# Pressure exerted by ideal gas on walls of container

All derivations I've come across are more or less similar to this:

All of them take total momentum change as $2mv_{x}$ and divide it by total time to get the force exerted on the wall. That seems to make a number of assumptions, the most obvious one being that the particle will collide again in the time it took for it to travel between the two walls i.e., that the particle won't collide with other particles on its way out.

So I decided to do something like this but the final expression doesn't seem to match.

Lets consider the x component of the force exerted by the particles on a differential are on the wall:

in $\mathrm dt$ time, a bunch of particles with momentum $(\rho\,\mathrm dA \,v\,\mathrm dt)v$ change their momentum to $-(\rho\,\mathrm dA\, v\,\mathrm dt)v$. Therefore, the impulse on the wall during this time will be $$\mathrm dJ=\rho\, \mathrm dA\,v^2\,\mathrm dt-(-\rho \,\mathrm dA\, v^2\,\mathrm dt)=2\rho \,\mathrm dA\,v^2\,\mathrm dt$$

The force exerted on the wall will be equal to $\frac {\mathrm dJ}{\mathrm dT}$ $$F=2\rho v^2\,\mathrm dA$$

Pressure exerted equals force divided by area, so $$P=2\rho v^2$$ But that doesn't seem to match with the correct solution. What is the mistake I am making?

I thought it might be that the momentum is $\left(\rho \,\mathrm dA\, v \frac{\mathrm dt}{2}\right)v$ but that doesn't make sense.

• Do you need to find the pressure exerted on the wall or want to derive the ideal-gas equation? If so, then the title is different from what you are asking. – user36790 Apr 10 '16 at 5:29
• Well, this is a part of derivation of $PV=nRT$. – Skawang Apr 10 '16 at 11:50

The net momentum of the particles in your volume $dA\; vdt$ is zero so on average only half of those particles will hit your wall and rebound. That is were your missing $\frac 12$ comes from.