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NOTE: Please Imagine A cube with 2 parllel faces I and II lying along the x axis.

In the derivation what we are thought

Change in the momentum of the molecule

= Final momentum - Initial momentum

= –mu1 – mu1 = –2mu1

During each successive collision on face I the molecule must travel a distance 2l from face I to face II and back to face I.

Time taken between two successive collisions is = 2l / u1

∴ Rate of change of momentum = Change in the momentum/Time taken

= – 2mu1/(2l/u1) = -2mu1^2/2l = – mu1^2/l

(i.e) Force exerted on the molecule = – mu1^2/l

Components of Velocity

∴ According to Newton’s third law of motion, the force exerted by the molecule,

= – (– mu1^22)/l = mu1^2/l

I think this is wrong my reason:

Assuming normal force equation F=ma, F=mdv/dt => F=dP/dt

in the derivation the total time taken between 2 successive collisions is given this is not the time taken to change the velocity of the particle it changes at the instant when the molecule hits the wall of the container and change in velocity with respect to time is acceleration so i think the above derivation feels wrong to me can any one help me out here am I missing something.

I have copied the derivation from somewhere else as I have very less time due to my exam :p.

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    $\begingroup$ Please do not post formulae as plain text, but use MathJax instead. $\endgroup$ – ACuriousMind Feb 9 at 10:49
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Indeed you are right but during concerned time period, a large number of molecules with nearly same speed would also be colliding wih the wall hence if we consider rightly that the particle gives an short lasting impulse to wall , a large no. of particles will do so (say 10) in time by which particle bounces back from other wall( in time t1) now force is a vector so we perform vector sum of these forces to get resultant felt by the wall: enter image description here Now if i replace the impulse by assuming momentum is changing gradually at constant rate such that total change in momentum is same as earlier , I get: enter image description here Again i do this simplification for all the ten particles within for time t1 and perform vector sum of each , the resultant felt by wall in this case is: enter image description here But this resultant is same as resultant of earlier impulses. Thus the wall feels sees same reality in two cases

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  • $\begingroup$ Can you explain the second graph in words $\endgroup$ – HOME WORK AND EXERCISES Feb 9 at 9:13
  • $\begingroup$ I have reframed the answer in a more comprehensive manner $\endgroup$ – Kutsit Feb 9 at 9:44
  • $\begingroup$ thank you that was great unfortunately due to my low rep i can't upvote. $\endgroup$ – HOME WORK AND EXERCISES Feb 9 at 9:53

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