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Greeting everyone,
The formula for pressure-volume work I have been given in the textbook I am using to learn general chemistry (Chemistry.A.Molecular.Approach.Global.Edition.4th.Edition, Nivaldo J.Tro) is as follows:

w = -PΔV where P is the external pressure. Why is the external pressure used if the force is from the expanding gas, not from the external atmosphere?

N.B: The system is a gas trapped in cylinder between a dead end and a piston. We assume that the piston is massless. We assume that the pressure is constant.

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  • $\begingroup$ Think about if the external pressure was $0$. What would the gas be doing work on? $\endgroup$ – Aaron Stevens Mar 30 at 13:44
  • $\begingroup$ @AaronStevens The piston. $\endgroup$ – Stooniel Schiffer Mar 30 at 14:00
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    $\begingroup$ @StoonielSchiffer But you said the piston is massless. $\endgroup$ – Bob D Mar 30 at 14:15
  • $\begingroup$ Yes, indeed. My mistake. Maybe I should learn about gases before diving further into the problem. $\endgroup$ – Stooniel Schiffer Mar 30 at 14:24
  • $\begingroup$ @AaronStevens Unfortunately the possible duplicate you are indicating contains part of the answer but something is missing. But the way SE works makes not very appealing to add a new answer there. $\endgroup$ – GiorgioP Mar 30 at 15:58
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For a massless piston, the force per unit area exerted by the gas on the piston is always equal to the external pressure.

If the expansion is reversible, the force per unit area exerted by the gas on the piston can be calculated from the ideal gas law (or other real-gas equation of state). This is only because, in a reversible process, the gas is close to being in thermodynamic equilibrium.

But the ideal gas law only applies to a gas that is in equilibrium (or nearly so). But, if the expansion is irreversible (i.e., very rapid), the ideal gas law does not correctly predict the force per unit area exerted by the gas on the piston because the gas is not nearly at equilibrium; in particular, viscous friction contributes to the force (which is not present in a reversible process). So, the only alternative is to use the external pressure (if it is known or specified).

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    $\begingroup$ Looks like you answered this in the possible duplicate as well haha $\endgroup$ – Aaron Stevens Mar 30 at 14:23
  • $\begingroup$ Ok. But if the force per unit is always equal, why does a mouvement occur? I mean, the force from the external enviromnent is then always equal to the force exerted by the gas on the piston, in terms of magnitude, but opposite in terms of sign. Which means that the forces cancel each other. (I know I am wrong but I cannot find where.) $\endgroup$ – Stooniel Schiffer Mar 30 at 14:33
  • $\begingroup$ @ChetMiller you are neglecting the role of piston-container friction, which is a second, independent problem in this story. $\endgroup$ – GiorgioP Mar 30 at 16:04
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    $\begingroup$ @GiorgioP The OP will never be able to understand this with piston friction included if he first of all doesn't even understand it without piston friction included. For a thread with addresses piston friction, see the following in Physics Forums: physicsforums.com/threads/… $\endgroup$ – Chet Miller Mar 30 at 16:12
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    $\begingroup$ @ChetMiller thanks for the link, at a first look (but it will take some time to read the whole thread) looks as a comprehensive discussion of the issue. Including the important issue of stating explicitly which is the system of interest. Comgratulations for that post. $\endgroup$ – GiorgioP Mar 30 at 16:28

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