# When deriving pressure, why do we take rebound time to be travelling time?

The derivation of the ideal gas law runs along the following lines. Imagining a container filled with some particles, by definition:

$P = F/A$

$F = dp/dt$

$F_{avg} = \Delta P/ \Delta t$

If we assume perfectly elastic collissions for particles rebounding off some container, then $| \Delta P| = 2mv$.

Now here's the sketchy part:

In the formula for $F_{avg}$, $\Delta t$ is supposed to be the time over which the rebound occurs. The derivation in my course uses $\Delta t$ = $2L/v$, where L is the distance from one wall to another. Why do we take time to be that required to traverse the distance from one wall to another and back and not, in some way, give a value for the time of collission.

• I didn't include the full derivation, because I didn't feel that understanding it is essential to how we frame time. – Muno Nov 9 '15 at 21:14

There is not just one particle in the the box or container. There will be many particles rebounding every time on the wall under consideration. That's why we used $F_{avg}$ and not just $F$. It's because $F$ means the force applied by just one single particle and $F_{avg}$ means the total force applied by all the particles over that period of time $\Delta t$. And the time taken by the particle to return again to the wall is given by $\Delta t$. So we don't need to find every particular force. And that's why we don't need rebound time.

• Ah, so our choice of $\Delta t$ is dependent on what we take $F_{avg}$ to be. What I now understand is that, in this derivation, we assume that by the time one particle returns for another collission (i.e. $\Delta t$ has elapsed), $N$ particles have made similar collissions. Hence, a change in momentum has taken place $N$ times. This assumption sets up, in the derivation, the grounds for eventually introducing $N$ into the equation for $F_{avg}$. Thanks. I hope I understood this. – Muno Nov 9 '15 at 22:34
• And thanks for making me turn some pages of my book and that's because of you i have revised that concept again and with better understanding now. :) – manshu Nov 10 '15 at 7:48

When deriving pressure, why do we take rebound time to be travelling time ... $\Delta t$ is supposed to be the time over which the rebound occurs.?

That $\Delta t$ is not the time over which the rebound occurs. $1/\Delta t$ is simply the number of collisions per second.

At each collision the change in momentum is $2mv$. As the frequency of collisions is $v/2L$ collisions/sec it means that the average change in momentum per second is the change in momentum for one collision multiplied with the number of collisions per second which is $(2mv)(v/2L)=mv^2/L$ (see Pressure and KMT.

The ideal gas law is the time averaged steady state of this system. Consider it on a very long timescale: if you wait several round trips, what is the average impulse imparted in $\Delta t$? This of course includes the interior transit time.

Another way of seeing this is that at any instant, most of the gas particles are not pushing on the balloon surface - they're stuck in the interior - and its only the ones near the surface that contribute to the pressure at any moment.