Editing Kinetic theory of gases (section)

== Equilibrium properties ==
=== Pressure and kinetic energy ===
<!-- This section is linked from [[Pressure]] -->
In the kinetic theory of gases, the [[pressure]] is assumed to be equal to the force (per unit area) exerted by the individual gas atoms or molecules hitting and rebounding from the gas container's surface.

Consider a gas particle traveling at velocity, <math display="inline">v_i</math>, along the <math>\hat{i}</math>-direction in an enclosed volume with [[characteristic length]], <math>L_i</math>, cross-sectional area, <math>A_i</math>, and volume, <math>V = A_i L_i</math>. The gas particle encounters a boundary after characteristic time
<math display="block"> t = L_i / v_i.</math>

The [[momentum]] of the gas particle can then be described as
<math display="block"> p_i = m v_i = m L_i / t .</math>

We combine the above with [[Newton's second law]], which states that the force experienced by a particle is related to the time rate of change of its momentum, such that
<math display="block">F_i = \frac{\mathrm{d}p_i}{\mathrm{d}t} = \frac{m L_i}{t^2}=\frac{m v_i^2}{L_i}.</math>

Now consider a large number, <math>N</math>, of gas particles with random orientation in a three-dimensional volume. Because the orientation is random, the average particle speed, <math display='inline'> v </math>, in every direction is identical
<math display="block">v_x^2 = v_y^2 = v_z^2.</math>

Further, assume that the volume is symmetrical about its three dimensions, <math>\hat{i}, \hat{j}, \hat{k}</math>, such that
<math display="block">\begin{align}
V ={}& V_i = V_j = V_k, \\
F ={}& F_i = F_j = F_k, \\
     & A_i=A_j=A_k.
\end{align}</math>
The total surface area on which the gas particles act is therefore
<math display="block">A = 3 A_i.</math>

The pressure exerted by the collisions of the <math>N</math> gas particles with the surface can then be found by adding the force contribution of every particle and dividing by the interior surface area of the volume,
<math display="block">P = \frac{N \overline{F}}{A}=\frac{NLF}{V} </math>
<math display="block"> \Rightarrow PV = NLF = \frac{N}{3} m v^2.</math>

The total translational [[kinetic energy]] <math>K_\text{t} </math> of the gas is defined as
<math display="block">K_\text{t} = \frac{N}{2} m v^2 ,</math>
providing the result
<math display="block">PV = \frac{2}{3} K_\text{t} .</math>

This is an important, non-trivial result of the kinetic theory because it relates pressure, a [[macroscopic]] property, to the translational kinetic energy of the molecules, which is a [[microscopic]] property.

The mass density of a gas <math>\rho </math> is expressed through the total mass of gas particles and through volume of this gas: <math> \rho = \frac {N m}{V}</math>. Taking this into account, the pressure is equal to
<math display="block">P = \frac{\rho v^2}{3} .</math>

Relativistic expression for this formula is <ref>{{Cite journal |last=Fedosin |first=Sergey G. | date=2021 |title= The potentials of the acceleration field and pressure field in rotating relativistic uniform system |journal= Continuum Mechanics and Thermodynamics |volume= 33|issue= 3|pages= 817–834|language=en |doi= 10.1007/s00161-020-00960-7|s2cid= 230076346 |arxiv=2410.17289 |bibcode= 2021CMT....33..817F}}</ref>

<math display="block"> P = \frac {2 \rho c^2 }{3} \left({\left(1 - \overline{v^2} / c^2\right)}^{-1/2} - 1 \right) , </math>
where <math> c </math> is [[speed of light]]. In the limit of small speeds, the expression becomes <math>P \approx \rho \overline{v^2}/3</math>.

