Théorie cinétique des gaz/Annexe/Démonstration de l'équation d'état du gaz parfait

Homogénéité et isotropie de la distribution des vitesses : $p(v)=4\pi {\left(\frac{a}{\pi }\right)}^{3/2}{v}^2{e}^{-a{v}^2}$ avec $a=\frac{m}{2kT}$

${\mathrm{d}}^{3}n=d^{*}{\mathrm{d}}^3\tau =\frac{N}{V}{\mathrm{d}}^3\tau$

On commence le calcul pour les molécules de vitesse v à dv près ${\mathrm{d}}^{4}n=\frac{N}{V}{\mathrm{d}}^3\tau p(v)\mathrm{d}v$

Parmi ces molécules, seules celles qui sont dans le bon angle solide comptent ${\mathrm{d}}^{6}n=\frac{N}{V}{\mathrm{d}}^3\tau p(v)\mathrm{d}v\frac{{\mathrm{d}}^2\mathrm{\Omega }}{4\pi }$

En sphérique ${\mathrm{d}}^3\tau =r^2\mathrm{d}r\mathrm{d}\theta \sin (\theta )\mathrm{d}\phi$

${\mathrm{d}}^2\mathrm{\Omega }=\frac{\cos (\theta )}{r^2}{\mathrm{d}}^{2}S$

$\begin{array}{cc}{\mathrm{d}}^{4}n& =\underset{\phi =0}{\overset{\phi =2\pi }{\int }}\underset{\theta =0}{\overset{\theta =\frac{\pi }{2}}{\int }}\underset{r=0}{\overset{r=v\mathrm{d}t}{\int }}\frac{N}{V}{r}^{2}p(v)\mathrm{d}v\frac{\cos (\theta )}{4\pi }\frac{{\mathrm{d}}^{2}S}{r^2}\sin (\theta )\mathrm{d}r\mathrm{d}\theta \mathrm{d}\phi \\ & =\frac{N}{V}p(v)\mathrm{d}v\frac{{\mathrm{d}}^{2}S}{4\pi }\underset{\phi =0}{\overset{\phi =2\pi }{\int }}\underset{\theta =0}{\overset{\theta =\frac{\pi }{2}}{\int }}\underset{r=0}{\overset{r=v\mathrm{d}t}{\int }}\cos (\theta )\sin (\theta )\mathrm{d}r\mathrm{d}\theta \mathrm{d}\phi \\ & =\frac{N}{V}p(v)\mathrm{d}v\frac{{\mathrm{d}}^{2}S}{4\pi }\cdot v\mathrm{d}t\cdot 2\pi \cdot \frac{1}{2}\\ & =\frac{N}{V}p(v)\mathrm{d}v\frac{{\mathrm{d}}^{2}S}{4}v\mathrm{d}t\end{array}$

On intègre sur les vitesses

$\begin{array}{cc}{d}^{3}n& =\underset{v=0}{\overset{v=+\mathrm{\infty }}{\int }}\frac{N}{V}p(v)v\frac{{\mathrm{d}}^{2}S}{4}\mathrm{d}t\mathrm{d}v\\ & =\frac{N{\mathrm{d}}^{2}S}{4V}\mathrm{d}t\underset{v=0}{\overset{v=+\mathrm{\infty }}{\int }}v4\pi {\left(\frac{a}{\pi }\right)}^{3/2}{v}^2{e}^{-a{v}^2}\mathrm{d}v\\ & =\frac{N{\mathrm{d}}^{2}S}{V}\mathrm{d}t{\left(\frac{a}{\pi }\right)}^{3/2}\pi \underset{v=0}{\overset{v=+\mathrm{\infty }}{\int }}{v}^3{e}^{-a{v}^2}\mathrm{d}v\\ & =\frac{N{\mathrm{d}}^{2}S}{V}\mathrm{d}t{\left(\frac{a}{\pi }\right)}^{3/2}\frac{\pi }{a}\underset{v=0}{\overset{v=+\mathrm{\infty }}{\int }}v{e}^{-a{v}^2}\mathrm{d}v\\ & =\frac{N{\mathrm{d}}^{2}S}{V}\mathrm{d}t\frac{\pi }{{\pi }^{3/2}}\frac{a^{3/2}}{2a^2}\\ & =\frac{N}{2V}\frac{1}{\sqrt{\pi a}}{\mathrm{d}}^{2}S\mathrm{d}t\end{array}$

