Why the round trip instead collision time?

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The assumptions that are being made here are perfectly elastic collisions with the walls where the atom will rebound with the same speed as it came it with. Its trying to calculate the average force based on the number of interactions per unit time with the wall, and is therefore fundamentally related to how long it takes for successive collisions between the walls. As it states above, $F = \Delta p/\Delta t$, and the only time that $\Delta p$ changes for an atom interacting with the wall is during a collision with the wall. To calculate the average force using the time in between collisions allows us to ignore the actual interaction time of a single collision. The assumption here is that there are many particles interacting with the walls at all times, so that the average force on the walls stays relatively consistent. If there are $N$ collisions in time $\Delta t$, the total momentum change is $2Nm\bar{v}_x$, for $N$ particles with an average velocity perpendicular to the wall $\bar{v}_x$. If $\Delta t = 2L/\bar{v}_X$, the average round trip time, then this gives an average force for $N$ molecules of:

$$\bar{F} = mN\bar{v}_x^2/L$$

The point of all of this is that its taking a statistical average of many collisions over a long time period, even when its considering the single particle case. Since for an ideal gas, we know that there will be a perfect rebound, we know precisely what $\Delta p$ should be for every collision even if we don't know how long each collision will take individually. But, we can calculate the number of collisions in a time $\Delta t$ in order to calculate the average force over the entire time period, which is what they did in the above example.

Also, note that it states that the gas is an ideal gas, meaning that the relationship $PV = nRT$ can be invoked. The pressure, $P$, is the force per unit area on the walls, and $V$ is the volume of the container. Since temperature is just related to the energy spread of the gas, we can determine the following: If the walls are infinitely far apart, then the volume, $V$ is infinite, therefore, $P = nRT/V \rightarrow 0$, and thus the force per unit area is zero. This is consistent with the fact that for infinitely spaced walls, $\Delta t$ between collisions $\rightarrow \infty$.

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Jye Quan
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Jye Quan

Updated on December 14, 2020

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  • Jye Quan
    Jye Quan almost 3 years

    enter image description here

    From Newton's second law, the $\Delta t$ is defined as the collision time, but why in this case, it can be assumed to be the value of time between successive collisions on 1 wall? If I have an infinitely large vessel, doesn't that make the $\Delta t$ infinitely large, and thus exert approximately 0N of force on the wall? What are the assumptions being made here?

