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I have seen the derivation of the following equation for a cubical container with $N$ molecules.

$$PV = \frac 13 mNu^2 $$

How is this equation valid even for spherical container or any other non cubical shaped container?

(I prefer rigorous equation-based proof but I'm okay even with analytical proof )

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  • $\begingroup$ With any non-cubic container, the problem would require some calculus. $\endgroup$ – Ivan Neretin Mar 1 '18 at 7:12
  • $\begingroup$ And I'm not even sure that it's valid.I was just told so. $\endgroup$ – Tilak Maddy Mar 1 '18 at 7:25
  • $\begingroup$ How would any of the molecules "know" the container isn't exactly cubic? Slightly more formal, assume the container deviates from a cube by dV in one place and -dV in another. This does not affect the equation, it turns out, regardless of the magnitude of dV. $\endgroup$ – MSalters Mar 1 '18 at 11:26
  • $\begingroup$ The equation tells us that energy represented as $pV$=(force/area) . volume = force . distance $\equiv$ Newton . metre = Joule is equal to the (kinetic) energy of the particles $mu^2$ = mass . velocity = kg . m /s$^2 $ = Joule. So this cannot depend on the shape of the container. $\endgroup$ – porphyrin Mar 1 '18 at 15:59

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