Consider a cubical box, L on each side. The box contains N molecules of ideal gas, each of mass m.

All collisions are elastic. The force exerted by one molecule when it collides with a wall of the box that is perpendicular to the x-axis is, from the impulse equation: $$F = \frac{2mv_x}{Δt} \tag{1}$$ Our single molecule collides with this wall once every: $$Δt = \frac{2L}{v_x} \tag{2}$$ This gives an average force of: $$F = \frac{mv_x^2}{L} \tag{3}$$ This is the force from a single molecule. The total force on the wall is the sum over all the molecules: $$\Sigma F = \frac{Σmv_x^2}{L} = \frac{mN}{L} \left[\frac{Σv_x^2}{N}\right ] \tag{4}$$

The term in square brackets represents the average value of $v_x^2$. The square root of this average is known as the root-mean-square (rms) average of $v_x$, so: $$\Sigma F = \frac{mN}{L} v_{x rms}^2 \tag{5}$$ By symmetry, all directions in the box are equivalent, so: $$ v^2 = v_x^2 + v_y^2 + v_z^2 = 3v_x^2 \tag{6}$$ This gives: $$\Sigma F = \frac{mN}{3L} v_{rms}^2 \tag{7}$$ Dividing by the area, $L^2$, of the wall gives the pressure: $$P = \frac{mN}{3L^3} v_{rms}^2 \tag{8}$$ which is: $$PV = \frac{N}{3} \left[mv_{rms}^2\right] = \frac{2N}{3} \left[½ mv_{rms}^2\right] $$ The term in square brackets is $K_av$, the average translational kinetic energy of the molecules in the box. Therefore $$PV = \frac{2N}{3} K_{av} $$

*Why is only $v_x^2$ being summed over in $(4)$ if it is the total force? Why not $$ \sum_{i} m_i (v_x^2)_i \tag?$$

  • 2
    $\begingroup$ The second quotes sentence says “of mass m” not $m_i$. They’re all the same. $\endgroup$ – Bob Jacobsen Jul 11 at 4:11

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