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Work done by system on the piston should be $F_{\text{sys}}\cdot(ds)$ and work done by the surrounding on the piston should be $F_{\text{sur}}\cdot(-ds)$, thus the net work is stored in the piston as its kinetic energy. The piston exerts equal and opposite force on the system and surrounding, therefore $dE_{\text{system}}$ should be equal to $dq - (p_{\text{sys not surr}})dV$. But is it not so? If no then how?

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  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    Commented Oct 4, 2023 at 19:34

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In thermodynamics the usual assumption is that the changes are quasi-static. This implies that, we push the piston slowly into a cylinder containing 'the system', so that the force that the surroundings exert on the piston is almost equal and opposite to the force that the system exerts on the piston. So the piston does not acquire significant kinetic energy.

Your second sentence follows from the First Law of Thermodynamics, as you are stating that the increase in the system's internal energy is equal to the heat flowing in minus the work done by the system on the piston (and hence on the surroundings).

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  • $\begingroup$ When a difference in pressure is created across a movable piston and the piston is nonconducting then the systems will never reach equilibrium. The piston will oscillate . If the difference is small then the oscillation will not be observable. When pressure is equal on both sides , velocity of the piston is not 0. This pushes the system towards non equilibrium again ( just like what happens in spring oscillations). So for quasi-static almost no oscillation is there. But what I want to say is that decrease in energy of a system always = dq - p( internal not external ) dv even when it is not qua $\endgroup$ Commented Oct 4, 2023 at 12:27
  • $\begingroup$ Sistatic process $\endgroup$ Commented Oct 4, 2023 at 12:27
  • $\begingroup$ From newton's laws of motion , dv(velocity of piston)= F(net)/M(mass of piston) * ds . And F(net)=F(sys)-F(surr). From newton's law of equal and opposite force, the piston exerts a force -F(sys). Thus decrease of energy from the system in the form of work = -F(sys) ds = -p(sys) dv(volume not velocity). $\endgroup$ Commented Oct 4, 2023 at 12:39
  • $\begingroup$ For example consider an adiabetic free expansion. According to the norm , no work is done because p(surr) = 0 and therefore change in internal energy is also 0 . But if no work is done , from where velocity (indicating its kinetic energy) of the piston comes? $\endgroup$ Commented Oct 4, 2023 at 12:46
  • $\begingroup$ According to me F also depends on mass of the piston . We know that F=da(momentum)/dt . This exchange of momentum becomes 0 if the piston is massless(apply the acurate equation of collision not the approximate one which applies only when the piston is very very heavy compared to the gas molecules). Thus no gas can exert pressure or force on a massless piston. This is why no work is done by a gas on massless piston or no piston. But anyways ... the kinetic theory behind pressure is almost true when piston is very heavy. $\endgroup$ Commented Oct 4, 2023 at 13:13

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