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Maxwell speed distribution 
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#1
Apr2211, 10:23 AM

P: 109

1. The problem statement, all variables and given/known data
Show that the number [tex]N(0,v)[/tex], of molecules of an ideal gas with speeds between 0 and v is given by [tex]N(0,v) = N \left[ erf(\xi)  \frac{2}{\sqrt{\pi}} \xi e^{\xi^2} \right][/tex] Where, [tex]erf(\xi) = \frac{2}{\sqrt{\pi}} \int_{0}^{\xi} e^{x^2} dx [/tex] And, [tex]\xi^2 = \left(\frac{mv^2}{2kT} \right)[/tex] 2. Relevant equations [tex]\frac{dN_v}{N} = \sqrt{\frac{2}{\pi}} \left( \frac{m}{kT}\right)^{\frac{3}{2}} v^2 e^{\frac{mv^2}{2kT}} \ dv [/tex] 3. The attempt at a solution Alright, so I managed to get to the following [tex]\frac{dN_v}{N} = \frac{4}{\sqrt{\pi}} x^2 e^{x^2} dx [/tex] Where, [tex]\alpha = \frac{m}{2kT}, \ x = \sqrt{\alpha}v [/tex] =========================\\=========================== So far so good, but now when checking the solution on the textbook it claims the following algebraic manipulation which I can't follow [tex]\frac{2N}{\sqrt{\pi}}\int_{0}^{\xi} x (2xe^{x^2} dx) = \frac{2N}{\sqrt{\pi}} x e^{x^2} _{\xi}^{0} \ + \ N \frac{2}{\sqrt{\pi}} \int_{0}^{\xi} e^{x^2} dx [/tex] What was done on the integral above, how can you "split" it like that? 


#2
Apr2211, 10:49 AM

P: 61

Integration by parts



#3
Apr2211, 11:00 AM

Sci Advisor
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P: 25,244

It's integration by parts. udv=d(uv)vdu with u=x and dv=(2x)e^(x^2)dx.



#4
Apr2211, 11:10 AM

P: 109

Maxwell speed distribution
Thanks for the highlight, Lx! 


#5
Apr2211, 11:15 AM

P: 61

No worries. It's not immidiately obvious, although the integral is written in a way which gives you a clue: x(2x) instead of 2x^2



#6
Apr2211, 11:29 AM

Sci Advisor
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P: 25,244




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