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I was drawing a free body diagram of the 3rd step of the Carnot cycle as shown below.

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The forces with - sign are reaction forces to the forces without the sign, according to Newton's 3rd law of motion. In this diagram, the forces applied on the piston from the gas are -Fpg and Fpres. Fpres is the pressure force due to collisions of gas particles on the piston, while -Fpg is originated from the piston which is pushing the gas; since the piston pushes the gas, the gas also pushes the piston in opposite direction. Since -Fpg and Fpres comes from different origins, so I think they're different force entities. So in the calculation of the work done on the piston by the gas, these forces are all to be included.

However, the thermodynamic textbook seems only taking Fpres into account so that the infinitesimal work is just PdV. Why is -Fpg not included in work calculation?

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So, first, the work done on an ideal gas is always $PdV$. I really have no idea how your question relates to what is essentially a fundamental law of ideal gasses: you do work on them by changing the volume that they are constrained to. Nonetheless, I will try to explain what I perceive to be your misunderstanding.

I'm not super sure what $F_{pg}$ is referring to, but I think you are meaning the downward force of the piston (due to gravity or some external force) and will assume as such. I will also assume that $F_{pres}$ refers to the upward force of the gas caused by pressure.

If this is the case, I think you are assuming that Newton's Third law states that every action has an equal and opposite reaction of the same nature. What opposes the downward force of the piston is the upward pressure by the gas.

Hopefully this helps. If not, I would suggest clarifying your question.

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  • $\begingroup$ Hello. Fpg is the force applied on the gas by the piston as the piston pushes the gas; for example, the gravity attracts the piston downward so the attracted piston has to push the gas. Then, due to the Newton's 3rd law, the gas gives the piston equal and opposite force, which is -Fpg. My question is why -Fpg is not included in the calculation of the work done on the piston by gas, although the force giver of -Fpg is the gas. $\endgroup$
    – Donggyu Jang
    Mar 19, 2017 at 9:16
  • $\begingroup$ Hm...I got some flash idea right now…This force may not be taken into account since the origin of this force is the piston; the work with –Fpg on the piston may be the work done by the piston itself via the reaction force. It is reminiscent to the situation that a person A pushes the hard wall. His upper body will be pushed back due to the reaction force from the wall, however, in this case, the wall doesn’t do work on A, no energy is drained from the wall, it is just that A works on himself via the reaction force. Could you tell me whether or not my reason is right? $\endgroup$
    – Donggyu Jang
    Mar 19, 2017 at 9:16
  • $\begingroup$ As I said, the gas does exert a force. It exerts a pressure force to counter the gravity force of the piston. $F_{pg} = - F_{press}$ $\endgroup$
    – Sam Spade
    Mar 21, 2017 at 1:48

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