Molecular kinetic interpretation of pressure
The derivation can be obtained without having to explicitly consider the time of impact. Suppose the molecules have kinetic energy $E=mc^2/2$ where velocity is $c$ and mass $m$. The molecules are in a fixed box of volume $v=xyz$. The molecule can move in any direction and so has components $u, \;v,\; w$ along each axis meaning that $c^2=u^2+v^2+w^2$ but as no direction is special (the gas fills the volume uniformly) then we can let $c^2=3u^2$. 

The pressure on the vessel's walls is due to the change in momentum and for a wall collision after moving along the $\pm x$ direction is $2mu$. A molecule will on average hit opposite walls $u/2x$ times per second when moving in the $\pm x$ direction. By Newtons second law the rate of change in momentum equals force, thus a molecule exerts a force $2mu \cdot u/2x=mu^2/x$ on the face of the container perpendicular to $x$ which has area $yz$. As pressure is force /area and as the collision area is $yz$ then $p=mu^2/(xyz)\equiv mu^2/V$. Finally, substituting for $u$ and allowing for $N$ molecules the pressure is $p=Nmc^2/3V$ i.e pressure is twice the kinetic energy divided by three times the volume, $\displaystyle p=\frac{2E}{3V}$