${\mathrm{d}}^6{\overrightarrow{p}}_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{o}\mathrm{i}}=\frac{N}{V}{\mathrm{d}}^3\tau p(v)\mathrm{d}v(-2mv\cos (\theta ){\overrightarrow{u}}_z)\frac{{\mathrm{d}}^2\mathrm{\Omega }}{4\pi }$

$\begin{array}{cc}{\mathrm{d}}^4{\overrightarrow{p}}_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{o}\mathrm{i}}& =\underset{\phi =0}{\overset{\phi =2\pi }{\int }}\underset{\theta =0}{\overset{\theta =\frac{\pi }{2}}{\int }}\underset{r=0}{\overset{r=v\mathrm{d}t}{\int }}\frac{N}{V}{\mathrm{d}}^3\tau p(v)\mathrm{d}v(-2mv\cos (\theta ))\frac{{\mathrm{d}}^2\mathrm{\Omega }}{4\pi }{\overrightarrow{u}}_z\\ & =\underset{\phi =0}{\overset{\phi =2\pi }{\int }}\underset{\theta =0}{\overset{\theta =\frac{\pi }{2}}{\int }}\underset{r=0}{\overset{r=v\mathrm{d}t}{\int }}-\frac{N}{V}p(v)\mathrm{d}v2mv\cos (\theta )\frac{1}{4\pi }{r}^2\sin (\theta )\frac{{\mathrm{d}}^{2}S\cos (\theta )}{r^2}\mathrm{d}r\mathrm{d}\theta \mathrm{d}\phi {\overrightarrow{u}}_z\\ & =-\frac{Nmvp(v)}{2\pi V}\mathrm{d}v\underset{\phi =0}{\overset{\phi =2\pi }{\int }}\underset{\theta =0}{\overset{\theta =\frac{\pi }{2}}{\int }}\underset{r=0}{\overset{r=v\mathrm{d}t}{\int }}\cos (\theta )\sin (\theta ){\mathrm{d}}^{2}S\mathrm{d}r\mathrm{d}\theta \mathrm{d}\phi {\overrightarrow{u}}_z\\ & =-\frac{Nmvp(v){\mathrm{d}}^{2}S\mathrm{d}v}{2\pi V}2\pi v\mathrm{d}t\frac{1}{3}{\overrightarrow{u}}_z\\ & =-\frac{Nm{\mathrm{d}}^{2}S\mathrm{d}t}{3V}{v}^{2}p(v)\mathrm{d}v{\overrightarrow{u}}_z\end{array}$

On intègre sur les vitesses :

$\begin{array}{cc}{\mathrm{d}}^3{\overrightarrow{p}}_{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{o}\mathrm{i}}& =\underset{v=0}{\overset{v=+\mathrm{\infty }}{\int }}-\frac{Nm{\mathrm{d}}^{2}S\mathrm{d}t}{3V}{v}^{2}p(v)\mathrm{d}v{\overrightarrow{u}}_z\\ & =-\frac{Nm{\mathrm{d}}^{2}S\mathrm{d}t}{3V}\underset{v=0}{\overset{v=+\mathrm{\infty }}{\int }}{v}^{4}4\pi {\left(\frac{a}{\pi }\right)}^{3/2}{e}^{-a{v}^2}\mathrm{d}v{\overrightarrow{u}}_z\\ & =-\frac{Nm{\mathrm{d}}^{2}S\mathrm{d}t}{3V}4\pi {\left(\frac{a}{\pi }\right)}^{3/2}\frac{3}{8a^2}\sqrt{\frac{\pi }{a}}{\overrightarrow{u}}_z\\ & =-\frac{Nm{\mathrm{d}}^{2}S\mathrm{d}t}{2V}\frac{1}{a}{\overrightarrow{u}}_z\\ & =-\frac{Nm{\mathrm{d}}^{2}S\mathrm{d}t}{2V}\frac{2kT}{m}{\overrightarrow{u}}_z\\ & =-\frac{NkT}{V}{\mathrm{d}}^{2}S\mathrm{d}t{\overrightarrow{u}}_z\\ \end{array}$

$\overrightarrow{{\mathrm{d}}^2{F}_{{\mathrm{d}}^{2}S}}=-\frac{NkT}{V}{\mathrm{d}}^{2}S=-p{\mathrm{d}}^{2}S{\overrightarrow{u}}_z$ avec $p=\frac{NkT}{V}$