lighter the molecules of the gas, the more rapidly it effuses. Mathematically, this can be • View Available Hint(s) ? escape of m/s Submit Previous Answers expressed as effusion rate ox - X Incorrect; Try Again; 5 attempts remaining The relative rate of effusion can be expressed in terms of molecular masses ma and mp as
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