Editing Maxwell–Boltzmann distribution (section)

== Typical speeds ==
[[File:Stellar MB.png|alt=Solar Atmosphere Maxwell–Boltzmann Distribution.|thumb|400px|The Maxwell–Boltzmann distribution corresponding to the solar atmosphere. Particle masses are one [[proton mass]], {{math|1=''m''<sub>p</sub> = {{val|1.67|e=-27|u=kg}} ≈ {{val|1|ul=Da}}}}, and the temperature is the effective temperature of the [[Photosphere|Sun's photosphere]], {{math|1=''T'' = 5800 K}}. <math>\tilde{V}</math>, <math>\bar{V}</math>, and {{math|''V''<sub>rms</sub>}} mark the most probable, mean, and root mean square velocities, respectively. Their values are <math>\tilde{V}</math> ≈ {{val|9.79|u=km/s}}, <math>\bar{V}</math> ≈ {{val|11.05|u=km/s}}, and {{math|''V''<sub>rms</sub> ≈ {{val|12.00|u=km/s}}}}.]]
The [[expectation value|mean]] speed <math> \langle v \rangle</math>, most probable speed ([[Mode (statistics)|mode]]) {{math|''v''<sub>p</sub>}}, and root-mean-square speed <math display="inline">\sqrt{\langle v^2 \rangle}</math> can be obtained from properties of the Maxwell distribution.

This works well for nearly [[ideal gas|ideal]], [[noble gas|monatomic]] gases like [[helium]], but also for [[Molecule|molecular gas]]es like diatomic [[oxygen]]. This is because despite the larger [[heat capacity]] (larger internal energy at the same temperature) due to their larger number of [[Equipartition theorem|degrees of freedom]], their [[Translation (physics)|translational]] [[kinetic energy]] (and thus their speed) is unchanged.<ref>{{cite book | title = College Physics, Volume 1 | edition = 9th | first1 = Raymond A. | last1 = Serway | first2 = Jerry S. | last2 = Faughn | first3 = Chris | last3 = Vuille | name-list-style=amp | year = 2011 | isbn = 9780840068484 | page = 352 | publisher = Cengage Learning | url = https://books.google.com/books?id=HLxV-IKYO5IC&pg=PA352 }}</ref>

{{bulleted list
| The most probable speed, {{math|''v''<sub>p</sub>}}, is the speed most likely to be possessed by any molecule (of the same mass {{mvar|m}}) in the system and corresponds to the maximum value or the [[mode (statistics)|mode]] of {{math|''f''(''v'')}}. To find it, we calculate the [[derivative]] {{tmath|\tfrac{df}{dv},}} set it to zero and solve for {{mvar|v}}: <math display="block">
\frac{df(v)}{dv} = -8\pi \biggl[\frac{m}{2 \pi k_\text{B}T}\biggr]^{3/2} \, v \, \left[\frac{mv^2}{2k_\text{B}T}-1\right] \exp\left(-\frac{mv^2}{2k_\text{B}T}\right) = 0
</math> with the solution: <math display="block">
\frac{mv_\text{p}^2}{2k_\text{B}T} = 1; \quad v_\text{p} = \sqrt{ \frac{2k_\text{B}T}{m} } = \sqrt{ \frac{2RT}{M} }
</math>where:
*{{mvar|R}} is the [[gas constant]];
*{{mvar|M}} is molar mass of the substance, and thus may be calculated as a product of particle mass, {{mvar|m}}, and [[Avogadro constant]], {{math|''N''<sub>A</sub>}}: <math>M = m N_\mathrm{A}.</math>
For diatomic nitrogen ({{chem2|N2}}, the primary component of [[air]])<ref group="note">The calculation is unaffected by the nitrogen being diatomic. Despite the larger [[heat capacity]] (larger internal energy at the same temperature) of diatomic gases relative to monatomic gases, due to their larger number of [[Equipartition theorem|degrees of freedom]], <math>\frac{3RT}{M_\text{m}}</math> is still the mean [[Translation (physics)|translational]] [[kinetic energy]]. Nitrogen being diatomic only affects the value of the molar mass {{math|1=''M'' = {{val|28|u=g/mol}}}}.
See e.g. K. Prakashan, ''Engineering Physics'' (2001), [https://books.google.com/books?id=6C0R1qpAk7EC&pg=SA2-PA278 2.278].</ref> at [[room temperature]] ({{val|300|u=K}}), this gives
<math display="block">v_\text{p} \approx \sqrt{\frac{2 \cdot 8.31\, \mathrm{J {\cdot} {mol}^{-1} K^{-1}} \ 300\, \mathrm{K}}{0.028\, \mathrm{ {kg} {\cdot} {mol}^{-1} }}} \approx 422\, \mathrm{m/s}.</math>
| The mean speed is the [[expected value]] of the speed distribution, setting <math display="inline">b= \frac{1}{2a^2} = \frac{m}{2k_\text{B}T}</math>: <math display="block">\begin{align}
 \langle v \rangle
&= \int_0^{\infty} v \, f(v) \, dv \\[1ex]
&= 4\pi \left[ \frac{b}{\pi} \right]^{3/2} \int_0^\infty v^3 e^{-b v^2} dv
= 4\pi \left[ \frac{b}{\pi} \right]^{3/2} \frac{1}{2b^2} \\[1.4ex]
&= \sqrt{\frac{4}{\pi b}}
= \sqrt{ \frac{8k_\text{B}T}{\pi m}}
= \sqrt{ \frac{8RT}{\pi M}}
= \frac{2}{\sqrt{\pi}} v_\text{p}
\end{align}</math>
| The mean square speed <math>\langle v^2 \rangle</math> is the second-order [[Moment (mathematics)|raw moment]] of the speed distribution. The "root mean square speed" <math> v_\text{rms}</math> is the square root of the mean square speed, corresponding to the speed of a particle with average [[kinetic energy]], setting <math display="inline">b = \frac{1}{2a^2} = \frac{m}{2k_\text{B}T}</math>: <math display="block">\begin{align}
v_\text{rms}
& = \sqrt{\langle v^2 \rangle}
= \left[\int_0^{\infty} v^2 \, f(v) \, dv \right]^{1/2} \\[1ex]
& = \left[ 4 \pi \left (\frac{b}{\pi } \right)^{3/2} \int_{0}^{\infty} v^4 e^{-bv^2} dv\right]^{1/2} \\[1ex]
& = \left[ 4 \pi \left (\frac{b}{\pi}\right )^{3/2} \frac{3}{8} \left(\frac{\pi}{b^5}\right)^{1/2} \right]^{1/2}
= \sqrt{ \frac{3}{2b} } \\[1ex]
&= \sqrt { \frac{3k_\text{B}T}{m}}
= \sqrt { \frac{3RT}{M} }
= \sqrt{ \frac{3}{2} } v_\text{p}
\end{align}</math>
}}

