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I've read this in a book

During irreversible compression, large quantity of heat is produced due to friction, while the decrease in volume is less. Thus, increase in entropy due to rise in temperature exceeds the decrease in entropy due to decrease in volume.

I don't get this at all. Is the friction they're talking about the one between pistion and container? What if we consider an ideal condition with no dissipation?

I also want to know the physical interpretation of increase in entropy in adiabatic (irreversible) process. Shouldn't it be zero? This book tried to explain it but it was confusing and I'd love to get further insight on this.

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  • $\begingroup$ Difficult question because you can't really use thermodynamic equations to explain non equilibrium processes $\endgroup$
    – 666User666
    Oct 1 '20 at 12:45
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I agree that the paragraph is highly confusing. For one thing speaking of "heat" in the context of an adiabatic process requires an explanation as to where the heat is going. Clearly for an ideal gas in an adiabatic container the final temperature of the gas depends only on the total work done on gas, and irreversibility must be tied somehow to a difference in the applied external pressure and the internal pressure of the gas. For a reversible process the two would match each other during the compression. In the irreversible case there is a mismatch and some of the work done by the surroundings goes to waste (the applied pressure exceeds the internal pressure), so in the end less work is done (on the gas) than in the reversible case.

The proof that an irreversible adiabatic process has an associated nonzero entropy change starts from the observation that the terminal states for reversible and irreversible adiabatic processes are not the same (in this case for instance the volume is larger in the irreversible compression). To compute the entropy change for the irreversible adiabatic compression you need to identify a reversible process (of which there are many) between the same terminal states as the irreversible process. Evaluating the reversible process that will take the system between the end points of the irreversible process you find that an entropy change is required, even though no heat is exchanged in the actual irreversible adiabatic process and despite reversible adiabatic processes not generating a change in entropy.

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    $\begingroup$ The average internal pressure may exceed the external pressure (the pressure exerted by the inside piston face), but the internal gas pressure at the piston face must match the pressure exerted by the inside piston face, because the two constitute an action-reaction pair. $\endgroup$ Sep 30 '20 at 19:38
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The friction they are talking about is viscous friction of the gas itself. This comes into play when the gas is deformed rapidly, and is roughly proportional to the square of the rate of gas deformation.

Regarding entropy, the change is entropy is equal to the integral of dq/T only for a reversible process. In an irreversible process, in addition to this exchange of entropy with the surroundings, there is also entropy generated within the system itself.

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  • $\begingroup$ Alright. What if I consider a reversible adiabatic process with same initial and final state variables as the irreversible adiabatic process into consideration. Since Entropy is a state function change in entropy will be same for both the processes. The heat flow in this reversible adiabatic process is undeniably zero, then so is the entropy change. Being a state function, shouldn't it be zero for the irreversible adiabatic process, at least on paper? $\endgroup$
    – Eyy boss
    Sep 30 '20 at 17:36
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    $\begingroup$ It is impossible to define a reversible process between those same two states that is adiabatic. $\endgroup$ Sep 30 '20 at 18:13
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    $\begingroup$ If you want to get a better understanding of all of this in detail, see Bird, Stewart, and Lightfoot, Transport Phenomena, Chapter 11, Problem 11D.1, Equation of Entropy Change. There, they provide explicit expressions for the local rate of entropy generation during an irreversible process. $\endgroup$ Sep 30 '20 at 19:44
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    $\begingroup$ A point I would like to add is that you can't devise a reversible path for a irreversible process for a system with same surroundings which irreversible process occured. You would need to find to put the system in different surroundings for such processes. $\endgroup$
    – 666User666
    Oct 1 '20 at 12:44
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    $\begingroup$ That, of course, is correct. $\endgroup$ Oct 3 '20 at 22:35
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Here is the reasoning behind why friction produces heat.

The frictional forces arise from the interactions with the gaseous molecules and the surface of the container along with some resistance to motion caused from the interactions of the gas molecules. However, the reasons for which they arise are a little different (i.e. surfaces cause friction cause they are imperfect/not smooth; friction in liquids and gases arise from the interactions of the individual atoms).

Entropy is the amount of micro-states available for the molecules. So, even though the compression decreases the amount of micro-states (negative entropy), the heat introduced from friction causes a positive change in entropy, which has a greater magnitude than the decrease caused by compression. Therefore, the gas molecules have more available micro-states.

Entropy in reversible adiabatic processes is always zero. For irreversible adiabatic expansion, this is not the case as the gas does work and expands, causing the amount of micro-states available to increase. It is irreversible as the gas loses the ability to do work when the process is over and the gas won't spontaneously return to its compressed state.

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