Here is the extract from the book

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Now the weight $π$ is sought which can compress the air $ECDF$ into the space $eCDJ$, the velocities of the particles in either air (the natural and the compressed), of course, having been assumed the same; moreover, let $EC= 1$, and $eC = s$; but since the lid $EF$ is transferred to $eJ$, it suffers a greater pressure from the fluid in two ways: firstly, because the number of particles is now greater in proportion to the space in which they are contained, and secondly, because any particle repeats the impetus more often. In order to perform correctly the calculation of the increment which depends on the first cause, we shall consider the particles as resting, and we shall make $n$ the number of those which are adjacent to the lid in the position $EF$, and the equivalent number for the location of the lid at $ef$ will be $$n/\,\left(\frac{eC}{EC}\right)^{2/3} = \,\left(\frac{n}{s^{2/3}}\right)$$

My question is about the power $2/3$. How does it appear?


I would try something like this (sorry for my english) :

The particle density of the gas has increased $n'=n\frac{EC}{eC}=\frac{n}{s}=n{{s}^{-1}}$ and the distance between the molecules has decreased $\delta '=\frac{1}{n{{'}^{1/3}}}=\delta {{s}^{1/3}}$.

So, in a thick layer $\delta '$, we have $N'=An'\delta '=An\delta {{s}^{-1+1/3}}=N{{s}^{-2/3}}$

I am not sure at all ! Hope it can help...

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  • $\begingroup$ it seems to be correct! Can you explain further the formula of the decrease of distance between molecules? $\endgroup$ – Vaggelis Kyrilas Feb 3 '19 at 18:07
  • $\begingroup$ The typical distance $d$ between molecules is obtained by writing that in a side cube this typical distance, there is a roughly one molecule: $n{{d}^{3}}\approx 1$ or $d\approx \frac{1}{{{n}^{1/3}}}$ $\endgroup$ – Vincent Fraticelli Feb 3 '19 at 18:11

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