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

AI Thread Summary
The discussion highlights the differences between quasi-static and non-quasi-static processes in thermodynamics, particularly regarding work done by and on the system. In a quasi-static process, the gas performs work as it expands, maintaining equilibrium, while in a non-quasi-static process, the gas does not move fast enough to exert work effectively, leading to additional complexities such as friction and viscous effects. The second law of thermodynamics plays a crucial role, indicating that entropy changes in non-quasi-static processes require more work to account for irreversible effects. The equations governing these processes, such as the ideal gas law, apply primarily to reversible conditions, emphasizing the challenges in calculating work during irreversible processes. Understanding these distinctions is essential for accurate thermodynamic analysis and applications.
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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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.
 
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