How do derive the equation for relative velocity of gas molecules

Question

The relative velocity is $$v_{rel}=\sqrt2v_{mean}$$ where $v_{mean}=(\frac{8KT}{\pi m})^{\frac12}$$$v_{rel}=(\frac{16KT}{\pi \frac m2})^{\frac12}$$ because the reduced mass of two of the same particles is half the mass of one. In general: $$v_{rel}=(\frac{8KT}{\pi \mu})^{\frac12}$$

However, my question is: what is the need for reduced mass? If $v_{rel}=\sqrt2v_{mean}$ where $v_{mean}=(\frac{8KT}{\pi m})^{\frac12}$. Wouldn't that just leave you with: $v_{mean}=(\frac{16KT}{\pi m})^{\frac12}$ Where does reduced mass come into it?

Answer 1

Your equation $v_{rel}=(\frac{16KT}{\pi \frac m2})^{\frac12}$ is incorrect.

$v_{mean}=(\frac{16KT}{\pi m})^{\frac12}$ (your last equation) is also incorrect.

Instead $v_{rel}=(\frac{16KT}{\pi m})^{\frac12}$ and $v_{rel}=(\frac{8KT}{\pi \frac m2})^{\frac12}$ would both be correct.

The need for reduced mass is for when the particles are not the same mass, usually in the context of considering collisions between two different molecules or atoms.

$\mu = m_1m_2/(m_1 + m_2)$ by definition. If $m_1 = m_2$ then $\mu = m/2$

When there are two different masses, there are two different mean velocities.

To see where the equation with the reduced mass comes from, look at appendix 1.4 of Kinetic Theory of Gases, along with pages 1-30 to 1-34.