Editing Ideal gas law (section)

=== Theoretical ===

==== Kinetic theory ====
{{Main|Kinetic theory of gases}}

The ideal gas law can also be derived from [[first principles]] using the [[kinetic theory of gases]], in which several simplifying assumptions are made, chief among which are that the molecules, or atoms, of the gas are point masses, possessing mass but no significant volume, and undergo only elastic collisions with each other and the sides of the container in which both linear momentum and kinetic energy are conserved.

First we show that the fundamental assumptions of the kinetic theory of gases imply that
: <math>P = \frac{1}{3}nmv_{\text{rms}}^2.</math>

Consider a container in the <math>xyz</math> Cartesian coordinate system. For simplicity, we assume that a third of the molecules moves parallel to the <math>x</math>-axis, a third moves parallel to the <math>y</math>-axis and a third moves parallel to the <math>z</math>-axis. If all molecules move with the same velocity <math>v</math>, denote the corresponding pressure by <math>P_0</math>. We choose an area <math>S</math> on a wall of the container, perpendicular to the <math>x</math>-axis. When time <math>t</math> elapses, all molecules in the volume <math>vtS</math> moving in the positive direction of the <math>x</math>-axis will hit the area. There are <math>NvtS</math> molecules in a part of volume <math>vtS</math> of the container, but only one sixth (i.e. a half of a third) of them moves in the positive direction of the <math>x</math>-axis. Therefore, the number of molecules <math>N'</math> that will hit the area <math>S</math> when the time <math>t</math> elapses is <math>NvtS/6</math>.

When a molecule bounces off the wall of the container, it changes its momentum <math>\mathbf{p}_1</math> to <math>\mathbf{p}_2=-\mathbf{p}_1</math>. Hence the magnitude of change of the momentum of one molecule is <math>|\mathbf{p}_2-\mathbf{p}_1|=2mv</math>. The magnitude of the change of momentum of all molecules that bounce off the area <math>S</math> when time <math>t</math> elapses is then <math>|\Delta \mathbf{p}|=2mvN'/V=NtSmv^2/(3V)=ntSmv^2/3</math>. From <math>F=|\Delta \mathbf{p}|/t</math> and <math>P_0=F/S</math> we get

: <math>P_0=\frac{1}{3}nm v^2.</math>

We considered a situation where all molecules move with the same velocity <math>v</math>. Now we consider a situation where they can move with different velocities, so we apply an "averaging transformation" to the above equation, effectively replacing <math>P_0</math> by a new pressure <math>P</math> and <math>v^2</math> by the arithmetic mean of all squares of all velocities of the molecules, i.e. by <math>v_{\text{rms}}^2.</math> Therefore
: <math>P=\frac{1}{3}nm v_{\text{rms}}^2</math>
which gives the desired formula.

Using the [[Maxwell–Boltzmann distribution]], the fraction of molecules that have a speed in the range <math>v</math> to <math>v + dv</math> is <math>f(v) \, dv</math>, where
: <math>f(v) = 4\pi \left(\frac{m}{2\pi k_{\rm B}T}\right)^{\!\frac{3}{2}}v^2 e^{-\frac{mv^2}{2k_{\rm B}T}}</math>
and <math>k</math> denotes the Boltzmann constant. The root-mean-square speed can be calculated by
: <math>v_{\text{rms}}^2 = \int_0^\infty v^2 f(v) \, dv = 4\pi \left(\frac{m}{2\pi k_{\rm B}T}\right)^{\frac{3}{2}}\int_0^\infty v^4 e^{-\frac{mv^2}{2k_{\rm B}T}} \, dv.</math>

