?The average speed of molecules in an ideal gas is

Chapter 7, Problem 74

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The average speed of molecules in an ideal gas is

$\bar{v}=\frac{4}{\sqrt{\pi}}\left(\frac{M}{2 R T}\right)^{3 / 2} \int_{0}^{\infty} v^{3} e^{\left.-M v^{2} / 2 R T\right)} d v$

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

$\bar{v}=\sqrt{\frac{8 R T}{\pi M}}$

Equation Transcription:

${\overline{v}}^{\ }=\ \frac{4}{\sqrt{\pi{}}}(\frac{M}{2RT}{)}^{3/2}\int\limits_{0}^{\infty{}}{v}^{3}{e}^{-M{v}^{2}/2RT)}dv$

${\overline{v}}^{\ }=\ \sqrt{\frac{8RT}{\pi{}M}}$

Text Transcription:

v bar = 3/square root of pi (M/2RT)^3/2 integral from 0 to infinity v^3 e^-Mv^2/2RT) dv

v bar = square root 8RT/piM

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