ATKINS' PHYSICAL CHEMISTRY-ACCESS
ATKINS' PHYSICAL CHEMISTRY-ACCESS
11th Edition
ISBN: 9780198834700
Author: ATKINS
Publisher: OXF
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Chapter 1, Problem 1B.9P
Interpretation Introduction

Interpretation: The fraction of O2 molecules with the speeds in the range 100ms1 to 200ms1 in a gas at 300K and 1000K has to be calculated.

Concept introduction: The speeds of individual molecules spread across a wide range in a gas and the gas collisions ensure that their speeds change constantly.  The distribution of speeds is identified by recognizing that the molecules possess kinetic energy and the distribution of energy over the molecules is determined with the help of Maxwell Boltzmann distribution.

Expert Solution & Answer
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Answer to Problem 1B.9P

The fraction of molecules with the speeds in the range 100ms1 to 200ms1 in a gas at 300K and 1000K is shown below.

Velocity, vTemperature, TFraction
100ms1300K0.000344
110ms1300K0.000411
120ms1300K0.000482
130ms1300K0.000556
140ms1300K0.000634
150ms1300K0.000714
160ms1300K0.000797
170ms1300K0.000881
180ms1300K0.000965
190ms1300K0.00105
200ms1 300K0.001135
Velocity, vTemperature, TFraction
100ms11000K5.91156×105
110ms11000K7.12414×105
120ms11000K8.44087×105
130ms11000K9.85875×105
140ms11000K0.000113746
150ms11000K0.000129849
160ms11000K0.00014686
170ms11000K0.000164742
180ms11000K0.000183453
190ms11000K0.000202953
200ms1 1000K0.000223197

Explanation of Solution

The molecular distribution of speeds is given by,

    f(v)=4π(M2πRT)32v2eMv22RT

Where,

  • M is the molar mass.
  • R is the gas constant.
  • T is the temperature.
  • v is the speed.

It is given that,

  • M=32gmol1.
  • T=300K.

Take v=100ms1.

Substitute the values of M,T,v in the above equation to calculate the fraction of molecules with velocity, 100ms1.

  f(v)=4×3.14(32×103kgmol12×3.14×8.314JK1mol1×300K)32(100ms1)2e(32×103kgmol1)(100ms)22(8.314JK1mol1)(300K)=0.000344

Similarly the velocity values are taken upto 200ms1 in intervals of 10ms1 and calculate the fraction of molecules.  The calculated values are shown below.

Velocity, vTemperature, TFraction
100ms1300K0.000344
110ms1300K0.000411
120ms1300K0.000482
130ms1300K0.000556
140ms1300K0.000634
150ms1300K0.000714
160ms1300K0.000797
170ms1300K0.000881
180ms1300K0.000965
190ms1300K0.00105
200ms1 300K0.001135

It is given that,

  • M=32gmol1.
  • T=1000K.

Take v=100ms1.

Substitute the values of M,T,v in the above equation to calculate the fraction of molecules with velocity, 100ms1.

  f(v)=4×3.14(32×103kgmol12×3.14×8.314JK1mol1×1000K)32(100ms1)2e(32×103kgmol1)(100ms)22(8.314JK1mol1)(1000K)=5.91156×105

Similarly the velocity values are taken upto 200ms1 in intervals of 10ms1 and calculate the fraction of molecules.  The calculated values are shown below.

Velocity, vTemperature, TFraction
100ms11000K5.91156×105
110ms11000K7.12414×105
120ms11000K8.44087×105
130ms11000K9.85875×105
140ms11000K0.000113746
150ms11000K0.000129849
160ms11000K0.00014686
170ms11000K0.000164742
180ms11000K0.000183453
190ms11000K0.000202953
200ms1 1000K0.000223197