Using the integration formula
: <math>\int_0^\infty x^{2n}e^{-\frac{x^2}{a^2}} \, dx = \sqrt{\pi} \, \frac{(2n)!}{n!}\left(\frac{a}{2}\right)^{2n+1},\quad n\in\mathbb{N},\,a\in\mathbb{R}^+,</math>
it follows that
: <math>v_{\text{rms}}^2 = 4\pi\left(\frac{m}{2\pi k_{\rm B}T}\right)^{\!\frac{3}{2}}\sqrt{\pi} \, \frac{4!}{2!}\left(\frac{\sqrt{\frac{2k_{\rm B}T}{m}}}{2}\right)^{\!5} = \frac{3k_{\rm B}T}{m},</math>
from which we get the ideal gas law:
: <math>P = \frac{1}{3} nm\left(\frac{3k_{\rm B}T}{m}\right) = nk_{\rm B}T.</math>

==== Statistical mechanics ====
{{Main|Statistical mechanics}}

Let '''q''' = (''q''<sub>x</sub>, ''q''<sub>y</sub>, ''q''<sub>z</sub>) and '''p''' = (''p''<sub>x</sub>, ''p''<sub>y</sub>, ''p''<sub>z</sub>) denote the position vector and momentum vector of a particle of an ideal gas, respectively. Let '''F''' denote the net force on that particle. Then (two times) the time-averaged kinetic energy of the particle is:
: <math>\begin{align}
\langle \mathbf{q} \cdot \mathbf{F} \rangle
&= \left\langle q_{x} \frac{dp_{x}}{dt} \right\rangle +
\left\langle q_{y} \frac{dp_{y}}{dt} \right\rangle +
\left\langle q_{z} \frac{dp_{z}}{dt} \right\rangle\\
&=-\left\langle q_{x} \frac{\partial H}{\partial q_x} \right\rangle -
\left\langle q_{y} \frac{\partial H}{\partial q_y} \right\rangle -
\left\langle q_{z} \frac{\partial H}{\partial q_z} \right\rangle = -3k_\text{B} T,
\end{align}</math>
where the first equality is [[Newton's second law]], and the second line uses [[Hamilton's equations]] and the [[equipartition theorem]]. Summing over a system of ''N'' particles yields
: <math>3Nk_{\rm B} T = - \left\langle \sum_{k=1}^{N} \mathbf{q}_{k} \cdot \mathbf{F}_{k} \right\rangle.</math>

By [[Newton's third law]] and the ideal gas assumption, the net force of the system is the force applied by the walls of the container, and this force is given by the pressure ''P'' of the gas. Hence
: <math>-\left\langle\sum_{k=1}^{N} \mathbf{q}_{k} \cdot \mathbf{F}_{k}\right\rangle = P \oint_{\text{surface}} \mathbf{q} \cdot d\mathbf{S},</math>
where d'''S''' is the infinitesimal area element along the walls of the container. Since the [[divergence]] of the position vector '''q''' is
: <math>
\nabla \cdot \mathbf{q} =
\frac{\partial q_{x}}{\partial q_{x}} +
\frac{\partial q_{y}}{\partial q_{y}} +
\frac{\partial q_{z}}{\partial q_{z}} = 3,
</math>
the [[divergence theorem]] implies that
: <math>P \oint_{\text{surface}} \mathbf{q} \cdot d\mathbf{S}
= P \int_{\text{volume}} \left( \nabla \cdot \mathbf{q} \right) dV
= 3PV,</math>
where ''dV'' is an infinitesimal volume within the container and ''V'' is the total volume of the container.

Putting these equalities together yields
: <math>3 N k_\text{B} T = -\left\langle \sum_{k=1}^{N} \mathbf{q}_{k} \cdot \mathbf{F}_{k} \right\rangle = 3PV,</math>
which immediately implies the ideal gas law for ''N'' particles:
: <math>PV = Nk_{\rm B} T = nRT,</math>
where ''n'' = ''N''/''N''<sub>A</sub> is the number of [[mole (unit)|moles]] of gas and ''R'' = ''N''<sub>A</sub>''k''<sub>B</sub> is the [[gas constant]].