These notes I found online state that "work is entropy‐free, and no entropy is transferred with work." I take this to mean that entropy is not generated in a work process. Why is this? Why is no entropy created when work is done on a system?
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2$\begingroup$ When work is done on the system comprising my office, entropy increases rapidly and dramatically. :) $\endgroup$– Ross PresserOct 7, 2021 at 20:59
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2$\begingroup$ @RossPresser It is all reversible, until published. $\endgroup$– Roger V.Oct 8, 2021 at 13:35
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4 Answers
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Entropy is generated in work processes involving viscous dissipation of mechanical energy such as rapid expansion or compression, stirring, etc.
answered Oct 7, 2021 at 12:29
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1$\begingroup$ In its current form, this answer is more of a comment than an answer. $\endgroup$ Oct 8, 2021 at 0:05
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$\begingroup$ Thanks for the answer. Can I simply interpret this to mean that a quasi-static expansion/compression without dissipation (ex. friction) generates no entropy? $\endgroup$– bakerOct 8, 2021 at 5:08
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$\begingroup$ yes, with no sliding friction and without viscous friction, there is no entropy generation in quasi-static expansion/compression, provided the heat transfer is also quasi-static (negligible heat conduction within the system). $\endgroup$ Oct 8, 2021 at 12:48
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Because the entropy in this notes is defined in terms of the heat transfer: $$ dS=\left(\frac{\delta Q}{T}\right)_{int, rev}, $$ which is the increase of the internal energy not associated with work.
Remarks
- As pointed by @ChetMiller in their answer and in the comments, the statement applies to the entropy transfer, but not to the entropy generation, which can result from work in an irreversible process (note that the entropy above is defined for a reversible process).
- There are different ways to define entropy, notably the axiomatic definition in thermodynamics (the one above), and the statistical physics definition as the logarithm of the number of microstates (Boltzmann entropy). See, e.g., this answer.
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$\begingroup$ This applies only to a reversible path. For an irreversible path involving non-quasistatic work, entropy is generated. $\endgroup$ Oct 7, 2021 at 12:30
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$\begingroup$ @ChetMiller Yes, indeed. But here is a statement made in specific context, linked in the OP. $\endgroup$– Roger V.Oct 7, 2021 at 12:32
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$\begingroup$ The statement in the notes is very confusing and seems to apply only to entropy transfer, not entropy generation. Of course, the only mechanism for entropy transfer is by heat flow. The authors of the text should have done a better job of explaining this. Otherwise, the rest of their text is very well presented and correct. $\endgroup$ Oct 7, 2021 at 12:42
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$\begingroup$ @ChetMiller Indeed. And the OP also mixes entropy transfer and entropy generation. $\endgroup$– Roger V.Oct 7, 2021 at 12:44
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1$\begingroup$ @baker If a process is reversible (quasistatic) than no entropy is generated: if it increased, we would not be able to reverse it. Energy is not the same as entropy: the energy transferred to the system in a reversible process can be recovered by reversing it. $\endgroup$– Roger V.Oct 7, 2021 at 15:36
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The notes should read “reversible work”, which is a useful idealization because it lets us ignore entropy generation. (All real processes generate entropy, often with a rate dependence that overly complicates introductory problems.)
Work doesn’t transfer entropy because it raises the energy of all particles in a system equally, with no dispersion. In this way, it can be contrasted with heating, which widens the energy spectrum (and does transfer entropy). Reversible work further doesn’t generate entropy because it’s performed at the limit of slowness, with no gradients (in force, pressure, voltage, magnetic field, chemical potential, or surface tension, for example).
answered Oct 7, 2021 at 16:27
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$\begingroup$ Thanks for the answer. I understand why a temperature gradient (i.e. a temperature difference between the surroundings and the system) would generate entropy. But why would something like a pressure gradient (i.e. a pressure difference between the surroundings and the system) generate entropy? $\endgroup$– bakerOct 8, 2021 at 7:25
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$\begingroup$ It’s the shifting of volume between the system and surroundings, driven by the pressure difference, that generates entropy; no entropy is generated by a static pressure difference alone. Broadly, the boundary movement accelerates particles that then randomly decelerate to produce heating. $\endgroup$ Sep 13, 2022 at 15:48
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$\begingroup$ @Chemomechanics If it raises the energy of all particles in the system equally, then for isothermal expansion of an ideal gas, entropy change should be zero, which is not the case. Can we attribute the change in entropy to the heat absorbed (gas cools a little bit when expands)? $\endgroup$ Sep 14, 2022 at 22:26
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$\begingroup$ Yes; because isothermal expansion of an ideal gas is isenergic (i.e., occurring at constant energy), the process requires heating to exactly offset the work done on the surroundings. That's why the entropy change isn't zero—heat transfer is entropy transfer. $\endgroup$ Sep 14, 2022 at 22:33
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From an atomistic and quantum perspective,
- the entropy of a system is defined (the "Gibbs definition") in terms of the probability $P_i$ that a randomly selected microsystem from that system is found in the $i$th of the discrete quantum stationary states available for a microsystem to be in, as $$S = -Nk_{\textrm{B}}\sum_{\textrm{All }i}\left(P_i\ln\left(P_i\right)\right)$$ where $N$ is the total number of microsystems in the system;
- heat is defined as a transfer of energy into or out of the system which proceeds by some of the microsystems making transitions between different quantum stationary states: it's possible for this to change (some of) the $P_i$ values, and therefore to change the entropy of the system;
- work is defined as a transfer of energy into or out of the system which proceeds by leaving the microsystems in the same quantum stationary states, but altering the potential field in such a way that the characteristic energy of each quantum stationary state changes: since the microsystems all stay in the same quantum stationary states, the $P_i$ values don't change, and neither does the entropy of the system.
This atomistic definition of entropy is equivalent to the macroscopic definition of entropy given by @RogerVadim if the probability distribution over the quantum stationary states is given by the Boltzmann distribution.
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1$\begingroup$ Very nice explanation. The third bullet becomes more clear if we just consider energy shift $+\epsilon$ for all the microstates (this is just a toy example). The probability becomes: $$P_i ' = \frac{e^{-\beta E_i '}}{Z'} = \frac{e^{-\beta (E_i + \epsilon)}}{e^{-\beta \epsilon}Z} = \frac{e^{-\beta E_i }}{Z} = P_i $$ $\endgroup$ Sep 13, 2022 at 11:25