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I was reading the derivation of the average translational kinetic energy of an ideal gas in Sears and Zemansky's University Physics. This derivation uses a cylinder with height $|v_{x}|dt$ and base area $A$ and states that the total change in momentum $dP_{x}$ can be found by multiplying the change in momentum of an individual molecule with the number of molecules found in this cylinder. This got me wondering just how small the $dt$ in this derivation is. If $dt$ is infinitesimal, then how can it be larger than the time it takes for an individual collision between the wall and an ideal gas molecule? In other words, how can it describe successive collisions? Does this mean that $dt$ for an individual collision equals 0?

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  • $\begingroup$ maybe this will help en.wikipedia.org/wiki/Differentiable_function . The physics assumption in differentiation is that there exists such a continuous function for the molecular momentum.. $\endgroup$
    – anna v
    Apr 29, 2022 at 3:39
  • $\begingroup$ @annav, but for there to be a change in momentum, I thought that the molecules had to collide for a time dt. In this derivation, dt is given to the time between successive collisions, and the time of each collision is assumed to be 0. $\endgroup$
    – Piksiki
    Apr 29, 2022 at 13:40
  • $\begingroup$ It is the mathematics of continuous functions. dt means a very small increment approaching zero but not zero. $\endgroup$
    – anna v
    Apr 29, 2022 at 19:39

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$|vx|dt$ has units of length squared and cannot be a height. Do you mean height $vdt = dx$ or what? Also, if $P$ is momentum why is the change in momentum $dPx$ instead of $dP$. Please clarify your question.

Regardless, the assumption is that although the volume element has small dimensions, it contains many gas molecules. That is, for the small volume element the gas is considered a continuous medium. But, the volume element is sufficiently small such that the properties of the gas (temperature, pressure, etc.) are the same throughout the volume element.

This sort of treatment is common. For example, consider a rod (a rigid body) of constant cross sectional area. We treat the mass of the rod within a small differential length $dx$ as$\mu dx$ where $\mu$ is the linear density- mass per unit length- of the rod, even if $dx$ is very small; although at the microscopic level the rod is not a continuum of mass but a collection of discrete molecules. But, $dx$ is sufficiently small such that that positions $x$ and $x + dx$ along the rod are both at $x$. For example, we calculate the moment of inertia about an axis through an end of the rod as $\int_{0}^{L} \mu x^2dx = {ML^2 \over 3}$ where $L$ is the length of the rod.

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  • $\begingroup$ So in this example, dt is small such that the gas molecules are assumed to strike the wall at the same time? But I was thinking that for there to be a change in momentum, the gas molecules need to collide with the wall for a time dt? In this approximation the collision takes 0 time. $\endgroup$
    – Piksiki
    Apr 29, 2022 at 13:19
  • $\begingroup$ I need to see the complete derivation, but I think what is meant is that moving at $v_x$ within a time dt all the molecules move a distance $dx=v_xdt$ and within $dt$ all strike the wall. $\endgroup$
    – John Darby
    Apr 29, 2022 at 13:35
  • $\begingroup$ The molecules next to the wall strike the wall in $o$ time; those molecules at a distance $dx = v_xdt$ strike the wall after dt; those molecules at some intermediate distance strike the wall within $[0, dt]$ at a time depending on how far they were initially from the wall. Hope this helps. $\endgroup$
    – John Darby
    Apr 29, 2022 at 13:57
  • $\begingroup$ But how long do these collisions last? For the momentum to change by dP, the individual collisions needs to last for some time dt, but what is dt in this case? $\endgroup$
    – Piksiki
    Apr 29, 2022 at 14:10
  • $\begingroup$ I think $dt$ is the time over which all collisions are considered, but when a collision does occur it is instantaneous. I need to see the complete discussion of the derivation you are referencing to be sure; can you add it to your question? $\endgroup$
    – John Darby
    Apr 29, 2022 at 14:33

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