Intuitive understanding of work done in quasistatic and non-quasistatic process Work by definition is given by,
$W=\vec{F}.\vec{dx}$.
If $\theta=0$, we can obtain:
$dW=PdV$
So, according to me, it should hold for all the processes be it quasistatic or non-quasistatic.
But I have read that $dW=PdV$ is only valid for quasistatic processes. Why isn't it valid for non-quasistatic processes? Can anyone please, give any intuitive explanation? And what's the expression for $dW$ in non-quasistatic process?
 A: The behavior of a gas experiencing a non-quasistatic process is very different from that of the same gas experiencing a quasi static process.  In a non-quasistatic process, the gas deformation is so rapid that the inertia of the gas is important; this causes the pressure and the temperature to be non-uniform within the gas.  In addition, there are viscous stresses present in the gas which are proportional to the rate at which the gas is deforming, and also affect the pressure distribution.  As a result of all this, the gas pressure at the boundary (e.g., at a piston face) where the work is being done is not equal to the average gas pressure within the cylinder, and an equation of state like the ideal gas law can not be used to determine the pressure at the work boundary.  Even under these extreme circumstances, the gas pressure at the piston face is still (by Newton's 3rd law) equal to the force per unit area exerted by the piston on the gas $P_{ext}$.  We just can't use the ideal gas law to calculate this pressure.  In the case of a non-quasistatic process, all we can do to determine the amount of work done is to externally impose the force exerted on the gas at the piston face.
A: In a quasistatic process, the difference between the external pressure and the internal pressure is infinitesimal for each infinitesimal change in the volume of the system.
$P_{ext}=P_{int} \pm dP $,
As you know that the term for work in terms of pressure is given by, $dW=PdV$. This means that work obtained will be maximum if the pressure is maximum for each infinitesimal change in volume. 
In a reversible process, the work done in each step obtained is maximum since the external pressure in only infinitesimally greater (or smaller) than the internal pressure.
This enables us to connect the internal pressure and external pressure using the ideal gas law, which in turn enables us to derive work in terms of volume change, without knowing the pressure.
For non-reversible process, such thing is not possible. This process is instantaneous. Internal pressure won't have enough time to become almost equal to external pressure. Only, external pressure is a way to find the work.
For non-reversible process,
$W=P_{ext}(V_f-V_i)$
