[NOTES/SM-04006] Maxwell Distribution of speeds in an ideal gas

Maxwell's distribution of speed

Given this distribution function, we can compute distribution function for speed \(v\):

\begin{align}
\eta(v)dv = &\int_0^{2\pi}d\phi\int_0^\pi d\theta
\eta(\vec{v})dv^2\sin\theta du\\
=&4\pi v^2\eta(\vec{v})dv\\
=& N (4\pi v^2) \left(\frac{m}{2\pi kT}\right)^{3/2}
\exp\left(-\frac{mv^2}{2kT}\right) dv.\label{EQ02}
\end{align}

Velocity and speed distribution of molecules

From \eqRef{EQ02}, we can now write an expression for the number of molecules having {\tt velocity} in the range \(\vec{v}+d\vec{v}\) is given by

\begin{eqnarray}
\boxed{n(\vec{v}) d^3v = N V \exp\Big(-\frac{m}{2kT} (v_x^2+v^2_y+v^2_z) \Big)} d^3v.
\end{eqnarray}
Multiplying the expression in \eqRef{EQ02} by total number \(N\) of molecules, we get the probable number of molecules having {\tt speed} between $v$ and $v+dv$ as

\begin{equation}
\boxed{n(v) dv = (4\pi v^2)N\left(\frac{m}{2\pi kT}\right)^{3/2}
\exp\left(-\frac{mv^2}{2kT}\right) dv.}
\end{equation}

Most probable speed

The most probable speed, \(v_m\), is computed by finding the maximum of \(n(v)\) and we have

$$
v_m=\left(\frac{2kT}{m}\right)^{1/2}
$$

Average speed

The average speed $\bar{v}$ is given by

\begin{align*}
\bar{v}=&\frac{1}{N}\int_0^\infty 4\pi v^2\eta(v_xv_yv_z)dv\\
=&\frac{4}{\sqrt{\pi}}\left(\frac{m}{2kT}\right)^{3/2} \int_0^\infty
\exp\left(-\frac{mv^2}{2kT}\right)dv\\
=&\left(\frac{8kT}{\pi m}\right)^{1/2}\\
\bar{v}=&\frac{1}{\sqrt{\pi}} v_m\qquad(\vec{v}/v_m)\approx
\frac{1}{1.113}
\end{align*}

An Example

Average speed of $O_2$ molecule at $T=300^0K$, $V=100 cc$, $p=1$atm
\begin{align*}
m(O_2) = &16 \text{amu} \times 2 =32 \text{amu}\\
=&32\times1.66\times10^{-2}\text{kg}\\
m=&5.3\times10^{-26}\text{kg}\\
\bar{v} =&445 \text{m/sec\ (~1000 miles/hour)}.
\end{align*}