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In an adiabatic irreversible compression process, suppose the pressure $P_{ext}$ compresses the piston and does some work say 50J. (50J of energy is lost by surrounding). Since the pressure $P_{int}$ is less than the $P_{ext}$, the work done by system on surrounding will be -20J(suppose). Negative sign shows that 20J of energy is lost my surrounding. But initially we saw that surrounding lost 50J! How to tackle this discrepancy? I am very confused.

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  • $\begingroup$ The internal force per unit area that the gas exerts on the piston is not less than the external pressure (say, for a massless, frictionless piston). By Newton's 3rd law, the force the gas exerts on the inside piston face is equal and opposite to the force that the inside piston face exerts on the gas. So it is actually incorrect to say that the internal pressure (actually, more precisely, internal compressional stress) is less than the external pressure. $\endgroup$ Aug 12, 2021 at 15:36
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    $\begingroup$ What about a practical process in which a gas beaker compresses suddenly due to the excessive external pressure? And according to the 3 law of Newton, force on gas due to piston= force on piston due to gas. We can not this: force on piston due to gas= force on piston due to surrounding. Because it will leas to a static equilibrium which is not the case I mentioned above $\endgroup$ Aug 12, 2021 at 15:52
  • $\begingroup$ Could you verify this statement for me, is it always that work done by gas on environment= work done by gas on environment (due to pressures being equal across boundary)? @ChetMiller $\endgroup$ Aug 12, 2021 at 15:56
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    $\begingroup$ @Buraian In the case of an irreversible deformation, calling the force per unit area exerted by the gas on the inside face of the piston (i.e., its surroundings) the pressure is a bit imprecise, because the compressional stress in an irreversible deformation is not isotropic, and pressure is an entity which is isotropic. The viscous stresses are not isotropic, and add (or subtract) from what one would interpret as the pressure. If you are asking whether the work done by the compressional force per unit area of the gas on its environment is equal to the environment work on the gas, then yes. $\endgroup$ Aug 12, 2021 at 16:30
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    $\begingroup$ @Buraian You need to decide what your system is, and, in particular, where your system ends and the surroundings begin. Does your system include only the gas (and the piston is part of the surroundings), or does your system also include the piston, and the surroundings begins at the outside face of the piston? $\endgroup$ Aug 12, 2021 at 20:43

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The force that the gas exerts on the inside face of the piston must be equal in magnitude and opposite in direction to the force the inside face of the piston exerts on the gas. So the work the gas does on the piston must always be equal and magnitude and opposite in sign to the work the inside face of the piston (i.e., the surroundings) does on the gas.

The problem is that, for an irreversible compression or expansion, one cannot use the ideal gas law to calculate the force the gas exerts on the piston or the work done by the gas, because the ideal gas law is not valid for an irreversible compression or expansion. The ideal gas law only applies at thermodynamic equilibrium, which is not present in an irreversible compression or expansion. In the irreversible case, you will learn when you study fluid mechanics, that there are viscous stresses in the gas over and above that from the ideal gas behavior which affects the force per unit area at the piston face (as well as throughout the gas).

In the irreversible case, if we want to calculate the work, our only option is to use the external force per unit area exerted by the surroundings on the gas at the inside face of the piston (if this is known or specified).

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