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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?

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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.

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    $\begingroup$ Beautiful answer.. $\endgroup$
    – Mitchell
    Commented Sep 10, 2017 at 9:44
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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,

$W=P_{ext}(V_f-V_i)$

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  • $\begingroup$ External pressure doesn't need to be constant during the whole process. So, shouldn't be the total work an integral of PdV (where P is external pressure)? I am talking about the very last line you wrote. $\endgroup$
    – Kawsar Ahmed
    Commented Jul 17, 2017 at 16:09
  • $\begingroup$ External pressure need not be constant for the whole process. It depends on the process. Reversible processes are not real, where non-reversible processes are real. Putting a weight on the piston will make the external pressure constant and the process will reach equilibrium when the internal pressure equals the external. For non-reversible processes the P-V digram gives us a straight line parallel to the volume axis. For reversible processes, we know the relation between P and V (Ideal gas equation), which gives us the area under the curve and V axis. The rest is mathematics. $\endgroup$
    – Mitchell
    Commented Jul 17, 2017 at 16:22
  • $\begingroup$ Integration is just like adding. The external pressure changes according to the change in volume of system so that the difference between the internal and external pressure is infinitesimal. Its just like putting a lot of light weight pebbles on the piston and removing them one by one. Such is the case of reversible process. Due to existence of the ideal gas law, we can perform the integration and get the desired formula for work. $\endgroup$
    – Mitchell
    Commented Jul 17, 2017 at 16:22
  • $\begingroup$ Why should external pressure be constant during non-quasistatic process? $\endgroup$
    – Kawsar Ahmed
    Commented Jul 17, 2017 at 17:35
  • $\begingroup$ It doesnt have to be constant for the entire process. You can break the whole process into parts where the pressure in constant and then separately calculate and add the work done in these parts to get the net work. Check this out : web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node5.html $\endgroup$
    – Mitchell
    Commented Jul 17, 2017 at 17:40

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