=== Temperature and kinetic energy ===
Rewriting the above result for the pressure as <math display="inline">PV = \frac{1}{3}Nmv^2 </math>, we may combine it with the [[ideal gas law]]
{{NumBlk||<math display="block"> PV = N k_\mathrm{B} T ,</math>|{{EquationRef|1}}}}
where <math> k_\mathrm{B}</math> is the [[Boltzmann constant]] and <math> T</math> is the [[Thermodynamic temperature|absolute]] [[temperature]] defined by the ideal gas law, to obtain
<math display="block">k_\mathrm{B} T = \frac{1}{3} m v^2, </math>
which leads to a simplified expression of the average translational kinetic energy per molecule,<ref>The average kinetic energy of a fluid is proportional to the [[Root-mean-square speed|root mean-square velocity]], which always exceeds the mean velocity - {{usurped|1=[https://web.archive.org/web/20071011213546/http://www.mikeblaber.org/oldwine/chm1045/notes/Gases/Kinetic/Gases08.htm Kinetic Molecular Theory]}}</ref>
<math display="block"> \frac{1}{2} m v^2 = \frac{3}{2} k_\mathrm{B} T.</math>
The translational kinetic energy of the system is <math>N</math> times that of a molecule, namely <math display="inline"> K_\text{t} = \frac{1}{2} N m v^2 </math>. The temperature, <math> T</math> is related to the translational kinetic energy by the description above, resulting in
{{NumBlk||<math display="block"> T = \frac{1}{3} \frac{m v^2}{k_\mathrm{B} } </math>|{{EquationRef|2}}}}
which becomes
{{NumBlk||<math display="block"> T = \frac{2}{3} \frac{K_\text{t}}{N k_\mathrm{B} }. </math>|{{EquationRef|3}}}}
Equation ({{EquationNote|3}}) is one important result of the kinetic theory:
''The average molecular kinetic energy is proportional to the ideal gas law's absolute temperature''.
From equations ({{EquationNote|1}}) and ({{EquationNote|3}}), we have
{{NumBlk||<math display="block"> PV = \frac{2}{3} K_\text{t}.</math>|{{EquationRef|4}}}}
Thus, the product of pressure and volume per [[Mole (unit)|mole]] is proportional to the average
translational molecular kinetic energy.

Equations ({{EquationNote|1}}) and ({{EquationNote|4}}) are called the "classical results", which could also be derived from [[statistical mechanics]];
for more details, see:<ref>
[http://clesm.mae.ufl.edu/wiki.pub/index.php/Configuration_integral_%28statistical_mechanics%29 Configuration integral (statistical mechanics)] {{webarchive|url=https://web.archive.org/web/20120428193950/http://clesm.mae.ufl.edu/wiki.pub/index.php/Configuration_integral_%28statistical_mechanics%29 |date=2012-04-28 }}
</ref>

The [[equipartition theorem]] requires that kinetic energy is partitioned equally between all kinetic [[degrees of freedom]], ''D''. A monatomic gas is axially symmetric about each spatial axis, so that ''D'' = 3 comprising translational motion along each axis. A diatomic gas is axially symmetric about only one axis, so that ''D'' = 5, comprising translational motion along three axes and rotational motion along two axes. A polyatomic gas, like [[water]], is not radially symmetric about any axis, resulting in ''D'' = 6, comprising 3 translational and 3 rotational degrees of freedom.

Because the [[equipartition theorem]] requires that kinetic energy is partitioned equally, the total kinetic energy is
<math display="block"> K =D K_\text{t} = \frac{D}{2} N m v^2. </math>

Thus, the energy added to the system per gas particle kinetic degree of freedom is
<math display="block"> \frac{K}{ND} = \frac{1}{2} k_\text{B} T . </math>

Therefore, the kinetic energy per kelvin of one mole of monatomic [[ideal gas]] (''D'' = 3) is
<math display="block"> K = \frac{D}{2} k_\text{B} N_\text{A} = \frac{3}{2} R, </math>
where <math>N_\text{A}</math> is the [[Avogadro constant]], and ''R'' is the [[ideal gas constant]].

Thus, the ratio of the kinetic energy to the absolute temperature of an ideal monatomic gas can be calculated easily:
* per mole: 12.47&nbsp;J/K
* per molecule: 20.7&nbsp;[[yoctojoule|yJ]]/K = 129&nbsp;μeV/K

At [[Standard temperature and pressure|standard temperature]] (273.15&nbsp;K), the kinetic energy can also be obtained:
* per mole: 3406&nbsp;J
* per molecule: 5.65&nbsp;[[zeptojoule|zJ]] = 35.2&nbsp;meV.