    • Michael Seifert
      Michael Seifert almost 8 years
      If you have an infinitely long vessel, then it has infinite volume $V$. If we also assume finite temperature $T$, then the ideal gas law implies that $P = NkT/V \to 0$, i.e., there is zero force exerted on the wall. So that limit actually makes sense.
    • Max Payne
      Max Payne almost 8 years
      @Jye Quan I agree. I too had this question. In Newtonian mechanics, the $F_{avg} = \dfrac{\Delta p}{\Delta t}$. Here, time interval is the time taken to change the momentum. Which is certainly not equal to the time interval between successive collisions. I posted this question but realized that it had been asked just 4 days ago!
  • Max Payne
    Max Payne almost 8 years
    how long it takes for successive collisions between the walls is not related to how long particle takes to change its momentum, in my opinion. Although it will be very impractical to measure the latter, wouldn't it be inaccurate to consider the former?
  • tmwilson26
    tmwilson26 almost 8 years
    @TimKrul You are correct here. For any given event, it will be difficult to measure the force for any individual event. The first part of my answer assumes that there will be many particle collisions in a given time period, and knowing the total momentum change (which we know for an ideal gas in a container as the question states) we can calculate the average force over any given time period. Things get tricky for an infinite volume of course, but in this limit there should be no collisions with the walls, and therefore no force.
  • tmwilson26
    tmwilson26 almost 8 years
    @TimKrul to answer your question, it is accurate for a statistical average of many particle collisions, but not accurate for a single collision with the wall.
  • tmwilson26
    tmwilson26 almost 8 years
    I've also edited my answer to hopefully clarify what I meant.
  • Max Payne
    Max Payne almost 8 years
    Thank you! As you say "it is accurate for a statistical average of many particle collisions" seems quite reasonable, but i think it should be stated as an assumption of the kinetic theory.
  • Jye Quan
    Jye Quan almost 8 years
    Thank you @tmwilson26 for your reply. So to sum up, we are taking the total momentum change over Δt, where Δt is the fixed time interval, instead of doing it for individual particles. Hence, assumptions made here, I assume ( no pun intended ), is that the momentum of gas particles are distributed such that if a graph of momentum against time is plotted, you might get something like a sine curve, and therefore the Δt is taken as the period, where there is an overall change in momentum. And the Δt is the time between collisions. This will be consistent with the above arguments. Am I right?
  • Jye Quan
    Jye Quan almost 8 years
    @tmwilson26. One more thing is, for the equation P=nRT/V, where V approaches ∞ will result in P=0, what if the number of gas particles, n, also increases up to ∞? Gas particles still have to travel back and forth between parallel faces, assuming they do not collide with each other. And in this case, ∞/∞ will be undefined, right?
  • tmwilson26
    tmwilson26 almost 8 years
    @JyeQuan Thats not quite right, the momentum distribution should resemble a Boltzmann distribution, and if you have an ideal gas in a small container, there should be collisions happening at every wall in 3 dimensions, so the momentum distribution shouldn't change that much.
  • tmwilson26
    tmwilson26 almost 8 years
    @JyeQuan The force on a single wall is related to the total number of collisions in a time period $\Delta t$. On average, each particle in the gas will only collide once with the wall for every round trip time, and on average the momentum change twice the mass times the average velocity. This gives you a relationship to calculate the average force $N\Delta p/\Delta t$.
  • tmwilson26
    tmwilson26 almost 8 years
    @JyeQuan Clarifying your second comment, that is correct that it would not be defined in that case. But you could consider scaling things up. If you knew the pressure at a certain $n$ and $V$, if they increase at the same rate, the pressure remains fixed as they approach infinity.
  • tmwilson26
    tmwilson26 almost 8 years
    @JyeQuan If there is non-zero pressure, then $F$ is non-zero as well because $F = PA$ where $A$ is a chosen area of the wall (also non-zero). $L$ is increasing, but $N$ is increasing at least as fast as $L$ depending on the geometry. In fact, if the area of the side wall increases towards infinity, the force will go to infinity as well at non-zero pressure.
  • Jye Quan
    Jye Quan almost 8 years
    @tmwilson26 I think I finally got it. Thanks again!
  • tmwilson26
    tmwilson26 almost 8 years
    @JyeQuan No problem at all. I'm happy to help :)
  • Jye Quan
    Jye Quan almost 8 years
    @tmwilson26 I think there is a misunderstanding about my statement on the momentum graph. What I actually mean is if you draw a graph of a single particle traveling back and forth, you will get something like a square wave. But since in Δt, each particle collides with 1 wall once, then particles could have a range of collision times, adding individual graphs together to get a curved graph with constant period. Momentum against time.
  • tmwilson26
    tmwilson26 almost 8 years
    @JyeQuan A single particle will certainly have a square-wave like momentum graph in 1-D. However, every particle does not collide once during every period, but there will be on average $N$ collisions for $N$ particles in a time $\Delta t$. Since the gas is an ideal gas, it has a Maxwell-Boltzmann momentum distribution. The faster particles will collide more often than the slowest particles. The momentum distribution also shouldn't change with time, because for any collision on one wall, there are corresponding collisions on the opposite wall to balance things out.
  • Admin
    Admin almost 8 years
    @JyeQuan-Now I hope that you find it clear that why P=ML^-1T^-2,Then as I have given the correct answer to your question,so please accept my answer and give me a bounty as soon as you set a bounty,OK?
  • Antonios Sarikas
    Antonios Sarikas over 3 years
    @tmwilson26 What if between the round trip molecules collide therefore $\Delta t \neq \frac{2l}{u_x}$?
  • tmwilson26
    tmwilson26 over 3 years
    @adosar I think its best to think of things in terms of the average number of collisions per second with the wall. If it collides with another molecule, it may get turned around and collide with the wall again a little sooner, but the other one would then take longer, and the average value would be the same overall.