Work done while compressing an ideal gas (the physical significance of $\int \mathrm dp\,\mathrm dV$) Today in our chemistry class we derived the pressure-volume work done on an ideal gas. Our assumption was that $$p_\mathrm{ext}=p_\mathrm{int}+\mathrm dp$$ so that all the time the system remains (approximately) in equilibrium with the surrounding and the process occurs very slowly (it's a reversible process). Now
$$\begin {align}
W_\mathrm{ext}&=\int p_\mathrm{ext}\,\mathrm dV\\
\Rightarrow W_\mathrm{ext}&=\int (p_\mathrm{int}+\mathrm dp)\,\mathrm dV\\
W_\mathrm{ext}&=\int p_\mathrm{int}\,\mathrm dV
\end{align}$$
(Since $\mathrm dp\,\mathrm dV$ is very small $\Rightarrow \int \mathrm dp\,\mathrm dV =0$, 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 $W_\mathrm{ext}-W_\mathrm{int}=\int \mathrm dp\,\mathrm dV$. What does $ \int \mathrm dp\,\mathrm dV$ 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]

EDIT
I posted a similar on Maths SE to realize the mathematical significance of the term $\int \mathrm dp\,\mathrm dV$. 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 $\int \Delta p\,\mathrm dV$)
   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 incorporate the frictional force in it? 
 A: Let me try to convince you that $ \int dPdv$ is almost negligible. As you have said, $P_{ext} = P_{int} + dP$ but what $dP$ really is? Well I think it is better to assume $dP$ as very small number and hence just adding it to $P_{int}$ will give a value bigger than $P_{int}$ at any moment no matter whatever $P_{int}$ is. So, in this sense $dP$ is just acting as constant. Let's see what this angle of thinking about $dP$ can lead to $$ W_{ext} = \int_{V_i}^{V_f} (P_{int}+dP)dV$$
$$ W_{ext} = \int_{V_i}^{V_f} P_{int}dV + \int_{V_i}^{V_f}dPdV$$ Now, let's just focus on the $dP$ part $$ X= dP\int_{V_i}^{V_f}dV$$
as $dP$ is constant. $$X= dP (V_i - V_f)$$ 
We agreed that $dP$ is a very small number and hence if we multiply it with any other thing no matter what the result will be very very small and therefore $X$ will be a very small number. $$ W_{ext} = \int_{V_i}^{V_f} P_{int}dV + X$$ Now, we can neglect $X$ and hence write $$ W_{ext} = \int_{V_i}^{V_f} P_{int}dV = W_{int}$$. Your argument that $ \int dPdV$ is negligible is quite sloppy as the integral adds many many pieces of small things ($f(x)dx$ is a very small number as $dx$ is very very small but adding many many of them would produce a different result).
Even in mechanics, when calculate gravitational potential energy we take the working force to be just a little more than $mg$ and hence calculate the work done just plugging the work with $mg$, however, the actual force is more than that.
I said that your argument was sloppy because it’s a matter of hyperreal numbers that when and when we cannot consider something negligible, your argument is vey all right if we just accept the rules of differentials.
A: 
...$W_{ext}-W_{int}=\int dPdV$. 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".
A: Your exact question is what is the physical interpretation of 
 $$\int_{V_i}^{V_f}dPdV$$
I will try to explain this without using mathematics. 
Suppose you have a cylinder with piston at its one end which is free to move and the cylinder is filled with compressed gas at pressure $P_{int}$. 
Your task here is to keep the piston stationary. You will have to apply exactly same pressure at its other end to keep it stationary and hence maintaining the thermodynamic state of gas at its initial condition. 
$dP$, work done by you on the gas and work done by the gas on you are all zero in this case. This is the equilibrium state. 
However, if your task was to slowly push the piston inwards further compressing the gas, you will have to increase the pressure applied by you on piston. The piston's acceleration will be dictated by how much you increased external pressure. 
Let's assume external pressure is increased by an amount $\delta$. 
One assumption in your derivation is that the process occurs very slowly meaning the piston's acceleration is almost zero. 
Even if we assume that piston is not accelerating at all, we still need to increase the external pressure. 
Why? 
Because there is friction between piston and cylinder wall in real life scenario and additional pressure $\delta$ is used to overcome this friction. 
And, the work done by $\delta$ is $$\int_{V_i}^{V_f}\delta.dV$$
And $\delta$ is your $dP$ .
Therefore, $\int_{V_i}^{V_f}dPdV$ represents nothing energy lost to over any dissipative  force present in the system due to  irreversibility of the process. 
Since one of the assumption in your derivation is process is irreversible hence there is no friction and $\int_{V_i}^{V_f}\delta.dV$ is Zero. 
