In our previous posts, we derived the mean and root-mean-square speeds from the Maxwell-Boltzmann distribution (MBd hereafter). Here, we will attempt to derive the most probable speed that can be attained by the gas molecules following the MBd.
The most probable speed is defined as the speed at which the distribution graph reaches its peak.
Or,
The maximum speed attainable by the gas molecules.
Table of Contents
Deriving the most probable speed
The distribution function of MBd is expressed as follows,
\begin{equation}
f(v) = 4\pi\left(\frac{m}{2\pi k_BT}\right)^{3/2}v^2\exp\left[\frac{-mv^2}{2k_BT}\right]
\end{equation}The most probable speed would correspond to the maximum value of this function. To obtain the maximum value, we require the derivative of the function to be equal to zero. We now differentiate the function with respect to the speed $v$,
\begin{equation}
\frac{df(v)}{dv} = 4\pi\left(\frac{m}{2\pi k_BT}\right)^{3/2}\frac{d}{dv}\left(v^2\exp\left[\frac{-mv^2}{2k_BT}\right]\right)
\end{equation}To simplify our calcluations we make the following substitution,
b = \frac{m}{2k_BT}With this, we can rewrite equation $(4)$ as,
\begin{equation*}
\begin{aligned}
\frac{df(v)}{dv} &= 4\pi\left(\frac{b}{\pi}\right)^{3/2} \frac{d}{dv}v^2\exp\left[-bv^2\right]\\
\frac{df(v)}{dv} &= 4\pi\left(\frac{b}{\pi}\right)^{3/2}\left(\frac{d}{dv}v^2\exp\left[-bv^2\right] + v^2\frac{d}{dv}\exp\left[-bv^2\right]\right)\\
\frac{df(v)}{dv} &= 4\pi\left(\frac{b}{\pi}\right)^{3/2}\left(2v\exp\left[-bv^2\right]+ v^2\left[\exp\left[-bv^2\right]\frac{d}{dv}(-bv^2)\right]\right)\\
\frac{df(v)}{dv} &= 4\pi\left(\frac{b}{\pi}\right)^{3/2} \left(2v\exp\left[-bv^2\right] + v^2\exp\left[-bv^2\right]\left(-2bv\right)\right)\\
\frac{df(v)}{dv} &= 4\pi\left(\frac{b}{\pi}\right)^{3/2}\left(2v\exp\left[-bv^2\right] - 2bv^3\exp\left[-bv^2\right]\right)
\end{aligned}
\end{equation*}We now simplify the last expression obtained,
\begin{equation}
\frac{df(v)}{dv} = 4\pi\left(\frac{b}{\pi}\right)^{3/2}2v\exp\left[-bv^2\right]\left(1 - bv^2\right)
\end{equation}We note that the factored out expressions, or the exponential function cannot be equal to zero. Hence, we have only two possible solutions,
v = 0 \qquad\text{or}\qquad1-bv^2=0Since $v=0$ will give the minimum value of the distribution, so we solve the second expression to obtain the maxmimu of the distribution function,
\begin{equation}
\begin{aligned}
1 - bv^2&=0\\
bv^2 &= 1\\
\frac{m}{2k_BT}v^2 &= 1\\
v^2 &= \frac{2k_BT}{m}
\end{aligned}
\end{equation}Taking the square root,
\begin{equation}
v_\text{p} = \sqrt{\frac{2k_BT}{m}}
\end{equation}Equation $(5)$ represents the expression for the most probable speed in the Maxwell-Boltzmann distribution.
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