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The potential for a thermodynamic process to be reversed in time. Alternatively, a quantification of how far an irreversible process is from being reversible, which relies on a comparison to a corresponding theoretical reversible process.
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Is thermodynamic reversibility a function of path?
With reversibility the external control parameters are the ones that also describe the state of the system under control at any given instant, and that the relationship among the various intensive and … Reversibility and quasi-static are not the same. There are systems that in the words of Bridgman are completely surrounded by irreversibility independently of the speed of the process. …
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Reversible process in General Physics
Any inelastic deformations and ferromagnetic cycles are always irreversible processes.
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Are work and heat state variables (or functions) for reversible processes?
Callen is talking about a reversible process of a system connected to a single thermal reservoir and a single work reservoir. Since there is only one temperature at which the reversible thermal exchan …
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Issue with the work of a reversible process
In general, the 1st law would say that $dU=\delta Q +\delta L$ but you have postulated a purely mechanical energy exchange, that is one with $dU=\delta L$. This is called an adiabatic process.
If yo …
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Hysteresis and dissipation
Bridgman in "The Thermodynamics of Plastic Deformation and
Generalized Entropy", REVIEWS OF MODERN PHYSICS VOLUME 22. NUMBER 1 JANUARY, 1950, is discussing specifically stress-strain hysteretic cycles …
0
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How do I prove reversible transformations are necessarily quasistatic without using entropy?
The simplest way to formulate reversibility mathematically is having all relationships among the thermodynamic variables be piecewise differentiable functions, no rates, no gradient, "no nothing", only …
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How can the Entropy (change in entropy) of reversible and irreversible process be the same? ...
The issue is not that when comparing two processes, say $\mathfrak P_r$ and $\mathfrak P_i$, both starting at the equilibrium state $\mathfrak S_0$ and both ending at the another equilibrium sate $\ma …
3
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Defining reversibility without resorting to entropy
Reversible process means that given the outside controllable mechanical, electrical, magnetic, chemical, etc., macroscopic parameters $\hat x_1,\hat x_1,\hat x_2,...,\hat x_n$ of the surroundings and …
10
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Example of an irreversible process using this formal definition
The most obvious example is magnetic hysteresis. The current value of magnetization $M=M(H_{ext})$ does not just depend on the current bias field but it also depends on the history of how the sample h …
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How can an irreversible isothermal process exist?
Whether you can or cannot specify a single temperature is not related to being reversible or not. There are inherently irreversible processes irrespective of its velocity, for example, the hysteretic …
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How to know when a reversible process between end states exists?
The answer to this question is negative in the sense that not all systems allow for reversible processes at all, not even in principle. The most famous examples maybe hysteretic ferromagnetism and hys …
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Increase of entropy when two reservoirs are in thermal contact
Denote the transported entropy by $S_A$ from body $A$ at temperature $T_A$. The same entropy falls from temperature $T_A$ to temperature $T_B$ and the thermal work being dissipated is exactly $\Delta …
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Entropy maximum postulate and reversibility
I think this is what Callen has in mind: he takes one finite body isolated from everything else and assumes that it is separated by internal constraints so that the several homogeneous parts that main …
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Use of Clausius theorem to prove entropy inequality in Fermi's Thermodynamics
All mysteries go away if you realize that when you say $dQ$ (or $\delta Q$) heat is supplied from the reservoir whose temperature is $T_e$ then along with said heat the reservoir supplies $dS_e=\frac{ …
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How does Fermi jump to this conclusion in Clausius inequality?
When you reverse a reversible process then $T>0$ still holds but $\delta Q$ becomes $-\delta Q$ and the Clausius inequality must hold for these heat exchanges:
$\int \frac{-\delta Q}{T} \le 0$.
Compar …