At higher temperatures (typically thousands of kelvins), vibrational modes become active to provide additional degrees of freedom, creating a temperature-dependence on ''D'' and the total molecular energy. Quantum [[statistical mechanics]] is needed to accurately compute these contributions.<ref>{{cite book |last1=Chang |first1=Raymond | last2=Thoman | first2=John W. Jr. |title=Physical Chemistry for the Chemical Sciences |date=2014 |publisher=University Science Books |location=New York |pages=56–61}}</ref>

=== Collisions with container wall ===
For an ideal gas in equilibrium, the rate of collisions with the container wall and velocity distribution of particles hitting the container wall can be calculated<ref name="OCW">{{cite web |title=5.62 Physical Chemistry II |url=https://ocw.mit.edu/courses/chemistry/5-62-physical-chemistry-ii-spring-2008/lecture-notes/29_562ln08.pdf |website=MIT OpenCourseWare}}</ref> based on naive kinetic theory, and the results can be used for analyzing [[Effusion#Physics in Effusion|effusive flow rate]]s, which is useful in applications such as the [[Gaseous diffusion#Technology|gaseous diffusion]] method for [[Isotope separation#Diffusion|isotope separation]].

Assume that in the container, the number density (number per unit volume) is <math>n = N/V</math> and that the particles obey [[Maxwell-Boltzmann distribution|Maxwell's velocity distribution]]:
<math display="block">f_\text{Maxwell}(v_x,v_y,v_z) \, dv_x \, dv_y \, dv_z = \left(\frac{m}{2 \pi k_\text{B} T}\right)^{3/2} e^{- \frac{mv^2}{2k_\text{B}T}} \, dv_x \, dv_y \, dv_z</math>

Then for a small area <math>dA</math> on the container wall, a particle with speed <math>v</math> at angle <math>\theta</math> from the normal of the area <math>dA</math>, will collide with the area within time interval <math>dt</math>, if it is within the distance <math>v\,dt</math> from the area <math>dA</math>. Therefore, all the particles with speed <math>v</math> at angle <math>\theta</math> from the normal that can reach area <math>dA</math> within time interval <math>dt</math> are contained in the tilted pipe with a height of <math>v\cos (\theta) dt</math> and a volume of <math>v\cos (\theta) \,dA\,dt</math>.

The total number of particles that reach area <math>dA</math> within time interval <math>dt</math> also depends on the velocity distribution; All in all, it calculates to be:<math display="block">n v \cos(\theta) \, dA\, dt \times\left(\frac{m}{2 \pi k_\text{B}T}\right)^{3/2} e^{- \frac{mv^2}{2k_\text{B}T}} \left( v^2 \sin(\theta) \, dv \, d\theta \, d\phi \right).</math>

Integrating this over all appropriate velocities within the constraint <math>v > 0</math>, <math display="inline">0 < \theta < \frac{\pi}{2}</math>, <math>0 < \phi < 2\pi</math> yields the number of atomic or molecular collisions with a wall of a container per unit area per unit time:
<math display="block">J_\text{collision} = \frac{\displaystyle\int_0^{\pi/2} \cos(\theta) \sin(\theta) \, d\theta}{\displaystyle\int_0^\pi \sin(\theta) \, d\theta}\times n \bar v
= \frac{1}{4} n \bar v
= \frac{n}{4} \sqrt{\frac{8 k_\mathrm{B} T}{\pi m}}. </math>

This quantity is also known as the "impingement rate" in vacuum physics. Note that to calculate the average speed <math>\bar{v}</math> of the Maxwell's velocity distribution, one has to integrate over <math>v > 0 </math>, <math>0 < \theta < \pi </math>, <math>0 < \phi < 2\pi</math>.

The momentum transfer to the container wall from particles hitting the area <math>dA</math> with speed <math>v</math> at angle <math>\theta</math> from the normal, in time interval <math>dt</math> is:
<math display="block">[2mv \cos(\theta)]\times n v \cos(\theta) \, dA\, dt \times\left(\frac{m}{2 \pi k_\text{B}T}\right)^{3/2} e^{- \frac{mv^2}{2k_\text{B}T}} \left( v^2 \sin(\theta) \, dv \, d\theta \, d\phi \right).</math>
Integrating this over all appropriate velocities within the constraint <math>v > 0</math>, <math display="inline">0 < \theta < \frac{\pi}{2}</math>, <math>0 < \phi < 2\pi</math> yields the [[pressure]] (consistent with [[Ideal gas law]]):
<math display="block">P = \frac{\displaystyle 2\int_0^{\pi/2} \cos^2(\theta) \sin(\theta) \, d\theta}{\displaystyle \int_0^\pi \sin(\theta) \, d\theta}\times n mv_\text{rms}^2
= \frac{1}{3} n mv_\text{rms}^2
= \frac{2}{3} n\langle E_\text{kin}\rangle
= n k_\mathrm{B} T </math>
If this small area <math>A</math> is punched to become a small hole, the [[Effusion#Physics in Effusion|effusive flow rate]] will be:
<math display="block">\Phi_\text{effusion} = J_\text{collision} A= n A \sqrt{\frac{k_\mathrm{B} T}{2 \pi m}}. </math>

Combined with the [[ideal gas law]], this yields
<math display="block">\Phi_\text{effusion} = \frac{P A}{\sqrt{2 \pi m k_\mathrm{B} T}}. </math>

The above expression is consistent with [[Graham's law]].

