Editing Kinetic theory of gases (section)

=== 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>.