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Consider a perfectly insulating cylinder enclosing a gas at pressure same as external atmospheric pressure $P_o$. Now a mass $M$ is placed on the piston. What will be the work done on the gas when the gas comes again in equilibrium. Now according to my sources if the piston is displaced by distance $h$ the work done on gas is $Mgh$. But I think it should be different than this. As the process is irreversible the internal pressure will not vary uniformly. We have to calculate the work by using external pressure on the piston. So the work done on gas should be $$\int(P_o + \frac{Mg}{A}) dV $$ This expression takes into account work done by force due to mass and also due to atmosphere but Mgh doesn't consider the work done by atmosphere. So where am I wrong here?

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  • $\begingroup$ What work does the atmosphere do on the gas without the weight? $\endgroup$ – Bob D Mar 30 at 7:13
  • $\begingroup$ The key quesiton is what is the system on which work is done ?. Gas, o gas+mass ? $\endgroup$ – GiorgioP Mar 30 at 8:03
  • $\begingroup$ @GiorgioP the actual question is to find final temp given initial temp. So we would require to find work done on gas. Not gas+mass $\endgroup$ – ATHARVA Mar 30 at 8:57
  • $\begingroup$ @Bob D $\int P_{o} dV$ but if mass is not placed it will remain in equilibrium. $\endgroup$ – ATHARVA Mar 30 at 8:59
  • $\begingroup$ In my judgment, you are absolutely correct. The work done by the surroundings on the gas is exactly as you have written it (assuming a massless piston), except for the sign. It should be $-(P_0+\frac{Mg}{A})(V_f-V_i)$ $\endgroup$ – Chet Miller Mar 30 at 12:02

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