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 19.68E
Interpretation Introduction

Interpretation:

The average displacement of one atom of He in Ne in given time intervals is to be calculated. Also the total distance travelled by He in given time intervals are to be calculated.

Concept introduction:

The average displacement of the molecule in three dimensions is given by,

(3-Ddisplacement)avg2=6Dt

Where,

D is the diffusion coefficient.

t is the time.

Displacement is a vector quantity while distance is scalar quantity. Distance can be calculated with the help of average velocity.

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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?
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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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