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A process in a closed system is reversible if the entropy change is $dS = \frac{dQ}{T}$.

A process is quasistatic if a process is infinitely slowly.

Now, if a process is reversible, this means that we are always in equilibrium and this can only be the case, if we do this process very slowly. Thus, any reversible process is quasi-static.

Although I think I understand the basic definition, I don't see why we need the concept of a quasistatic process? Which equations or concepts in thermodynamics actually rely on this kind of process?

Wikipedia lists: isobaric, isochoric, isothermal processes. So is any such process necessarily quasistatic? What about adiabatic processes?

So far I did not get why it is important in thermodynamics to look at such processes and which equations hold only for such processes?

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In classical thermodynamics the equations are valid only for thermodynamic equilibrium. That means that your system must have his variables well defined all the time. For instance, if you have a gas the temperature must be well defined and the same across the gas. In order for this to be valid when you have a process, the process must be slow enough so that different parts of the gas keep the same temperature. That is why any reversible process is necessarily a quasistatic one (but, some quasistatic processes are irreversible).

In short, when you assume a reversible process you are assuming a quasistatic one, thus each time you use the thermodynamic equations for whatever reversible process you are implicitly assuming "quasistaticity". In particular the ones you asked, such as isobaric, isochoric, isothermal processes (if they are reversible).

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  • $\begingroup$ Well, I still don't get what this concept actually tells me then. Let's consider an isothermal process that is not quasistatic: You have a box with a gas in it and then you open this box. This one is certainly not quasistatic. But we can still apply the equations for isothermal processes and classical thermodynamics or is there now a problem with this( I would think so, as there is certainly no thermodynamic equilibrium when we open the box) ? Is quasistaticity equivalent to being in the thermodynamic equilibrium? $\endgroup$ – Tokoyo Jan 1 '15 at 18:24
  • $\begingroup$ The isothermal process you describe is irreversible. See en.wikipedia.org/wiki/Free_expansion. In this case you cannot dS=dQ/T to find the change in entropy. However, you can imagine a quasistaic reversible process that brings the system from the initial state to the final state, and compute the change in entropy this way. $\endgroup$ – Wolphram jonny Jan 1 '15 at 18:30
  • $\begingroup$ aha, so it is really that thermodynamics only holds in thermal equilibrium which can only be achieved by quasistatic processes? otherwise, we have to consider a process that substitutes our non quasistatic process? $\endgroup$ – Tokoyo Jan 1 '15 at 18:33
  • $\begingroup$ exactly!.....,, $\endgroup$ – Wolphram jonny Jan 1 '15 at 18:36
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Although I think I understand the basic definition, I don't see why we need the concept of a quasistatic process? Which equations or concepts in thermodynamics actually rely on this kind of process?

'Quasistatic process' means that the system studied (like gas in a cylinder) undergoes changes so slow that at any time, thermodynamic variables like pressure, temperature, internal energy and entropy have definite value and provide meaningful description of the system. Consequently, equations like

$$ dU = TdS - pdV $$

and

$$ dS=dQ/T $$

are valid for such processes. That's why the concept of quasistatic process is useful.

The fact that the process is quasistatic and the above equations are valid do not guarrantee that the process is thermodynamically reversible. Whether it is depends not merely on what happens to the system, but also what happends to other systems that it interacts with.

For example, consider the system studied is at temperature $T_1$ and is giving off heat to another, colder system at $T_2$ through poorly conducting heat bridge.

-------------------            heat ---->             -------------------
| system 1 at T_1 |===================================| system 2 at T_2 |
-------------------       poorly conducting rod       -------------------

The process of heat exchange will be quasistatic, the above equation for entropy change will apply to both systems individually:

$$ dS_1 = dQ_1/T_1 $$ $$ dS_2 = dQ_2/T_2 $$ while $dQ_2 > 0$ and $dQ_1=-dQ_2$.

But the process is fundamentally irreversible, as heat is flowing from warmer to colder body and entropy in the end will increase.

So quasistaticity only guarrantees applicability of thermodynamic quantities and relations between them; it does not by itself guarrantee the process is reversible.

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Quasi/static state process - is characterized by infinite slow change of parameter. it can also be termed sd a succession of equilibrium state. example: bad smell originating from dust bin within the house and the smell moves slowly such that it is felt uniformly everywhere within the house.

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  • $\begingroup$ WRONG!!!!!!!!!!!!!!!!!!!!!!!!!! $\endgroup$ – Tokoyo Feb 22 at 18:08

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