- #1
RagincajunLA
- 19
- 0
I was wondering if there is a mathematical difference between the RMS free path and the mean free path of molecules in an ideal gas. For example, When I calculate the mean free path, I use use the average velocity and the scattering rate which is a function of the average velocity. I then multiply the two and the average velocities cancel to produce the mean free path.
When I go to calculate the RMS free path, I assume I use the RMS velocity and the RMS scattering rate which is based on the RMS velocity. When I multiply these two, the RMS velocity cancels and I am left with the value as the mean free path. Is this correct? I assumed they would be different. My equations are...
When I go to calculate the RMS free path, I assume I use the RMS velocity and the RMS scattering rate which is based on the RMS velocity. When I multiply these two, the RMS velocity cancels and I am left with the value as the mean free path. Is this correct? I assumed they would be different. My equations are...
[itex]A[/itex] = [itex]\bar{v}[/itex][itex]\tau[/itex]
[itex]\tau[/itex] = [itex]\frac{1}{\sqrt{2}n \pi d^{2} \bar{v} }[/itex]
[itex]A_{rms}[/itex] = [itex]v_{rms}[/itex][itex]\tau_{rms}[/itex]
[itex]\tau_{rms}[/itex] = [itex]\frac{1}{\sqrt{2}n \pi d^{2} v_{rms} }[/itex]
[itex]\tau[/itex] = [itex]\frac{1}{\sqrt{2}n \pi d^{2} \bar{v} }[/itex]
[itex]A_{rms}[/itex] = [itex]v_{rms}[/itex][itex]\tau_{rms}[/itex]
[itex]\tau_{rms}[/itex] = [itex]\frac{1}{\sqrt{2}n \pi d^{2} v_{rms} }[/itex]