Published by Patrick Mutisya · 8 days ago
The kinetic theory treats a gas as a large number of tiny particles (molecules or atoms) moving randomly and colliding elastically with the walls of their container. From this microscopic picture we can derive the macroscopic quantity pressure and obtain the useful relation
\$pV = \frac{1}{3} N m \langle c^{2} \rangle\$
where:
| Symbol | Meaning |
|---|---|
| \$p\$ | Pressure of the gas |
| \$V\$ | Volume occupied by the gas |
| \$N\$ | Number of molecules in the volume \$V\$ |
| \$m\$ | Mass of a single molecule |
| \$\langle c^{2} \rangle\$ | Mean‑square speed of the molecules |
| \$\langle c_{x}^{2} \rangle\$ | Mean square of the \$x\$‑component of velocity |
Consider a cubic container of side \$L\$ (so \$V=L^{3}\$) with a single molecule of mass \$m\$ moving only in the \$x\$‑direction with speed \$c_{x}\$. The molecule collides elastically with the two opposite walls perpendicular to the \$x\$‑axis.
\$F = \frac{\Delta p}{\Delta t}= \frac{2 m c{x}}{2L/c{x}} = \frac{m c_{x}^{2}}{L}.\$
\$p = \frac{F}{A}= \frac{m c{x}^{2}}{L^{3}} = \frac{m c{x}^{2}}{V}.\$
For \$N\$ identical molecules moving only in \$x\$, we replace \$c{x}^{2}\$ by the average \$\langle c{x}^{2} \rangle\$ and obtain
\$pV = N m \langle c_{x}^{2} \rangle \qquad (1)\$
In a real gas molecules move in all three orthogonal directions. Because the gas is isotropic (no preferred direction), the mean square speeds in each direction are equal:
\$\langle c{x}^{2} \rangle = \langle c{y}^{2} \rangle = \langle c_{z}^{2} \rangle.\$
The total mean‑square speed is the sum of the three components:
\$\langle c^{2} \rangle = \langle c{x}^{2} \rangle + \langle c{y}^{2} \rangle + \langle c{z}^{2} \rangle = 3\langle c{x}^{2} \rangle.\$
Substituting \$\langle c_{x}^{2} \rangle = \langle c^{2} \rangle/3\$ into equation (1) gives the familiar kinetic‑theory result:
\$pV = \frac{1}{3} N m \langle c^{2} \rangle.\$
The ideal‑gas equation is \$pV = N k{\mathrm{B}} T\$, where \$k{\mathrm{B}}\$ is Boltzmann’s constant and \$T\$ is absolute temperature. Equating the two expressions for \$pV\$ yields
\$\$\frac{1}{3} N m \langle c^{2} \rangle = N k_{\mathrm{B}} T \;\;\Longrightarrow\;\;
\langle c^{2} \rangle = \frac{3 k_{\mathrm{B}} T}{m}.\$\$
This shows that the average kinetic energy per molecule, \$\tfrac{1}{2} m \langle c^{2} \rangle\$, is directly proportional to temperature:
\$\frac{1}{2} m \langle c^{2} \rangle = \frac{3}{2} k_{\mathrm{B}} T.\$
Find the pressure exerted by \$N = 2.5\times10^{25}\$ molecules of nitrogen (\$\mathrm{N_2}\$, molar mass \$28\;\text{g mol}^{-1}\$) in a \$0.010\;\text{m}^{3}\$ container if the mean‑square speed is \$\langle c^{2} \rangle = (500\;\text{m s}^{-1})^{2}\$.
\$m = \frac{28\times10^{-3}\;\text{kg mol}^{-1}}{N_{\!A}} = \frac{28\times10^{-3}}{6.022\times10^{23}} \approx 4.65\times10^{-26}\;\text{kg}.\$
\$\$p = \frac{1}{3}\frac{N m \langle c^{2} \rangle}{V}
= \frac{1}{3}\frac{(2.5\times10^{25})(4.65\times10^{-26})(500^{2})}{0.010}
\approx 9.2\times10^{4}\;\text{Pa}.\$\$
The result is close to atmospheric pressure (≈ \$1.0\times10^{5}\;\text{Pa}\$), illustrating how molecular motion underlies everyday gas behaviour.