   Chapter 7.8, Problem 62E

Chapter
Section
Textbook Problem

# The average speed of molecules in an ideal gas is v ¯ = 4 π ( M 2 R T ) 3 / 2 ∫ 0 ∞ v 3 e − M v 2 / ( 2 R T ) d v where M is the molecular weight of the gas, R is the gas constant, T is the gas temperature, and v is the molecular speed. Show that v ¯ = 8 R T π M

To determine

To show:

That the average speed of molecule is v¯=8RTπM.

Explanation

Given:

v¯=4π(M2RT)320v3eMV2(2RT)dv

For an ideal gas.

Formula used:

Integration by parts: 0tf(x)g'(x)dx=[f(x)g(x)]0t0tf'(x)g(x)dx

v¯=4π(M2RT)320v3eMV2(2RT)dv

Let suppose K=M2RT

Now that since M, R and T are all positive K is negative.

Therefore,0v3eKv2dv=limt0tv3eKv2dv

Now take,

f(v)=v2g(v)=eKv22K

Then

f(v)=2vg(v)=2KveKv22K=veKv2

Thus,0tv3eKv2dv=0tv2(veKv2)dv

=0tf(v)g(v)dv=[f(v)g(v)]0t0tf(v)g(v)dv

Now

0tf(v)g(v)dv=[v2eKv22K]0t0t2veK

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