Kinetic Theory

A cube with the volume $V$ is filled with a gas containing $N$ particles. We define then the particle density as $$n = \frac{N}{V}$$ The number of particles colliding per time with the walls of the cube is then given as $$N' = n_x v_x A\Delta t$$ Currently, we concentrate only on particles traveling in $x$-direction. During every collision, the momentum $$\Delta |\vec{p}| = 2m\Delta v$$ is transferred to the wall. From that, the force can be calculated according to: $$F = N'\frac{\Delta |\vec{p}|}{\Delta t} = 2mn_xAv_x^2$$ Dividing this equation by the area $A$ leads to the pressure $$p = \frac{F}{A} = mn_xv_x^2$$ If all directions are taken into account, the average of the squares is given as $$\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} = \frac{1}{3}\overline{v^2}$$ This gives the total pressure: $$p = \frac{1}{3}mn\overline{v^2}$$ Multiplying both sides with the volume $V$ and expanding it with the factor $1/2$ results in $$pV = \frac{2}{3}N\frac{1}{2}m\overline{v^2}$$ Comparing this with the formula for the kinetic energy, the following equation gives a correlation between the product $pV$ and the kinetic energy of the gas particles: $$\boxed{pV = \frac{2}{3}N\overline{E_\mathrm{kin}}}$$ By inserting the ideal gas equation, this equation can be rewritten as $$\boxed{\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_BT}$$ This form shows that the kinetic energy of the gas particles is directly connected to the temperature of the gas.