= Ideal gas pressure from wall collisions
{title2=$P=Nm\langle v_x^2\rangle/V$}
Consider a particle of <mass> $m$ in a rigid rectangular box of length $L$ along $x$ and wall <surface area> $A$, so $V=AL$ is the <volume (thermodynamics)>. Choose $+x$ toward the right wall. For an incoming normal <velocity> $v_x=v>0$, an <elastic collision> reverses the normal <momentum>:
$$
p_{x,\rm in}=mv,\qquad p_{x,\rm out}=-mv,\qquad
\Delta p_{x,\rm particle}=-2mv,\qquad
J_{x,\rm wall}=+2mv.
$$
The <impulse> delivered to the wall points outward, toward $+x$, by <Newton's third law>. The particle's opposite <momentum> change is not the wall's <impulse>. A tangential <velocity> component is unchanged by the ideal smooth-wall <collision>.
\Image[/ideal-gas-wall-momentum.png]
{title=Schematic ideal gas model: an elastic bounce transfers outward momentum to a wall; this is not a real experimental setup}
Ignoring intervening particle-particle <collisions> in this counting model, the particle travels a normal distance $2L$ before hitting the same wall again. Its hit rate is $|v_x|/(2L)$; multiplying by the <impulse> per hit gives the time-averaged normal <force> and its contribution to <pressure>:
$$
\overline F_{\perp,1}=(2m|v_x|)\frac{|v_x|}{2L}
=\frac{mv_x^2}{L},\qquad
P_1=\frac{\overline F_{\perp,1}}{A}=\frac{mv_x^2}{V}.
$$
For example, stipulate $m=4\times10^{-26}\,\mathrm{kg}$, $|v_x|=500\,\mathrm{m\,s^{-1}}$, $L=0.10\,\mathrm m$, and $A=0.010\,\mathrm{m^2}$. Then $V=0.0010\,\mathrm{m^3}$, the <impulse> per hit is $4\times10^{-23}\,\mathrm{N\,s}$, the hit rate is $2500\,\mathrm{s^{-1}}$, and $P_1=10^{-17}\,\mathrm{Pa}$. This is one particle's tiny time-averaged contribution to one wall's <pressure>, not an <isotropic> equilibrium <gas> by itself.
For $N$ particles of equal <mass>, add their <force> contributions and define $\langle v_x^2\rangle=N^{-1}\sum_{i=1}^Nv_{x,i}^2$. In the <gas>'s rest frame, <isotropy> gives $\langle v_x^2\rangle=\langle v_y^2\rangle=\langle v_z^2\rangle=\langle v^2\rangle/3$, where $v^2=v_x^2+v_y^2+v_z^2$. Hence
$$
P=\frac{Nm\langle v_x^2\rangle}{V}
=\frac{Nm\langle v^2\rangle}{3V},\qquad
\frac12m\langle v^2\rangle=\frac32k_BT
\quad\Longrightarrow\quad PV=Nk_BT.
$$
The final equality uses the classical <equipartition theorem>. The <kinetic pressure of an isotropic gas> therefore recovers the <ideal gas law> from <momentum> transfers. Doubling the normal speed would double both the <impulse> per hit and the hit rate, giving four times the <pressure> contribution; <pressure> measures a mean squared speed. Interparticle <elastic collisions> redistribute <velocities> and help establish <isotropy>; the individual round-trip times need not remain $2L/|v_x|$, but the equilibrium <momentum flux> result survives. https://openstax.org/books/university-physics-volume-2/pages/2-2-pressure-temperature-and-rms-speed[OpenStax University Physics Volume 2, section 2.2: wall-collision pressure derivation]{ref}
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