Pressure is the number of collisions with the container per unit area. Imagine a particle in a box of length L$L$. Assuming for the moment that it only moves in the x$x$-direction, whenever it collides with a wall of the container the wall will gain momentum from the particle: $$Δp = p_{final, x} - p_{initial, x} = p_{final, x} - (- p_{final, x}) = 2p_{final, x} = 2mu_x$$ where \begin{align} \Delta p &= p_{\mathrm{final}, x} - p_{\mathrm{initial}, x} \\ &= p_{\mathrm{final}, x} - (- p_{\mathrm{final}, x})\\ &= 2p_{\mathrm{final}, x}\\ &= 2mu_x, \end{align} where $u_x$ is the x$x$-component of velocity.
The particle hits a specific wall every $ Δt = 2L/u_x$$\Delta t = \frac{2L}{u_x}$, since it travels the length of the box twice. Force can be defined as momentum per unit time, therefore $$ F = \frac{Δp}{Δt} = \frac{mu_x^2}{L}$$ For $$ F = \frac{\Delta p}{\Delta t} = \frac{mu_x^2}{L}$$ For N$N$ number of particles, the expression becomes $$ F = \frac{Nm\overline {u_x^2}}{L}$$ where $$ F = \frac{Nm\overline {u_x^2}}{L},$$ where $\overline{u_x^2}$ is the mean square velocity of the N$N$ particles.
Now, we can extend the situation so that our particle can move in three directions. Since the box is cubic, the x$x$-, y$y$-, and z$z$-components are equivalent: $$ \overline{u_x^2} = \overline{u^2} / 3$$ Therefore $$ \overline{u_x^2} = \frac{\overline{u^2} }{ 3}$$ Therefore, $$ F = \frac{Nm\overline {u^2}}{3L}$$ The $$ F = \frac{Nm\overline {u^2}}{3L}.$$ The area of any wall is L^2$L^2$, so $$ P = \frac{F}{L^2} = \frac{Nm\overline {u^2}}{3L^3} = \frac{Nm\overline {u^2}}{3V}$$ This \begin{align} p &= \frac{F}{L^2}\\ &= \frac{Nm\overline {u^2}}{3L^3}\\ &= \frac{Nm\overline {u^2}}{3V} \end{align} This leads to the expression $$ PV = \frac{1}{3}Nm\overline{u^2} $$ $$pV = \frac{1}{3}Nm\overline{u^2}$$
See also http://en.wikipedia.org/wiki/Kinetic_theoryWikipedia.
