- #1
AStaunton
- 100
- 1
maxwell Boltzmann (very sorry!)
I attempted to post this already but the latex code didn't work out and also the edit command didn't seem to work...so I am reposting, very sorry for the duplicate but do not know how to remove original!:
trying to show v_average=sqrt(8kT/(pi)m) using max boltz. distrib.:
f(v)=4\pi(\frac{m}{2\pi mkT})^{\frac{3}{2}}v{}^{2}e^{-\frac{mv^{2}}{2kT}}
to start with the formula is:
\langle v\rangle=\int_{0}^{\infty}vf(v)dv
to solve this need substitution x=v^2 and this gives dx=2vdv which gives 1/2dx=vdv and so leads to:
4\pi(\frac{m}{2\pi kT})^{\frac{3}{2}}\frac{1}{2}\int_{0}^{\infty}xe^{-mx/kT}dx
can now integrate the above by parts:
u=x dv=e^-mx/kT
du=dx v=-(kT/m)e^-mx/kT
the uv expression evaluates to 0 and so the final part of interest I am left with is:
\frac{kT}{m}\int_{0}^{\infty}e^{-mx/kT}dx=-\frac{k^{2}T^{2}}{m^{2}}e^{-mx/kT}|_{0}^{\infty}=-\frac{k^{2}T^{2}}{m^{2}}
not forgetting the constant term that was brought outside integral at very start:
(2\pi(\frac{m}{2\pi kT})^{\frac{3}{2}})(-\frac{k^{2}T^{2}}{m^{2}})=-\sqrt{\frac{kT}{2m\pi}}
this is my final answer and is in disagreement with the answer I should have gotten for two reasons:
the sign is clearly wrong and also instead of sqrt2 on the bottom I should have sqrt8 on the top:
\sqrt{\frac{8kT}{m\pi}}
Can anyone suggest where problem is, as I have rechecked a few times and I keep coming out with the same answer.
I attempted to post this already but the latex code didn't work out and also the edit command didn't seem to work...so I am reposting, very sorry for the duplicate but do not know how to remove original!:
trying to show v_average=sqrt(8kT/(pi)m) using max boltz. distrib.:
f(v)=4\pi(\frac{m}{2\pi mkT})^{\frac{3}{2}}v{}^{2}e^{-\frac{mv^{2}}{2kT}}
to start with the formula is:
\langle v\rangle=\int_{0}^{\infty}vf(v)dv
to solve this need substitution x=v^2 and this gives dx=2vdv which gives 1/2dx=vdv and so leads to:
4\pi(\frac{m}{2\pi kT})^{\frac{3}{2}}\frac{1}{2}\int_{0}^{\infty}xe^{-mx/kT}dx
can now integrate the above by parts:
u=x dv=e^-mx/kT
du=dx v=-(kT/m)e^-mx/kT
the uv expression evaluates to 0 and so the final part of interest I am left with is:
\frac{kT}{m}\int_{0}^{\infty}e^{-mx/kT}dx=-\frac{k^{2}T^{2}}{m^{2}}e^{-mx/kT}|_{0}^{\infty}=-\frac{k^{2}T^{2}}{m^{2}}
not forgetting the constant term that was brought outside integral at very start:
(2\pi(\frac{m}{2\pi kT})^{\frac{3}{2}})(-\frac{k^{2}T^{2}}{m^{2}})=-\sqrt{\frac{kT}{2m\pi}}
this is my final answer and is in disagreement with the answer I should have gotten for two reasons:
the sign is clearly wrong and also instead of sqrt2 on the bottom I should have sqrt8 on the top:
\sqrt{\frac{8kT}{m\pi}}
Can anyone suggest where problem is, as I have rechecked a few times and I keep coming out with the same answer.