Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 5 - Gases - Active Learning Questions - Page 233: 18

Answer

a. The average kinetic energy of the Xe atoms of container A and B will be equal as temperature is constant. b. The force with which the Xe atoms collide with the container wall of A and B will be equal as mass and temperature is same. c. The root mean square velocity of the Xe atoms as mass and temperature is same. d. The collision frequency of the Xe atoms (with other atoms) of container A will be less than B because it depends on density i,e volume of the container. e. Pressure depends on temperature and volume. So, the pressure of the Xe sample of container A will be less than B.

Work Step by Step

a. Average kinetic energy $〈ϵ〉=1/2 KT$ so it depends on temperature only. So, only temperature must be known to calculate $〈ϵ〉$. The average kinetic energy of the Xe atoms of container A and B will be equal as temperature is constant. b. Average force of each impact on the wall is $〈F_W 〉=(1/3 m〈c^2 〉)/l$ and $⟨c^2 ⟩=3RT/M$. So, mass of the molecule, temperature of the gas sample must be known. The force with which the Xe atoms collide with the container wall of A and B will be equal as mass and temperature is same. c. Root mean square velocity of gas is $c_rms=√(3RT/M)$. So, mass of the molecule and temperature of the gas sample must be known. The root mean square velocity of the Xe atoms as mass and temperature is same. d. Average number of collisions per second with other gas molecules is $Z=1/2 √2 πσ^2 √(8kT/πm) N ̃^2$ where $σ$ is collision diameter, m is the mass of the molecule, $N ̃ $ is $molecules/m^3.$ So, mass of the molecule, temperature of the gas sample, size of the molecule, and density of the gas sample must be known. The collision frequency of the Xe atoms (with other atoms) of container A will be less than B because it depends on density i,e volume of the container. e. As pressure of the gas is $p=((1/3) Nm⟨c^2 ⟩)/V$ where $⟨c^2 ⟩=3RT/M$. So, we can write $p=RT/V$. Pressure depends on temperature and volume. So, the pressure of the Xe sample of container A will be less than B.
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