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I was studying the derivation of the relation between pressure ($P$) and root mean square speed ($v_{\text {RMS}}$) of an ideal gas from Fundamentals of Physics by Halliday, Resnick, Walker. (The same derivation can be found in Wikipedia page on kinetic theory of gases under "Equilibrium properties" section.)

One of the steps in the derivation process bothers me, which is illustrated below:

When a gas molecule collides with the wall of the container perpendicular to the x axis and bounces off in the opposite direction with the same speed (an elastic collision), the change in momentum is given by: $${\Delta p=p_{i,x}-p_{f,x}=p_{i,x}-(-p_{i,x})=2p_{i,x}=2mv_{x}}$$

where p is the momentum, i and f indicate initial and final momentum (before and after collision), x indicates that only the x direction is being considered, and v is the speed of the particle (which is the same before and after the collision).

The particle impacts one specific side wall once every $${\displaystyle \Delta t={\frac {2L}{v_{x}}}}$$ where L is the distance between opposite walls.

So far no problem, but now:

The force due to this particle is $$F={\frac {\Delta p}{\Delta t}}={\frac {mv_{x}^{2}}{L}}$$

  • How/why did they substitute the time period between two successive collisions for a given molecule with time for which the force is applied?

I might be wrong, but aren't the two things (time period between two successive collision and of the collison) different and hence shouldn't be substituted for one another?

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    $\begingroup$ Why are you typing every equation twice? Could you fix that? $\endgroup$
    – Semoi
    Commented Jul 13, 2020 at 17:48
  • $\begingroup$ Short answer: it is an approximation, namely it is implicitly assumed that the force is equally distributed between collisions. It makes sense because one is analyzing a small enough patch of the wall where there won't be many more hits within $\Delta t$ $\endgroup$
    – ohneVal
    Commented Jul 13, 2020 at 18:08

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I believe you look at the equations too closely. By taking a step back and thinking about the force we arrive at the following conclusion: The mean force is given by the product of

  1. the average momentum transferred (to the wall) per collision, $\bar p = 2 m \bar v$, and
  2. the average number of collisions (with the wall) per unit time interval. Doing the calculation (see below) we obtain $\bar N = \frac{1}{6}n\bar v A$, where $n$ is the density.

Hence, thinking in these terms and using "unit time intervals" your problem disappears and we conclude that $$ \bar F = \bar N \cdot \bar p = \frac{1}{6}n\bar v A \cdot 2 m \bar v = \frac{1}{3} n A m \bar v^2 $$ The pressure is straight forward, $\bar P= \bar F / A$.

Note that for the proper calculation we have to take into account that the momentum is not directed perpendicular to the wall. Instead, we have to consider a distribution of directions and calculate the mean momentum transfer according to this distribution. In doing so we find that the above stated result remains valid, if we replace $\bar{v}^2 = \langle v \rangle^2$ by $\langle v^2 \rangle$.

Calculation:

  1. We consider one wall of a 3D box.
  2. Let's assume that each particle travels only in one of the six possible direction ($\pm x, \pm y, \pm z$). If we have in total $N$ particles, then on average $N/6$ travel towards the particular wall chosen in step 1.
  3. Let's consider the unit time interval $dt$. Within this time interval only those particle which are at distance $ds = v dt$ to the wall are able to reach the wall. Hence, these particle occupy a volume $dV = A ds$.

Putting these three ideas together we conclude that the average number of collisions with the wall is given by $\bar N = \frac{n}{6}(A ds)$, where $n$ is the number density.

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The aim of this part of the derivation is not to find the mean force that the molecule exerts on the wall during a collision, but the mean force it exerts over the whole time that the molecule is bouncing back and forth. That's why we divide the change in momentum per collision with the wall by the time between collisions with this wall (for example the time interval from mid-collision to mid-collision).

You might think of this graphically. A graph of force exerted on the wall by the molecule would be a series of tall spikes at regular intervals. But we're not interested in the individual spikes. We're interested in the mean force, and this would be a low horizontal line on the graph. The area under this line over a given time interval, $\Delta t$, would be equal to the sum of the areas under the spikes, which would equal the sum of the molecule's momentum changes, $\Sigma \Delta \vec p$, due to collisions with the wall over time $\Delta t$. So the mean force would be $\frac {\Sigma \Delta \vec p}{\Delta t}$.

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As you've pointed out, the collision time $t_c$ and time between collisions $t_b$ need not be the same.

Nonetheless, the molecule still bounces back and forth between the walls with a period of $T = t_b + t_c$.

But, what is the average force, $F_{avg}$, experienced by the wall during the whole time in general? Even when the molecule isn't touching anything at all? (zero force)

Well, if $F_{col}$ is the average force experienced by the wall during a collision, then we can say $$F_{avg} = \frac{F_{col}\times t_c}{t_b + t_c} = \frac{2mv}{T}$$ And if the walls are perfectly rigid such that $t_c = 0$ and $T = t_b + t_c = 2L/v$, then we can even write $$F_{avg} = \frac{2mv}{2L/v} = \frac{mv^2}{L} = \frac{2E}{L}$$ And if this $F_{avg}$ is applied over some wall area $A$, then we can also write $$P = \frac{F_{avg}}{A} = \frac{2E}{L\times A} = \frac{2E}{V} \Longleftrightarrow PV = 2E$$ thus deriving the pressure characteristic in question.

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