- #1
Je m'appelle
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Homework Statement
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]
Homework 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]
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]
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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?