Regarding the molecular kinetic interpretation of pressure, we arrive at a point in the $F=\frac{\Delta P}{\Delta t}=\frac{2mvx}{\Delta t}$, that is to say, the force exerted in the shock on the wall is calculated by dividing the change in the amount of movement of the wall by that time, which I understand should be the time of duration of the shock, but nevertheless the time used is the time between two shocks, the time in which the molecule travels the distance of the edge of the vessel for one side and returns...this does not contradict the definition of impulse that refers to the duration of the shock??
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$\begingroup$ Essentially, we are talking about the time-averaged force. $\endgroup$– Ivan NeretinMay 20, 2018 at 18:15
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$\begingroup$ The theory works as well for hard-sphere molecules, where the contact time is zero. Force is mass times acceleration, and the latter is ($2 v$ per (time interval that in average contains exactly one wall collision)) $\endgroup$– KarlMay 20, 2018 at 19:15
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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, $\displaystyle p=\frac{2E}{3V}$
