External pressure vs internal pressure If I have understood this correctly; when we calculate the work done by a system or on the system we can use external pressure. This is because this pressure will stay constant compared to the internal pressure that will change during the process - so the calculation will be easier. Is this correct? 
What will the difference between the (V,p) and (V,Pex) diagrams be? I guess that the difference is that the (V,P) diagram will have an curved graph while the (V,Pex) will have a straight graph that is horizontal. Is this correct? 
 A: If the external pressure is constant, you will get a straight graph parallel to the V axis in the p-V-diagram. Yes, the external pressure is constant if it is related to atmospheric pressure.
The internal pressure will give you a curved graph in the p-V diagram. The area between this curved graph and the other graph in p-V diagram will give you the work that is done to the system.
A: Consider the case of a block moving on a frictionless surface with a force $\mathbf F$ acting on it for a displacement $\mathbf d$ then work done by the external force is $W = \mathbf F \cdot \mathbf d$. Now you may ask why didn't we considered the internal force for this? The reason being that they both represent the same exchange of energy i.e., from the pusher to the block. 

Now consider the case of a force pushing a piston which itself is pushing a gas. Here you may see that for every instance (other than the equilibrium state) $P_{ext} = P_{int} + \Delta P$ i.e., they don't form a action reaction pair and unless the piston is at rest at start and end this $\Delta P$ dissipates some of the this energy. So to find out the work done on the system by the external force we consider $\int P_{ext} \cdot dV$ and not $\int P_{int} \cdot dV$.
