Pressure is the number of collisions with the container per unit area. Imagine a particle in a box of length $L$. Assuming for the moment that it only moves in the $x$-direction, whenever it collides with a wall of the container the wall will gain momentum from the particle: 
\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 $x$-component of velocity.

The particle hits a specific wall every $\Delta t = \frac{2L}{u_x}$, since it travels the length of the box twice. Force can be defined as momentum per unit time, therefore 
$$ F = \frac{\Delta p}{\Delta t} = \frac{mu_x^2}{L}$$ 
For $N$ number of particles, the expression becomes 
$$ F = \frac{Nm\overline {u_x^2}}{L},$$ 
where $\overline{u_x^2}$ is the mean square velocity of the $N$ particles.

Now, we can extend the situation so that our particle can move in three directions. Since the box is cubic, the $x$-, $y$-, and $z$-components are equivalent: 
$$ \overline{u_x^2} = \frac{\overline{u^2} }{ 3}$$ 
Therefore, 
$$ F = \frac{Nm\overline {u^2}}{3L}.$$ 
The area of any wall is $L^2$, so 
\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 
$$pV = \frac{1}{3}Nm\overline{u^2}$$ 

See also [Wikipedia](http://en.wikipedia.org/wiki/Kinetic_theory).