To calculate the velocity distribution of particles hitting this small area, we must take into account that all the particles with <math>(v,\theta,\phi)</math> that hit the area <math>dA</math> within the time interval <math>dt</math> are contained in the tilted pipe with a height of <math>v\cos (\theta) \, dt</math> and a volume of <math>v\cos (\theta) \, dA \, dt</math>; Therefore, compared to the Maxwell distribution, the velocity distribution will have an extra factor of <math>v\cos \theta</math>:
<math display="block">\begin{align}
f(v,\theta,\phi) \, dv \, d\theta \, d\phi
&= \lambda v\cos{\theta} \left(\frac{m}{2 \pi k T}\right)^{3/2} e^{- \frac{mv^2}{2k_\mathrm{B} T}}(v^2\sin{\theta} \, dv \, d\theta \, d\phi)
\end{align}</math>
with the constraint <math display="inline">v > 0</math>, <math display="inline">0 < \theta < \frac{\pi}{2}</math>, <math>0 < \phi < 2\pi</math>. The constant <math>\lambda</math> can be determined by the normalization condition <math display="inline">\int f(v,\theta,\phi) \, dv \, d\theta \, d\phi=1</math> to be <math display="inline">4/\bar{v} </math>, and overall:
<math display="block">\begin{align}
f(v,\theta,\phi) \, dv \, d\theta \, d\phi
&= \frac{1}{2\pi} \left(\frac{m}{k_\mathrm{B} T}\right)^2e^{- \frac{mv^2}{2k_\mathrm{B} T}}
(v^3\sin{\theta}\cos{\theta} \, dv \, d\theta \, d\phi) \\
\end{align};\quad v>0,\, 0<\theta<\frac \pi 2,\, 0<\phi<2\pi</math>

=== Speed of molecules ===
From the kinetic energy formula it can be shown that
<math display="block">v_\text{p} = \sqrt{2 \cdot \frac{k_\mathrm{B} T}{m}},</math>
<math display="block">\bar{v} = \frac {2}{\sqrt{\pi}} v_p = \sqrt{\frac {8}{\pi} \cdot \frac{k_\mathrm{B} T}{m}},</math>
<math display="block">v_\text{rms} = \sqrt{\frac{3}{2}} v_p = \sqrt{{3} \cdot \frac {k_\mathrm{B} T}{m}},</math>
where ''v'' is in m/s, ''T'' is in kelvin, and ''m'' is the mass of one molecule of gas in kg. The most probable (or mode) speed <math>v_\text{p}</math> is 81.6% of the root-mean-square speed <math>v_\text{rms}</math>, and the mean (arithmetic mean, or average) speed <math>\bar{v}</math> is 92.1% of the rms speed ([[isotropy|isotropic]] [[Maxwell–Boltzmann distribution#Distribution for the speed|distribution of speeds]]).

See:
* [[Average]],
* [[Root-mean-square speed]]
* [[Arithmetic mean]]
* [[Mean]]
* [[Mode (statistics)]]

=== Mean free path ===
{{Main|Mean free path}}
In kinetic theory of gases, the [[Mean free path#Kinetic theory|mean free path]] is the average distance traveled by a molecule, or a number of molecules per volume, before they make their first collision. Let <math> \sigma </math> be the collision [[Cross section (physics)#Collision among gas particles|cross section]] of one molecule colliding with another. As in the previous section, the number density <math> n </math> is defined as the number of molecules per (extensive) volume, or <math> n = N/V </math>. The collision cross section per volume or collision cross section density is <math> n \sigma </math>, and it is related to the mean free path <math>\ell</math> by<math display="block">\ell = \frac {1} {n \sigma \sqrt{2}} </math>

Notice that the unit of the collision cross section per volume <math> n \sigma </math> is reciprocal of length.