Explanation Given Info: A tabulated integral is given as ∫ 0 ∞ x n e − α x d x = n ! α n + 1 where n is a positive integer and α is a positive constant. The expression for f ( v ) is, f ( v ) = 4 π ( m 2 π k T ) 3 2 v 2 e − m v 2 2 k T f ( v ) is the probability per unit speed interval(distribution function). m is mass v is molecular speed. k is Boltzmann constant. T is absolute temperature. Substitute 4 π ( m 2 π k T ) 3 2 v 2 e − m v 2 2 k T for f ( v ) in ∫ 0 ∞ v f ( v ) d v . ∫ 0 ∞ v f ( v ) d v = ∫ 0 ∞ v [ 4 π ( m 2 π k T ) 3 2 v 2 e − m v 2 2 k T ] d v = ∫ 0 ∞ 4 π ( m 2 π k T ) 3 2 v 3 e − m v 2 2 k T d v = 4 π ( m 2 π k T ) 3 2 ∫ 0 ∞ v 3 e − m v 2 2 k T d v Thus, ∫ 0 ∞ v f ( v ) d v = 4 π ( m 2 π k T ) 3 2 ∫ 0 ∞ v 3 e − m v 2 2 k T d v (I) Substitute v 2 for x

University Physics with Modern Physics (14th Edition)

14th Edition

ISBN: 9780321973610

Author: Hugh D. Young, Roger A. Freedman

Publisher: PEARSON

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Maxwell speed distribution

Maxwell-Boltzmann distribution or commonly known as Maxwell speed distribution, explains the division of the energy levels of the molecules of an ideal gas on the basis of the statistical theory. In 1859, Scottish physicist James Clerk Maxwell establishe…

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Chapter 18, Problem 18.78P

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The ∫0∞v f(v)dv integral and compare it to vav .

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Chapter 18, Problem 18.77P

Chapter 18, Problem 18.79P

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Chapter 18 Solutions

University Physics with Modern Physics (14th Edition)

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The ∫ 0 ∞ v f ( v ) d v integral and compare it to v a v .

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The ∫0∞v f(v)dv integral and compare it to vav .

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Chapter 18 Solutions