I read in the book the derivation of pv=nrt, they relate the change of momentum and consider time it take for 1 collision. However, if considering a cubical volume expand at the speed that is faster than the speed of the gas molecule so that the gas molecule wont collide with the wall(i don't know if this is possible or not). At that particular instant the volume may have a volume of V while the gas havent touches the wall of the container(in this case work is done by external sources in expanding the volume of the cube). At this time, is it obey pv=nRT and can take p1v1=p2v2?(cause if this can be happen and no collision occur at the wall the pressure exert on the wall is 0)?

Dear  chua ,

if you are considering the gas to expand slowly than container than it actually is like as if gas is in free space as no molecule can collide to wall. it is like an expandable space. and the pv =nrt holds for that . so it shall continue to hold. why it holds in space  when it is derived in a container  ?  because we assume that at infinity there is a container and apply the results.  Now in your case no one can apply work continuously to the wall  as energy is finite and so wall shall meet the molecule at infinity and so it is actually not a feasible sitiuation.

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dear chua kok tong,

ACTUALLY THIS IS A HYPOTHETICAL SITUATION

BUT STILL IN MY OPINION ALL YOUR SUPPOSITIONS ARE CORRECT AND THE CONDITION SEEMS TO BE TRUE...

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AMAN BANSAL

Anyway the P at the formula is the pressure exert by gas on the container?