thermodynamics - Work done while compressing an ideal gas (The physical significance of $int dPdV$)

Today in our chemistry class we derived the pressure-volume work done on an ideal gas. Our assumption was that

$$P_{ext}=P_{int}+dP$$ so that all the time the system remains (approximately) in equilibrium with the surrounding and the process occurs very slowly (its a reversible process). Now $$W_{ext}=\int P_{ext}dV$$ $$\Rightarrow W_{ext}=\int (P_{int}+dP) dV$$ $$W_{ext}=\int P_{int}dV$$ (Since $dPdV$ is very small $\Rightarrow \int dPdV =0$, though it is an approximation I guess?)

Now, the question is:

• In the case of (say) pushing a book the force on the book and that on the pusher form action reaction pair hence their work shows the same energy transfer but such isn't the case here and hence their work done does not represent the same energy transfer. So what does it represent? As in non-approximate case $W_{ext}-W_{int}=\int dPdV$. What does $\int dPdV$ mean physically ?

[Note that I ain't equalizing the case of book with that of gas but giving (a kind of analogy or something) with respect to which I want the answerer to compare/contrast the compressing situation]

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EDIT

I posted a similar on Maths SE to realize the mathematical significance of the term $\int dPdV$. I got this answer over there. Though it mostly satisfies what I wanted to know but states that

The last term ( _{I believe is referring to $\int \Delta PdV$} ) is then the energy “lost” e.g. by friction, that is, it is not reversible.

Now I'm wondering how does this external pressure term incorporates the frictional force in it?

Answer

...$W_{ext}-W_{int}=\int dPdV$. What does $\int dP dV$ mean physically?

Note that in the "non-approximate" case, we have assumed that $P_{ext}\neq P_{int}$. More precisely $P_{ext}-P_{int}=dP$. Now let's assume that the ideal gas is stored in a container with a movable piston(of a finite mass $m$, but ignore gravity) of area $A$ on top. For now, let's assume that there is no friction. So to do external work, you(or rather, surroundings) are applying a pressure $P_{ext}$(which corresponds to a force $F_1=P_{ext}A$) and the gas is doing internal work by applying a pressure $P_{int}$(which corresponds to a force $F_2=P_{int}A$).

Now let's analyze the forces on the piston. So piston has an upward force of $F_2$(applied by the gas) and a downward force $F_1$ applied by the surroundings. So in this case the net force in the downward direction is,

$$dF_{net}=m(da_{net})=F_1-F_2=P_{ext}A-P_{int}A=(P_{ext}-P_{int})A=dP×A$$

$$\therefore dK = Fds=dP(Ads)=dPdV$$

where $dK$ is the infinitesimal change in the kinetic energy of the piston, and $dV=Ads$ is the infinitesimal change in the volume.

There you have it. You see, there is an infinitesimally small(yet non-zero) net force on the piston which gives an infinitesimally small(yet non-zero) acceleration to the piston. And this infinitesimal acceleration increases the speed of the piston from $0$ to some infinitesimally small velocity. And thus the piston gains an infinitesimal amount of kinetic energy. And the $\int dPdV$ term accounts for this change in kinetic energy.

I know the last paragraph is heavily populated with "infinitesimals", but it is just to show you the insignificance of the motion of the piston. Now what if friction would have been present? In that case, the piston won't move in the first place. But if we also assume that the force due to friction is infinitesimally small, then yeah, the piston would move. But this time it would have a lower value of that infinitesimal acceleration. And, also, it will lose some of its kinetic energy in the form of heat(due to frictional losses).

Summary :- The $\int dPdV$ term accounts for the infinitesimal change in the kinetic energy of the piston.

I hope this is what you meant by "physical interpretation".