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Pressure by gas molecules of an ideal gas on the walls is P = nRT/V

But by derivations P=1/3 nmv², where v is velocity.

Are they equal, if yes, then it means that we can deduce volume of free path just by having R,T,m and velocity, but finding volume by just these values are not intuitively possible I believe.

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The "$n$"s mean different things. The $n$ in the first equation is the number of moles of gas present (a pure number). The second "$n$" is the density of particles (a number per unit volume).

So cancelling the $n$ and the $V$ in the equation $$ P= nRT/V = \frac 13 (n\times N_{\rm Avogadro}) m v^2/V $$ and using $R=k_{\rm Boltzmann} N_{\rm Avogardo} $ gives you $$ k_{\rm Boltzmann} T = \frac 2 3 (\frac 12 m v^2) $$ or
$$ KE= \frac 32 k_{\rm Boltzmann} T $$ which is an example of the equipartition theorem.

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  • $\begingroup$ Thanks. And can we equate them. I mean both the equations are talking about pressures right and they have same value ? $\endgroup$
    – user253164
    Feb 10 '20 at 19:31
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    $\begingroup$ Yes.They are equal provided you replace the second "n" by $n \times N_A/V $, where $n$ is the first "$n$" and $N_A= 6\times 10^{23}$ is Avagadro's number---but what do you mean by "volume of free path"? $\endgroup$
    – mike stone
    Feb 10 '20 at 19:35
  • $\begingroup$ the V in ideal gas equation stands for:- Volume of container-volume of gas molecules. $\endgroup$
    – user253164
    Feb 10 '20 at 19:44
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    $\begingroup$ I've added material to my answer. $\endgroup$
    – mike stone
    Feb 10 '20 at 19:56

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