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Case -1

In a free expansion of an ideal gas in a region or container, we cannot trace a pressure-volume path since the pressure, temperature and volume of the gas fluctuate unpredictably when the gas fills the region (container).

But since entropy is a state property, the difference in entropy must depend only on initial and final state. So we can replace the irreversible free expansion with a reversible process with same initial and final condition.

Consider an insulated container along with a thermal reservoir attached to it whose temperature can be varied by a knob. It's current temperature is T. A piston is also attached to it. Above the piston is kept a small container containing iron balls (to balance the pressure due to atoms below it). The initial state is the same as the initial state of a free expansion.

Removing the iron balls one by one, we increase the volume with an infinitesimally amount and also by keeping the temperature constant. Since it follows an isothermal path, the heat absorbed by gas from reservoir is equal to the work done by the system (positive).

W=Q

After the final condition by the reversible isothermal process is achieved,

Change in entropy = Q/T And so, Entropy increases

To sum it up:- we can write :-

To find the entropy change for an reversible process occuring in a closed system, we replace that process with any reversible process that connects the same initial and final states.

Case - 2

Since the process is reversible we can proceed in a reversible manner, i.e we can add more and more iron balls to decrease the volume and consequently the energy must be extracted from the gas and be given to the thermal reservoir, i.e the heat Q is negative. Thus entropy must decrease.

But this does not happen since the thermal reservoir's entropy increases and the system consisting of the gas and reservoir, the total entropy change is 0.

Note - By taking the gas as the system, we do not meet the requirement of the entropy's definition of having a Closed System.

To sum up this :-

If a process occurs in a closed system, the entropy of the system increases for irreversible processes and remains constant for reversible process. It never decreases.

I have explained here in short about what has been given in my textbook for entropy.

My Question - In case 2, we take the system to be the gas as well as the reservoir and so the total entropy change is 0 while in case 1, we take the system to be only the gas. If I take the system, in case 1, to be gas and reservoir, the entropy should be zero. Since heat is being removed from the reservoir thus Heat Q in the entropy formula for reservoir should be negative. And thus the total change should be zero in case 1 as well.

Why doesn't this happen?

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closed as unclear what you're asking by Mithoron, Mathew Mahindaratne, Buck Thorn, Todd Minehardt, airhuff Aug 28 at 3:32

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Some people define a "closed system" as one that can exchange neither heat nor work not mass with the broader surroundings. Others, like us engineers, define a "closed system" as on that can exchange both heat and work with its surroundings, but not mass. We engineers call a system that can exchange neither heat nor work with the broader surroundings an "isolated system." Which definition of a closed system are you subscribing to?

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  • $\begingroup$ Isolated system $\endgroup$ – Asad Ahmad Aug 25 at 14:03
  • $\begingroup$ In case 1, for the irreversible path, the system is isolated. But, when calculating the change in entropy using an alternative reversible path, the system no longer needs to be isolated. The alternative reversible path shows that, as expected, for the actual irreversible path in the isolated system, the entropy increases. In case 2, you carry out a reversible path in an isolated system and surroundings, and, as expected, the combined entropy change is zero. $\endgroup$ – Chet Miller Aug 25 at 14:25

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