Pressure on wall by gas molecules

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.

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.
• 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"? Feb 10 '20 at 19:35