Physical Chemistry
Physical Chemistry
2nd Edition
ISBN: 9781133958437
Author: Ball, David W. (david Warren), BAER, Tomas
Publisher: Wadsworth Cengage Learning,
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Chapter 19, Problem 19.29E
Interpretation Introduction

Interpretation:

The expression for ν2¯ is to be determined by using the form of G(ν). The square root of the corresponding answer gives the definition of νrms, the root-mean-square speed is to be shown.

Concept introduction:

A gas is made up of many atoms or molecules that move with very high speeds. The kinetic energy of gases is very high. Every molecule or atom present in a gas can have a different velocity. Therefore, root mean square speed, most probable velocity, and mean velocity are calculated for a gas.

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There are two particles, one is heavy and the other is light. The light particles diffuse faster than the heavy particles. This relationship is known as Graham’s Law of Effusion. Since both gases are at the same temperature, they must have the same average kinetic energy (½ mv2), where m is mass and v is the velocity (like speed). Since both gases have the same average kinetic energy, you can state that ½ mLvL2 = ½ mHvH2. Multiplying both sides by 2 gives you mLvL2 = mHvH2. Rearranging the equation to get both masses on the same side of the equation will give you mL/mH = VH2/VL2. In 3a and 3b, you probably noticed that the heavy gas particles took twice as long to diffuse as the light gas particles. This means that the light gas particles are moving twice as fast, VH/VL = ½. Therefore, VH2/VL2 = ¼. How many times heavier is the heavy gas compared to the light gas?  If the light gas was Ne, what would be a reasonable identity for the heavy gas?
c) This relationship is known as Graham's Law of Effusion. Since both gases are at te same temperature, they must have the same average kinetic energy (½ mv²), where m is mass and v is velocity (like speed). Since both gases have the same average kinetic energy, you can state that ½ muvL2 = v ². Multiplying both sides by 2 gives you m v 2 y ². Rearranging the equation to get H H LL H H 2 m = m both masses on the same side of the equation will give you mu/mH = V 2/VL2. In 3a and 3b, you probably noticed that the heavy gas particles took twice as long to diffuse as the light gas particles. This means that the light gas particles are moving twice as fast, VH/VL = ½. Therefore, V 2/VL2 = ¼. How many times heavier is the heavy gas compared to the light gas? d) If the light gas was Ne, what would be a reasonable identity for the heavy gas?
1.3 mole of helium gas at a temperature of 276 K is confined to a cubical container whose sides are 12 cm long.Find the mean summed kinetic energy of ALL the atoms in the container in [J].

Chapter 19 Solutions

Physical Chemistry

Ch. 19 - Prob. 19.11ECh. 19 - Interstellar space can be considered as having...Ch. 19 - Prob. 19.13ECh. 19 - SF6 is a gas at room temperature, 295K. What is...Ch. 19 - Prob. 19.15ECh. 19 - Prob. 19.16ECh. 19 - If relativistic effects were ignored, what...Ch. 19 - Prob. 19.18ECh. 19 - Prob. 19.19ECh. 19 - Prob. 19.20ECh. 19 - Prob. 19.21ECh. 19 - Prob. 19.22ECh. 19 - Prob. 19.23ECh. 19 - Prob. 19.24ECh. 19 - What is the ratio of vrms/vmostprob for any gas at...Ch. 19 - Prob. 19.26ECh. 19 - Prob. 19.27ECh. 19 - Prob. 19.28ECh. 19 - Prob. 19.29ECh. 19 - Prob. 19.30ECh. 19 - Prob. 19.31ECh. 19 - The previous exercise gives an expression for...Ch. 19 - Prob. 19.33ECh. 19 - Prob. 19.34ECh. 19 - Prob. 19.35ECh. 19 - What must the pressure be if the mean free path of...Ch. 19 - Prob. 19.37ECh. 19 - Prob. 19.38ECh. 19 - Prob. 19.39ECh. 19 - Explain why the molecular diameter for argon, at...Ch. 19 - Prob. 19.41ECh. 19 - Prob. 19.42ECh. 19 - Prob. 19.43ECh. 19 - A 1.00-mol sample of Xe gas is kept at a...Ch. 19 - Prob. 19.45ECh. 19 - Prob. 19.46ECh. 19 - Prob. 19.47ECh. 19 - Prob. 19.48ECh. 19 - Prob. 19.49ECh. 19 - Consider a gas mixture containing equal...Ch. 19 - The inverse of the collision rate, 1/z, is the...Ch. 19 - Prob. 19.52ECh. 19 - Prob. 19.53ECh. 19 - Prob. 19.54ECh. 19 - Prob. 19.55ECh. 19 - Estimate the rate at which Hg effuses out a hole...Ch. 19 - Prob. 19.57ECh. 19 - Knudsen effusion cells are used to determine vapor...Ch. 19 - Prob. 19.59ECh. 19 - Prob. 19.60ECh. 19 - Prob. 19.61ECh. 19 - Prob. 19.62ECh. 19 - Prob. 19.63ECh. 19 - Prob. 19.64ECh. 19 - Prob. 19.65ECh. 19 - Prob. 19.66ECh. 19 - Prob. 19.67ECh. 19 - Prob. 19.68ECh. 19 - Prob. 19.69ECh. 19 - Prob. 19.70ECh. 19 - Prob. 19.71ECh. 19 - Prob. 19.72ECh. 19 - Prob. 19.73E
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