I'm just a beginner in thermodynamics, so I was confused abt the method by which work done is measured in thermodynamics. I'm not confused by the different sign conventions in physics and chemistry, I'm confused that the value of work done is coming different if I consider different forces. I've taken a case of isothermal irreversible expansion of an ideal gas in a container with a massless and frictionless piston . I've explained my problem furthur in it
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2 Answers
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For an irreversible expansion, the force exerted by the gas on the piston is not described by the ideal gas law. The ideal gas law applies only to thermodynamic equilibrium (or for a reversible process, in which the gas passes through a continuous sequence of thermodynamic equilibrium states). For your irreversible expansion, the force exerted by the gas on the piston is less than predicted by the ideal gas law and, for a massless frictionless position, according to Newton's 2nd law, is equal to the external pressure times the cross sectional area of the piston.
answered Jan 20, 2022 at 4:32
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Your P-V diagram is correct, but the work done by the surrounding gas is only a small fraction of the total work done. The majority of the work is done by whatever force is holding the rod and piston in place against the internal pressure. This force times the piston area will have to equal the internal pressure $P$ at all times, for the forces to balance.
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$\begingroup$ Yeah, u r right. Can we also think that as there is some change in volume meaning change in some height, therefore the atmospheric pressure is also not exactly constant at this time so that can also account for this.(if we forget that the piston even has a mass so piston will be rest at the interface) $\endgroup$ Commented Jan 20, 2022 at 4:30