# Difference between work done during quasi-static and non quasi-static expansion

happyparticle
I'm wondering what's the difference between work done on quasi-static and non quasi-static expansion.
In a quasi-static process, the gas inside the system must do a work to "extend".
However, in a non quasi-static process, where the gas inside the system doesn't move fast enough to "push" the "edge" to do a work, what happen exactly?
Is the work on the system in a quasi-static process lower than in a non quasi-static process, since there's no work done by the system, but only to the system?
Another way I see it is since ##w = p\delta v## for a similar ##\delta v## p must be greater for a greater work, but I'm not convinced.

Gold Member
However, in a non quasi-static process, where the gas inside the system doesn't move fast enough to "push" the "edge" to do a work, what happen exactly?
Is the work on the system in a quasi-static process lower than in a non quasi-static process, since there's no work done by the system, but only to the system?
This would be explained by the second law of thermodynamics:
https://en.wikipedia.org/wiki/Second_law_of_thermodynamics#Introduction said:
For an actually possible infinitesimal process without exchange of mass with the surroundings, the second law requires that the increment in system entropy fulfills the inequality
$$\displaystyle \mathrm {d} S>{\frac {\delta Q}{T_{\text{surr}}}}$$
This is because a general process for this case (no mass exchange between the system and its surroundings) may include work being done on the system by its surroundings, which can have frictional or viscous effects inside the system, because a chemical reaction may be in progress, or because heat transfer actually occurs only irreversibly, driven by a finite difference between the system temperature (T) and the temperature of the surroundings (Tsurr).
Combining this with the concept of enthalpy:
$$\partial W = \partial Q - dU$$
$$\partial W = TdS - dU$$
Now imagine a reversible process where a heat transfer is made to only change the internal energy, thus ##\partial Q = dU## or ##\partial W = 0##.

But with an irreversible process, we would measure that ##T_{surr}dS## is greater than the expected, ideal, ##\partial Q## and thus ##T_{surr}dS > dU## and the actual work ##\partial W## is also greater than zero. To get the ##dU## desired, you will have to put in some extra work to take into account, for example, the frictional or viscous effects.

But @Chestermiller understands this better than me and would probably explain it better than me.

Mentor
The equation of state for a gas, such as the ideal gas law or the Van Der Waals equation is valid only for thermodynamic equilibrium or for a reversible (quasi static) process, which is comprised of a continuous sequence of thermodynamic equilibrium states. The reason for this is that in an irreversible (non-quasistatic) process, there are viscous stresses present in the rapidly deforming gas which affect the forces exerted by the gas. For an irreversible (non-quasistatic) process, we can only calculate the work done by the gas by imposing (i.e., controlling) the forces acting on the gas (and from Newton's 3rd law)the forces imposed by the gas on its surroundings) manually. This can be done by imposing an external medium at constant pressure or by embedding a force transducer into the piston face, and using a feedback system.

• vanhees71