Is it right to consider an average force over time rather than an impulse? In a proof of the pressure exerted on a surface by a gas, the average force exerted by a particle on the surface $$ \vec{F} = \frac{\Delta \vec{p}}{\Delta t}$$
is used, where $\Delta t$ is the time needed for the particle to come back after every elastic collision, and $\Delta \vec p$ the variation of momentum of the surface due to one collision.
Is it right to do so? If so, why?
From what I understand, Newton's principle is not applicable in this case and I don't see why it is allowed to average such "instantaneous" force.
The proof where the fact is used can be found in a Wikipedia article on kinetic theory of gases.
 A: By definition, the time-average of any time-varying quantity $f(t)$  over a time interval $\Delta t$ is given by
$$
\langle f \rangle = \frac{1}{\Delta t} \int f(t) dt.
$$
In particular, the time-average of a time-varying force $\vec{F}(t)$ is 
$$
\langle \vec{F} \rangle = \frac{1}{\Delta t} \int \vec{F}(t) \, dt.
$$
But according to Newton's Second Law, $\vec{F} = d \vec{p} / dt$, and so we have
$$
 \int \vec{F}(t) \, dt = \int \frac{d\vec{p}}{dt} \, dt = \Delta \vec{p}.
$$
Thus, the statement
$$
\langle\vec{F}\rangle = \frac{\Delta \vec{p}}{\Delta t}
$$
is true so long as we only care about the time-averaged force on the molecule (or equivalently, by Newton's Third Law, on the wall of the container.) 
The reason we only care about the time-averaged force for this derivation is simply because we are looking at the combined force from many skillions of molecules, each hitting the walls at slightly different times.  The combined effect of all of these molecules' impacts gets "smeared out" over time, results in an approximately constant force on the wall.  If we could measure the force on the wall with enough time resolution, we might be able to measure the small fluctuations due to the impacts of individual gas molecules;  but usually we only care about the average force over times that are a good deal longer than the average $\Delta t$, and so it's fine to just talk about the time-averaged force.
A: Consider a particle bouncing back and forth between the two walls, assuming elastic collision with a constant speed equal to v. The velocity diagram can be drawn as such, check the attached image.  where $T$ is the time between collisions, while $t_c$ is the collision time or the time taken to revert the direction of the velocity. Thus from $[0,T]$, we have 0 force, the average force from $[T,T+t_c]$ is roughly $\dfrac{2mv}{t_c}$. Thus technically we need to be provided with $t_c$ if we want to approximate the force.
This is not the end of the story, now consider a uniform distribution of particles bouncing across the walls of our 1D box, all moving in the same direction. Let there be N such particles indexed as such. Based on our previous analysis, we have the force on each wall, due to particle $\#i$ is given by
$$F_i=\dfrac{2mv_i}{t_c},$$
where all the masses are the same, and the index in the velocity is just used to distinguish the particles, the magnitude is the same throughout. Since the problem has a natural repeating unit of $2T+2t_c\approx 2T$, we want to use this interval to define the average. Since all the particles are uniformly distributed and moving along the same direction, by the time a particle comes back to make a second collision, all other particles would have collided with the wall exactly one, thereby imparting some force. Now if $N$ is large enough, and the particles are uniformly distributed, then $Nt_c\approx 2T$, when one particle is bouncing off the wall, no other particle is in contact with the wall (provided the gases are dilute enough). Thus now the average force is
$$<F>=\dfrac{1}{N}\sum_1^NF_i=\dfrac{2m}{t_c}\dfrac{1}{N}\sum_1^Nv_i=\dfrac{2Nmv}{2T}$$
where $T=\dfrac{L}{v}$ where $L$ is the length of the box. Substituting in we get
$$<F>=\dfrac{Nmv^2}{L}.$$
Note that even larger $N$, the approximation $Nt_c$ doesn't make sense, but that's okay, cause that means the density is large and ideal gas conditions are no longer valid.
PS- Even though this was a really simplified case, I don't think there is any way around the $Nt_c\approx 2T$ approximation.
