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Physical Chemistry

2nd Edition
Ball + 3 others
Publisher: Wadsworth Cengage Learning,
ISBN: 9781133958437

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Section
BuyFindarrow_forward

Physical Chemistry

2nd Edition
Ball + 3 others
Publisher: Wadsworth Cengage Learning,
ISBN: 9781133958437
Chapter 19, Problem 19.47E
Textbook Problem
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What is the total number of collisions per second for the gas system described in the previous exercise? (Note how this question is different from the earlier one.)

Interpretation Introduction

Interpretation:

The total number of collisions per second for the gas system described in the previous exercise is to be calculated.

Concept introduction:

The average collision rate is given by the formula,

z=2πρd2kTπm

Where,

ρ is the density.

T is the temperature.

m is the mass of an atom.

d is the collision diameter.

k is the Boltzmann constant

Explanation of Solution

The molar mass of the ozone molecule is 7.97×1026kg. The given value of collision diameter, density and temperature is 5.00A, 8.06×1019molecules/dm3 and 45.0°C respectively.

The average collision rate is given by the formula,

z=2πρd2kTπm …(1)

Where,

ρ is the density.

T is the temperature.

m is the mass of an atom.

d is the collision diameter.

k is the Boltzmann constant.

Substitute the value of collision diameter and temperature in equation (1).

z=2×3.14×8.06×1022m3×(5.00A×1010m1A)2(1.381×1023J/K)(228K)3.14×7.97×1026kgz=1265.42×102m1×5.61×10115.00×1013s1z=1.42×107s1

Thus, the average collision rate is 1.42×107s1.

The number of collision per molecule is calculated as follows:

1.42×107s16.022×1023molecules=2.35×1017s1(molecules)1

Thus, the number of collisions per molecule is 2.35×1017s1(molecules)1

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Chapter 19 Solutions

Physical Chemistry
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