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Chapter 1 Solutions

ATKINS' PHYSICAL CHEMISTRY-ACCESS

Ch. 1 - Prob. 1A.3BECh. 1 - Prob. 1A.4AECh. 1 - Prob. 1A.4BECh. 1 - Prob. 1A.5AECh. 1 - Prob. 1A.5BECh. 1 - Prob. 1A.6AECh. 1 - Prob. 1A.6BECh. 1 - Prob. 1A.7AECh. 1 - Prob. 1A.7BECh. 1 - Prob. 1A.8AECh. 1 - Prob. 1A.8BECh. 1 - Prob. 1A.9AECh. 1 - Prob. 1A.9BECh. 1 - Prob. 1A.10AECh. 1 - Prob. 1A.10BECh. 1 - Prob. 1A.11AECh. 1 - Prob. 1A.11BECh. 1 - Prob. 1A.1PCh. 1 - Prob. 1A.2PCh. 1 - Prob. 1A.3PCh. 1 - Prob. 1A.4PCh. 1 - Prob. 1A.5PCh. 1 - Prob. 1A.6PCh. 1 - Prob. 1A.7PCh. 1 - Prob. 1A.8PCh. 1 - Prob. 1A.9PCh. 1 - Prob. 1A.10PCh. 1 - Prob. 1A.11PCh. 1 - Prob. 1A.12PCh. 1 - Prob. 1A.13PCh. 1 - Prob. 1A.14PCh. 1 - Prob. 1B.1DQCh. 1 - Prob. 1B.2DQCh. 1 - Prob. 1B.3DQCh. 1 - Prob. 1B.1AECh. 1 - Prob. 1B.1BECh. 1 - Prob. 1B.2AECh. 1 - Prob. 1B.2BECh. 1 - Prob. 1B.3AECh. 1 - Prob. 1B.3BECh. 1 - Prob. 1B.4AECh. 1 - Prob. 1B.4BECh. 1 - Prob. 1B.5AECh. 1 - Prob. 1B.5BECh. 1 - Prob. 1B.6AECh. 1 - Prob. 1B.6BECh. 1 - Prob. 1B.7AECh. 1 - Prob. 1B.7BECh. 1 - Prob. 1B.8AECh. 1 - Prob. 1B.8BECh. 1 - Prob. 1B.9AECh. 1 - Prob. 1B.9BECh. 1 - Prob. 1B.1PCh. 1 - Prob. 1B.2PCh. 1 - Prob. 1B.3PCh. 1 - Prob. 1B.4PCh. 1 - Prob. 1B.5PCh. 1 - Prob. 1B.6PCh. 1 - Prob. 1B.7PCh. 1 - Prob. 1B.8PCh. 1 - Prob. 1B.9PCh. 1 - Prob. 1B.10PCh. 1 - Prob. 1B.11PCh. 1 - Prob. 1C.1DQCh. 1 - Prob. 1C.2DQCh. 1 - Prob. 1C.3DQCh. 1 - Prob. 1C.4DQCh. 1 - Prob. 1C.1AECh. 1 - Prob. 1C.1BECh. 1 - Prob. 1C.2AECh. 1 - Prob. 1C.2BECh. 1 - Prob. 1C.3AECh. 1 - Prob. 1C.3BECh. 1 - Prob. 1C.4AECh. 1 - Prob. 1C.4BECh. 1 - Prob. 1C.5AECh. 1 - Prob. 1C.5BECh. 1 - Prob. 1C.6AECh. 1 - Prob. 1C.6BECh. 1 - Prob. 1C.7AECh. 1 - Prob. 1C.7BECh. 1 - Prob. 1C.8AECh. 1 - Prob. 1C.8BECh. 1 - Prob. 1C.9AECh. 1 - Prob. 1C.9BECh. 1 - Prob. 1C.1PCh. 1 - Prob. 1C.2PCh. 1 - Prob. 1C.3PCh. 1 - Prob. 1C.4PCh. 1 - Prob. 1C.5PCh. 1 - Prob. 1C.6PCh. 1 - Prob. 1C.7PCh. 1 - Prob. 1C.8PCh. 1 - Prob. 1C.9PCh. 1 - Prob. 1C.10PCh. 1 - Prob. 1C.11PCh. 1 - Prob. 1C.12PCh. 1 - Prob. 1C.13PCh. 1 - Prob. 1C.14PCh. 1 - Prob. 1C.15PCh. 1 - Prob. 1C.16PCh. 1 - Prob. 1C.17PCh. 1 - Prob. 1C.18PCh. 1 - Prob. 1C.19PCh. 1 - Prob. 1C.20PCh. 1 - Prob. 1C.22PCh. 1 - Prob. 1C.23PCh. 1 - Prob. 1C.24PCh. 1 - Prob. 1.1IACh. 1 - Prob. 1.2IACh. 1 - Prob. 1.3IA
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