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I’m having a hard time really getting how the relationship between pressure and square mean velocity of a gas is derived for one single reason...seems like I don’t really know what impulse is.

From what I know force is the rate of change of momentum I.e Impulse/Time of collision.

But in the various explanations I see, when a gas particle collides with the wall of the container, The time period considered is that between collisions and not the time of collision between that gas particle and the wall of the container.

I understand that that might be true by the formula since In this time between collisions the particle hits the wall once and thus changes momentum once in this time interval...But it just doesn’t seem right to me. Could we just then take any time interval during which the momentum of a particle changes once and then calculate force as this change In momentum divided by the arbitrarily chosen time interval ?

I’m really confused...help please.This is one of the explanations is got on the internet.

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    $\begingroup$ For purposes of indexing and for users accessing the website through mobile devices, please consider removing the image and typesetting its content in MathJax. Also, what are you asking specifically? The definition of impulse? The time between collisions? Try to clarify and this question will likely receive more attention. $\endgroup$
    – MannyC
    Feb 26 '20 at 22:50
  • $\begingroup$ Do you mean time between collisions of the particle with the wall multiple times vs say the time of collision of a single particle with the wall? They have nothing to do with one another $\endgroup$
    – Triatticus
    Feb 26 '20 at 22:53
  • $\begingroup$ I know...But Why is Time between collisions of the particle with the wall used in the formula for force instead of the time of collisions of the said particle with the wall ? Should the latter no be used instead ? $\endgroup$
    – user233118
    Feb 26 '20 at 22:57
  • $\begingroup$ The derivation did use the time over which the particle collides with the wall, that's the only time which would make sense $\endgroup$
    – Triatticus
    Feb 27 '20 at 0:03
  • $\begingroup$ Sadly the time used is really that between collisions...Unless I really have a serious problem with reading. $\endgroup$
    – user233118
    Feb 27 '20 at 0:09
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The interaction time during the collision turns out not to matter.

In most other "physics problems" calculations, we only care about the force during the interaction or collision. So the length of the collision is important.

But here we want the average force over time (meaning across multiple collisions). This average means we don't care about the forces in any single collision, just the total impulse from each, and how often they occur.

Let's imagine someone dribbling a soccer ball on their leg. The ball moves slightly, but overall is about the same height over time, suggesting that the average acceleration is zero and that the average force over time on the ball is zero.

Let's assume the ball has a weight of 5N, so the average force from the leg must be 5N. $F = \frac{\Delta p}{\Delta t}$ If the person is dribbling by hitting it once every second, we know the impulse delivered every second is sufficient to generate an average force of 5N over the entire second.

$$ \Delta p = 5\text{Ns}$$

The force during the interaction will be much higher, but isn't necessary to know.

In the gas calculation, we do the same math, but instead start with impulse information (because we know the change in momentum of the molecule) and the time information (because we know the speed and the size of the box), so we can calculate the average force.

Could we just then take any time interval during which the momentum of a particle changes once and then calculate force as this change In momentum divided by the arbitrarily chosen time interval ?

Yes, but for a continuous process (bouncing a soccer ball or molecules bouncing off walls), the time interval during which the momentum changes exactly once is not arbitrary. It's determined by the periodicity of the event. We can't use 1 second for the soccer ball if the person is dribbling the ball more slowly than that.

The Average value does depend on the chosen time interval...We do choose the time interval between two successive collisions, But why ?Like why is the average value over this time interval more legitimate than another one.

You can use any interval you want, but you have to count the total impulse delivered during that time. If there are two collisions in that interval, then each will presumably deliver half the impulse. Setting the interval to one in which there is only one collision is just a convenience. Twice as long, twice as many collisions, twice as much impulse. But total impulse divided by time remains constant.

Like when you say the average acceleration of the ball over time is zero...Why... ?

Because the ball remains in front of the dribbler. If the ball had some consistent acceleration, it would move away. It was just an explicit statement so that we could compare the weight of the ball (pushing it down) to the impulse delivered by the leg (pushing it up). Over the course of the session, these two must be almost exactly equal.

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  • $\begingroup$ Huge thanks for helping...I think I get this but something keeps troubling me.The Average value does depend on the chosen time interval...We do choose the time interval between two successive collisions, But why ?Like why is the average value over this time interval more legitimate than another one.Like when you say the average acceleration of the ball over time is zero...Why... ? $\endgroup$
    – user233118
    Feb 27 '20 at 0:42
  • $\begingroup$ Added a bit in the answer. $\endgroup$
    – BowlOfRed
    Feb 27 '20 at 5:43

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