Kinetic theory

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

${\displaystyle PV={\frac {1}{3}}Nmv_{\text{rms}}^{2}.}$

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

When a molecule bounces off the wall of the container, it changes its momentum ${\displaystyle \mathbf {p} _{1}}$ to ${\displaystyle \mathbf {p} _{2}=-\mathbf {p} _{1}}$. Hence the magnitude of change of the momentum of one molecule is ${\displaystyle |\mathbf {p} _{2}-\mathbf {p} _{1}|=2mv}$. The magnitude of the change of momentum of all molecules that bounce off the area ${\displaystyle S}$ when time ${\displaystyle t}$ elapses is then ${\displaystyle |\Delta \mathbf {p} |=2mvN'=NtSmv^{2}/(3V)}$. From ${\displaystyle F=|\Delta \mathbf {p} |/t}$ and ${\displaystyle P=F/S}$ we get

${\displaystyle P={\frac {1}{3}}Nm{\frac {v^{2}}{V}}.}$

We supposed that all molecules move with the same velocity ${\displaystyle v}$, but in fact they move with different velocities, so we replace ${\displaystyle v^{2}}$ in the equation by the arithmetic mean of all squares of all velocities of the molecules, i.e. by ${\displaystyle v_{\text{rms}}^{2}.}$ Therefore

${\displaystyle P={\frac {1}{3}}Nm{\frac {v_{\text{rms}}^{2}}{V}}}$

which gives the desired formula.

Using the Maxwell–Boltzmann distribution, the fraction of molecules that have a speed in the range ${\displaystyle v}$ to ${\displaystyle v+dv}$ is ${\displaystyle f(v)\,dv}$, where

${\displaystyle f(v)=4\pi \left({\frac {m}{2\pi kT}}\right)^{\!{\frac {3}{2}}}v^{2}e^{-{\frac {mv^{2}}{2kT}}}}$

and ${\displaystyle k}$ denotes the Boltzmann constant. The root-mean-square speed can be calculated by

${\displaystyle v_{\text{rms}}^{2}=\int _{0}^{\infty }v^{2}f(v)\,dv=4\pi \left({\frac {m}{2\pi kT}}\right)^{\frac {3}{2}}\int _{0}^{\infty }v^{4}e^{-{\frac {mv^{2}}{2kT}}}\,dv.}$

Using the integration formula

${\displaystyle \int _{0}^{\infty }x^{2n}e^{-{\frac {x^{2}}{a^{2}}}}\,dx={\sqrt {\pi }}\,{\frac {(2n)!}{n!}}\left({\frac {a}{2}}\right)^{2n+1},}$

it follows that

${\displaystyle v_{\text{rms}}^{2}=4\pi \left({\frac {m}{2\pi kT}}\right)^{\!{\frac {3}{2}}}{\sqrt {\pi }}\,{\frac {4!}{2!}}\left({\frac {\sqrt {\frac {2kT}{m}}}{2}}\right)^{\!5}={\frac {3kT}{m}},}$

from which we get the ideal gas law:

${\displaystyle PV={\frac {1}{3}}Nm\left({\frac {3kT}{m}}\right)=NkT.}$

Statistical mechanics

Let q = (q_x, q_y, q_z) and p = (p_x, p_y, p_z) 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:

${\displaystyle {\begin{aligned}\langle \mathbf {q} \cdot \mathbf {F} \rangle &=\left\langle q_{x}{\frac {dp_{x}}{dt}}\right\rangle +\left\langle q_{y}{\frac {dp_{y}}{dt}}\right\rangle +\left\langle q_{z}{\frac {dp_{z}}{dt}}\right\rangle \\&=-\left\langle q_{x}{\frac {\partial H}{\partial q_{x}}}\right\rangle -\left\langle q_{y}{\frac {\partial H}{\partial q_{y}}}\right\rangle -\left\langle q_{z}{\frac {\partial H}{\partial q_{z}}}\right\rangle =-3k_{\text{B}}T,\end{aligned}}}$

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

${\displaystyle 3Nk_{B}T=-\left\langle \sum _{k=1}^{N}\mathbf {q} _{k}\cdot \mathbf {F} _{k}\right\rangle .}$

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

${\displaystyle -\left\langle \sum _{k=1}^{N}\mathbf {q} _{k}\cdot \mathbf {F} _{k}\right\rangle =P\oint _{\text{surface}}\mathbf {q} \cdot d\mathbf {S} ,}$

where dS is the infinitesimal area element along the walls of the container. Since the divergence of the position vector q is

${\displaystyle \nabla \cdot \mathbf {q} ={\frac {\partial q_{x}}{\partial q_{x}}}+{\frac {\partial q_{y}}{\partial q_{y}}}+{\frac {\partial q_{z}}{\partial q_{z}}}=3,}$

the divergence theorem implies that

${\displaystyle P\oint _{\text{surface}}\mathbf {q} \cdot d\mathbf {S} =P\int _{\text{volume}}\left(\nabla \cdot \mathbf {q} \right)dV=3PV,}$

where dV is an infinitesimal volume within the container and V is the total volume of the container.

Putting these equalities together yields

${\displaystyle 3Nk_{\text{B}}T=-\left\langle \sum _{k=1}^{N}\mathbf {q} _{k}\cdot \mathbf {F} _{k}\right\rangle =3PV,}$

which immediately implies the ideal gas law for N particles:

${\displaystyle PV=Nk_{B}T=nRT,}$

where n = N/N_A is the number of moles of gas and R = N_Ak_B is the gas constant.