Starting with the Maxwell-Boltmann distribution expression from the previous post,
\begin{equation}
f(v)\,dv = 4\pi v^2\left(\frac{m}{2\pi k_BT}\right)^{3/2}\exp\left[\frac{-mv^2}{2k_BT}\right]\,dv
\end{equation}We determined the expression for mean speed to be,
\begin{equation}
\langle v\rangle = \sqrt{\frac{2k_BT}{m\pi}}
\end{equation}Table of Contents
Deriving the root mean square speed expression
The root-mean-square speed is the square root of the second-order moment of the mean square velocity,
\begin{equation}
v_{rms} = \sqrt{\lang v^2\rang}
\end{equation}We can work out the expression for $\langle v^2\rangle$ starting from equation $(1)$ as,
\begin{equation}
\lang v^2\rang = \int\limits_0^\infty v^2f(v)\,dv
\end{equation}Substituting equation $(1)$ in $(4)$ we get,
\begin{equation}
\begin{aligned}
\lang v^2\rang &= \int\limits_0^\infty v^2\cdot4\pi v^2\left(\frac{m}{2\pi k_BT}\right)^{3/2}\exp\left[\frac{-mv^2}{2k_BT}\right]\\
\lang v^2\rang &= 4\pi\left(\frac{m}{2\pi k_BT}\right)^{3/2}\int\limits_0^\infty v^4\exp\left[\frac{-mv^2}{2k_BT}\right]\,dv
\end{aligned}
\end{equation}Assuming, $$b = \frac{m}{2k_BT}$$ we can simplify the above expression as,
\begin{equation}
\begin{aligned}
\lang v^2\rang &= 4\pi\left(\frac{b}{\pi}\right)^{3/2} \int\limits_0^\infty v^4exp\left[-bv^2\right]\,dv\\
\lang v^2\rang &= 4\pi\left(\frac{b}{\pi}\right)^{3/2}A(v)
\end{aligned}
\end{equation}Solving for $A(v)$ we can obtain the expression,
\begin{equation}
A(v) = \int\limits_0^\infty v^3\cdot v\exp\left[-bv^2\right]\,dv
\end{equation}Using integration by parts on equation $(7)$,
\begin{equation*}
\begin{aligned}
A(v) &= v^3 \int\limits_0^\infty v\exp\left[-bv^2\right]\,dv - \int\limits_0^\infty \int v\exp\left[-bv^2\right]\,dv\left(\frac{d}{dv}v^3\right)\,dv\\
A(v) &= v^3\int\limits_0^\infty v\exp\left[-bv^2\right]\,dv - 3\int\limits_0^\infty v^2\left\{\int v\exp\left[-bv^2\right]\,dv\right\}\,dv\\
\end{aligned}
\end{equation*}\begin{equation}
A(v) = v^3 B(v) - 3\int\limits_0^\infty v^2 B(v)\,dv
\end{equation}Evaluating the expression for $B(v)$,
\begin{equation}
B(v) = \int v\exp\left[-bv^2\right]\,dv
\end{equation}Substituing $u = -bv^2$, we can reform the expression as,
\begin{equation*}
\begin{aligned}
u &= -bv^2\\
du &= -b\left(2v\,dv\right)\\
dv &= \frac{-1}{2bv}\,du
\end{aligned}
\end{equation*}Rewriting equation $(9)$ with the substitutions,
\begin{equation*}
\begin{aligned}
B(v) &= \int v\exp\left[u\right]\,\frac{-1}{2bv}\,du\\
B(v) &= -\frac{1}{2b}\int\exp[u]\,du\\
B(v) &= -\frac{1}{2b}\exp\left[u\right]
\end{aligned}
\end{equation*}Removing the substituion from $u = -bv^2$ we get,
\begin{equation}
B(v) = -\frac{1}{2b}\exp\left[-bv^2\right]
\end{equation}Substituting equation $(10)$ back in equation $(8)$ we get,
\begin{equation}
\begin{aligned}
A(v) &= \left.-\frac{v^3}{2b}\exp\left[-bv^2\right]\right|_0^\infty - 3\int\limits_0^\infty v^2 \left(-\frac{1}{2b}\exp\left[-bv^2\right]\right)\,dv\\
A(v) &= \left.-\frac{v^3}{2b}\exp\left[-bv^2\right]\right|_0^\infty + \frac{3}{2b}\int\limits_0^\infty v^2\exp\left[-bv^2\right]\,dv
\end{aligned}
\end{equation}The left expression in equation $(11)$ evaluates to $0$ under the limits of integration which simplifies the expression to,
\begin{equation}
A(v) = \frac{3}{2b}\int\limits_0^\infty v^2\exp\left[-bv^2\right]\,dv
