MBDistEPSolnME501F2011

# MBDistEPSolnME501F2011 - 1 ME 501 MB Speed Distribution...

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1 ME 501 MB Speed Distribution Example Problems Fall 2011 Answer b. 0.739 Solution: Given the Maxwell-Boltzmann speed distribution,  3/2 22 exp / 2 B B m fVd V V m V kTd V kT     We define 2 2 2 2 mp B BB B B Vm xV Vk T V x V dV x dV x dx mm m dV dx m   Therefore   exp / 2 2 exp 2 4 e x p e x p B B B m V fxd x V m V V xx d x m m m f x dx x x dx x x dx   

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2 The fraction of particles with speeds between 0 and V is thus given by    22 00 0 4 exp Vx x fVd V fxd x x x d x   We can evaluate this integral using integration by parts. 2 2 2 2 2 2 2 2 exp 2e x p 2 exp exp exp 1 exp exp exp 42 2 exp exp exp udv uv vdu ux v x du dx dv x x dx xx d x x x x d x x d x x d x x x x xd x x x      Therefore,     2 2 0 2 2 exp exp exp 2 exp x xV x f xd x x x x x x f x dx f V dV erf x x Now calculate the fraction of argon atoms at 300 K with 500 / Vm s .
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MBDistEPSolnME501F2011 - 1 ME 501 MB Speed Distribution...

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