LCPO CHEMISTRY W/MODIFIED MASTERING
LCPO CHEMISTRY W/MODIFIED MASTERING
8th Edition
ISBN: 9780135214756
Author: Robinson
Publisher: PEARSON
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Chapter 10, Problem 10.100SP
Interpretation Introduction

Interpretation:

The temperature when the He atoms have average speed as that of fastball that has a speed of 45 m/s is to be calculated.

Concept introduction:

In accordance with the kinetic-gas theory the relation between average speed, temperature, and molar mass is as follows:

u=3RTM

Where,

  • R is gas constant.
  • T represents temperature in kelvins.
  • M represents the molar mass in kg/mol.

Rearrange expression to calculate temperature as follows:

T=u2M3R

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Students have asked these similar questions
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?
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 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. If the light gas was Ne, what would be a reasonable identity for the heavy gas? Explain.

Chapter 10 Solutions

LCPO CHEMISTRY W/MODIFIED MASTERING

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