\end{equation}We can use the integration by parts on equation $(12)$,
\begin{equation*}
\begin{aligned}
A(v) &= \frac{3}{2b}\int\limits_0^\infty v\cdot v\exp\left[-bv^2\right]\,dv\\
A(v) &= \frac{3}{2b}\left\{v \int\limits_0^\infty v\exp\left[-bv^2\right]\,dv - \int\limits_0^\infty \int v\exp\left[-bv^2\right]\,dv\left(\frac{d}{dv}v\right)\,dv\right\}\\
\end{aligned}
\end{equation*}The expression inside the integrals is once again $B(v)$, we can resubstitute the values from equation $(10)$
\begin{equation}
A(v) = \frac{3}{2b}\left\{\left.-\frac{v}{2b}\exp\left[-bv^2\right]\right|_0^\infty + \frac{1}{2b} \int\limits_0^\infty \exp[-bv^2]\,dv\right\}
\end{equation}The expression on the left in equation $(13)$ becomes $0$ under the integration limits which simplifies equation $(13$) to,
\begin{equation}
\begin{aligned}
A(v) &= \frac{3}{4b^2}\int\limits_0^\infty\exp\left[-bv^2\right]\,dv\\
A(v) &= \frac{3}{4b^2}\,C(v)
\end{aligned}
\end{equation}Evaluating the expression for $C(v)$, we get
\begin{equation}
C(v) = \int\limits_0^\infty\exp\left[-bv^2\right]\,dv
\end{equation}Substituting $u = v\sqrt{b}$ we can reform the expression as,
\begin{equation*}
\begin{aligned}
u &= bv^2\\
du &= 2bv\,dv\\
dv &= \frac{du}{2bv}\\
dv &= \frac{du}{2b}\sqrt{\frac{b}{u}}
\end{aligned}
\end{equation*}Rewriting equation $(15)$ with the substitutions,
\begin{equation}
\begin{aligned}
C(v) &= \int\limits_0^\infty\exp\left[-u\right]\,\frac{du}{2b}\sqrt{\frac{b}{u}}\\
C(v) &= \frac{1}{2\sqrt{b}}\int\limits_0^\infty u^{-1/2}\exp\left[-u\right]\,du
\end{aligned}
\end{equation}The integration in eqaution $(16)$ is a special integral, the gamma function which takes the form of,
\Gamma(z) = \int\limits_0^\infty x^{z-1}\exp\left[-x\right]\,dxNow, wee can write equation $(16)$ as,
\begin{equation*}
\begin{aligned}
C(v) &= \frac{1}{2\sqrt{b}}\,\Gamma\left(\frac{1}{2}\right)\\
C(v) &= \frac{1}{2\sqrt{b}}\sqrt{\pi}\\
\end{aligned}
\end{equation*}\begin{equation}
C(v) = \sqrt{\frac{\pi}{4b}}\\
\end{equation}Substituting equation $(17)$ in equation $(14)$ we get,
\begin{equation}
\begin{aligned}
A(v) &= \frac{3}{4b^2}\sqrt{\frac{\pi}{4b}}\\
A(v) &= \frac{3}{8}\sqrt{\frac{\pi}{b^5}}
\end{aligned}
\end{equation}Substituting back equation $(18)$ in equation $(6)$ we finally get,
\begin{equation*}
\begin{aligned}
\lang v^2\rang &= 4\pi \left(\frac{b}{\pi}\right)^{3/2}\frac{3}{8}\sqrt{\frac{\pi}{b^5}}\\
\lang v^2\rang &= \frac{12}{8} \frac{\pi\cdot\pi^{1/2}}{\pi^{3/2}}\frac{b^{3/2}}{b^{5/2}}\\
\lang v^2\rang &= \frac{3}{2}\pi^{1+\frac{1}{2}-\frac{3}{2}}b^{\frac{3}{2}-\frac{5}{2}}\\
\lang v^2\rang &= \frac{3}{2}\pi^0b^{-1}
\end{aligned}
\end{equation*}\begin{equation*}
\begin{aligned}
\lang v^2\rang &= \frac{3}{2}b^{-1}\\
\lang v^2\rang &= \frac{3}{2}\left(\frac{m}{2k_BT}\right)^{-1}\\
\lang v^2\rang &= \frac{3}{2}\frac{2k_BT}{m}
\end{aligned}
\end{equation*}\begin{equation}
\lang v^2\rang = \frac{3k_BT}{m}
\end{equation}Equation $(19)$ is the expression for mean squre velocity. In order to obtain the root-mean square velocity, we substitute equation $(19)$ back in equation $(3)$ to get,
\begin{equation}
v_\text{rms} = \sqrt{\frac{3k_BT}{m}}
\end{equation}We have derived the expression for the root-mean-square speed from the Maxwell-Boltzmann distribution.
written in very efficient way
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