In summary, the typical speeds are related as follows:
<math display="block">v_\text{p} \approx 88.6\%\ \langle v \rangle < \langle v \rangle < 108.5\%\ \langle v \rangle \approx v_\text{rms}. </math>

The root mean square speed is directly related to the [[speed of sound]] {{mvar|c}} in the gas, by
<math display="block">c = \sqrt{\frac{\gamma}{3}} \ v_\mathrm{rms} = \sqrt{\frac{f+2}{3f}}\ v_\mathrm{rms} = \sqrt{\frac{f+2}{2f}}\ v_\text{p} ,</math>
where <math display="inline">\gamma = 1 + \frac{2}{f}</math> is the [[adiabatic index]], {{mvar|f}} is the number of [[degrees of freedom]] of the individual gas molecule. For the example above, diatomic nitrogen (approximating [[air]]) at {{val|300|u=K}}, <math>f = 5</math><ref group="note">Nitrogen at room temperature is considered a "rigid" diatomic gas, with two rotational degrees of freedom additional to the three translational ones, and the vibrational degree of freedom not accessible.</ref> and
<math display="block">c = \sqrt{\frac{7}{15}}v_\mathrm{rms} \approx 68\%\ v_\mathrm{rms} \approx 84\%\ v_\text{p} \approx 353\ \mathrm{m/s}, </math>
the true value for air can be approximated by using the average molar weight of [[Atmospheric chemistry|air]] ({{val|29|u=g/mol}}), yielding {{val|347|u=m/s}} at {{val|300|u=K}} (corrections for variable [[humidity]] are of the order of 0.1% to 0.6%).

The average relative velocity
<math display="block">\begin{align}
v_\text{rel} \equiv \langle |\mathbf{v}_1 - \mathbf{v}_2| \rangle
&= \int \! d^3\mathbf{v}_1 \, d^3\mathbf{v}_2 \left|\mathbf{v}_1 - \mathbf{v}_2\right| f(\mathbf{v}_1) f(\mathbf{v}_2) \\[2pt]
&= \frac{4}{\sqrt{\pi}}\sqrt{\frac{k_\text{B}T}{m}}
= \sqrt{2}\langle v \rangle 
\end{align}</math>
where the three-dimensional velocity distribution is
<math display="block"> f(\mathbf{v}) \equiv \left[\frac{2\pi k_\text{B}T}{m}\right]^{-3/2} \exp\left(-\frac{1}{2}\frac{m\mathbf{v}^2}{k_\text{B}T} \right). </math>

The integral can easily be done by changing to coordinates <math> \mathbf{u} = \mathbf{v}_1-\mathbf{v}_2 </math> and <math display="inline"> \mathbf{U} = \tfrac{1}{2}(\mathbf{v}_1 + \mathbf{v}_2).</math>