Exploring Work Done in Quasi-Static & Non Quasi-Static Expansion

In summary, a quasi-static process is one where the gas inside the system has to do a work to "extend". A non quasi-static process is where the gas inside the system doesn't move fast enough to "push" the "edge" to do a work, and what happen exactly is still up for debate. 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, the work done by the system is actually greater than zero.
  • #1
happyparticle
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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.
 
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  • #2
happyparticle said:
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.
 
  • #3
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.
 
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FAQ: Exploring Work Done in Quasi-Static & Non Quasi-Static Expansion

What is the difference between quasi-static and non-quasi-static expansion?

Quasi-static expansion refers to a process in which the system changes slowly enough that it can be considered in equilibrium at all times. Non-quasi-static expansion, on the other hand, occurs when the system changes rapidly and is not in equilibrium at all times.

How does work done in quasi-static expansion differ from work done in non-quasi-static expansion?

In quasi-static expansion, the work done is equal to the area under the curve on a pressure-volume graph. In non-quasi-static expansion, the work done is not equal to the area under the curve and must be calculated using other methods, such as integrating the product of pressure and volume over time.

What factors affect the amount of work done in quasi-static and non-quasi-static expansion?

The amount of work done in both types of expansion is affected by the initial and final volumes of the system, as well as the pressure applied. In quasi-static expansion, the rate at which the system changes also affects the amount of work done.

How does the first law of thermodynamics apply to quasi-static and non-quasi-static expansion?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted. In both types of expansion, the work done on or by the system is a form of energy transfer. However, in quasi-static expansion, the change in internal energy of the system is negligible, while in non-quasi-static expansion, the change in internal energy must be taken into account.

What are some real-world applications of quasi-static and non-quasi-static expansion?

Quasi-static and non-quasi-static expansion are commonly observed in various physical and chemical processes. Examples include the expansion of a gas in a piston, the compression of air in a car engine, and the expansion of a balloon when filled with air. These principles are also important in understanding the behavior of thermodynamic systems and in the design of efficient engines and machines.

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