pV work in reversible processes. Why is $p_{ext}=p_{internal}$ in a contraction? My textbook uses newtons third law to say to prove this. I think that's completely wrong. What I think happens is that we are doing an approximation that $p=p+dP$.
In case you haven't figured it out, my question is why do we say that in $-\int pdV$ during a contraction $p$ happens to be the pressure of the gas?
 A: It is certainly true if the boundary is moved slowly enough, that is, if the process is reversible. I'm not sure how you would use Newton's third law, but you can use the second: if the boundary is moved very slowly, then its velocity and acceleration are very close to zero, which means that the net force (and hence the pressure difference) must be very close to zero.
There is a more thermodynamic way to think about this, and it involves the very meaning of the word "reversible". If there were a finite pressure differential, then the boundary would move by itself in one direction but not in the other, which goes against the reversible nature of the movement.
Of course, this is an idealization, because truly reversible processes don't exist in nature. That is why we write $p_e = p_i + dp$. We only require the pressures to be very very close. In fact, if they were exactly equal, we couldn't get anything moving! A process in which things are moved very slowly (and in which the forces are almost zero) is called quasi-static, and this is a classic example.
