Is the kinetic gas equation valid for non-cubical containers?

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 )

• With any non-cubic container, the problem would require some calculus. – Ivan Neretin Mar 1 '18 at 7:12
• And I'm not even sure that it's valid.I was just told so. – Tilak Maddy Mar 1 '18 at 7:25
• 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. – MSalters Mar 1 '18 at 11:26
• 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. – porphyrin Mar 1 '18 at 15:59