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In kinetic theory of gases we know that the average number of collisions, $N_\mathrm{col}$, is given by

$$N_\mathrm{col} = \sqrt{2\pi\sigma^2 \overline{v}_\mathrm{rel}}\ N $$

where $N$ is the number of particles, and $\overline{v}_\mathrm{rel}$ its average relative velocity.

If we consider bimolecular collisions we should have divided the total number of collisions $N_\mathrm{col}$ by $2$. But instead we multiply it by $N/2$. Why?

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  • $\begingroup$ I guess when you refer to bimolecular collisions you are really meaning a collision between like atoms or molecules (A + A) instead of disimilar species (A + B). You have to take into account that you have N molecules of each kind and ieach of this molecules can collide with the rest of the molecules in the sample. $\endgroup$
    – PAEP
    Aug 6, 2023 at 16:33
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    $\begingroup$ 2 objects can only collide 1 way: a-b. 3 objects can collide 3 ways: a-b, a-c and b-c. 4 objects can collide 6 ways: a-b, a-c, a-d, b-c, b-d, and c-d. The more objects, the more ways they can collide, and the more collisions, so N is a factor. $\endgroup$ Aug 6, 2023 at 19:43
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    $\begingroup$ @paep I'm not sure I get the meaning of the angle signs. In MathJax They're set as relational operators. I'm confident that's not what you wanted to achieve. $\endgroup$ Aug 6, 2023 at 23:27
  • $\begingroup$ @DrMoishePippik So this means that using the factor 'N' along with 1/2 gives us an accurate measurement of the number of collisions $\endgroup$ Aug 7, 2023 at 1:35
  • $\begingroup$ Thanks for the tip @Martin-マーチン. I meant to write $\overline{v}_{rel}$. $\endgroup$
    – PAEP
    Aug 7, 2023 at 10:33

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I'm not sure if this answers your question, but in the special case where a collision/reaction combines two identical molecules $\mathrm{A+A} $ the number of indistinguishable combinations, i.e. the number of distinct A-A pairs of the two species must be calculated and this makes the bimolecular rate $\displaystyle \sim \frac{N_A(N_A - 1)}{2!}$ collisions. Normally, in dealing with chemical kinetics, $N_A$ is so vast that we can use $\displaystyle N_A^2/2$ without any error.

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  • $\begingroup$ This carries similarities to a single-rounded round-Robin tournament with $n$ participants, and $\frac{n(n-1)}{2}$ games (every participant plays only once with every other participant than him/herself). Note, the denominator I used here is $2$, not $2!$ (though indeed $2! = 2$ is true). $\endgroup$
    – Buttonwood
    Aug 7, 2023 at 8:45
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    $\begingroup$ Or the simple puzzle how many handshakes would occur if N friends met and all did handshaking with all others. $\endgroup$
    – Poutnik
    Aug 7, 2023 at 9:42
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    $\begingroup$ @Buttonwood only for interest, in a termolecular reaction the chance is $n(n-1)(n-2)/3!$, which is needed if a Monte Carlo method are used to calculate populations during reactions, rather than $n^3/6$, not that termolecular reactions actually ever occur except in a computer :) $\endgroup$
    – porphyrin
    Aug 7, 2023 at 11:05
  • $\begingroup$ @porphyrin Then (I now speculate) $2!$ is the more general, easier extendable approach to describe the situation with mathematics. $\endgroup$
    – Buttonwood
    Aug 8, 2023 at 7:02
  • $\begingroup$ @Buttonwood, yes I guess it is. $\endgroup$
    – porphyrin
    Aug 8, 2023 at 13:40

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