I know that $\Delta S \geq \int_a^b \frac{\delta Q}{T}$ holds in general. When the system (with its heat sources which interacts only with it) is isolated, I read that, since $\delta Q=0$, then for the equation above $\Delta S \geq 0$. But the $\delta Q$ in the integral are the amounts of heat that are exchanged between, for example, a gas and each of the heat sources, and T is the temperature of each source which of course is different for each one of them. Then why in an isolated system each $\delta Q$ must be zero? Isn't zero the $Q$ exchanged between the system and the outside?
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The heat sources are not part of your system; they are outside. – Chet Miller Dec 31 '21 at 16:25
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@ChetMiller I could be wrong, but on re-reading the OPs statements, particularly the last two sentences, the OP may be considering heat transfers occurring within the isolated system. – Bob D Dec 31 '21 at 22:04
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@BobD. First of all, Happy New Year. I agree. It seems to me, the OP does not recognize how important it is to precisely define exactly what constitutes his system and what constitutes the surroundings of his system. He seems to be flip-flopping between a system the includes the heat sources and one in which the heat sources are part of the surroundings. – Chet Miller Dec 31 '21 at 22:36
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1@ChetMiller That was my impression too. Anyway, Happy New Year to you too! I'm a Grandpa and I believe you are to (re: Grandpa Chets Entropy Recipe), so we both have a lot to be thankful for. And I am thankful for all the guidance you have given me. – Bob D Dec 31 '21 at 23:22
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Related: If some physical change occurs due to an increase in temperature, is the resulting entropy thermal or configurational? – Roger V. Apr 25 '22 at 07:21
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Then why in an isolated system each must be zero? Isn't zero the exchanged between the system and the outside?
By definition, an isolated system is one in which there is no mass transfer and no transfer of energy in the form of work or heat between the system and its surroundings. So, by definition, for an isolated system $\delta Q=0$
The fact that $\delta Q=$ does not mean there can be no entropy change in an isolated system. Although the entropy change is defined for a reversible transfer of heat you can devise any convenient reversible path between two equilibrium states and calculate the change in entropy of the system using the equation
$$\Delta S_{sys}=\int\frac{\delta Q_{rev}}{T}$$
because entropy is a state function independent of the path.
The classic example is that of an isolated system in which an ideal gas expands irreversibly into a vacuum. See the explanation in my answer here:Entropy change in the free expansion of a gas
Hope this helps.
Bob D
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Thanks for answering and happy New Year. I understand that in an isolated system of course the $\delta Q$ with the outside is zero. But how can we derive that $\Delta S \geq 0$ for an isolated system from Clausius integral? I'm asking this because if we have an isolated system (here the system includes the heat sources which exchange heat only with this particular system), then the $\frac{\delta Q}{T}$ refer to the heat sources of the system, they are not zero and they are not equal so we cannot cancel them out in the sum. – abc Jan 01 '22 at 11:22
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It appears all you are doing is including the thermal reservoirs, which are normally in the surroundings, in the system and then calling the system isolated. It's the same thing as calling the universe isolated. So now the system consists of two subsystems, the heat engine (gas) and thermal reservoirs. Then the Clausius inequality is applied to the heat transferred from one part of the system to the other. I really don't see any difference between the two scenarios. The total entropy change is still zero for a reversible process/cycle and greater than zero for an irreversible process/cycle. – Bob D Jan 01 '22 at 14:50
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So we can say that in an isolated system (gas+reservoir) Clausius integral is zero because the $T$ in $\delta Q/T$ for the reservoir is the temperature of the reservoir, and not the temperature of the gas. Instead for the gas, the $T$ in $\delta Q/T$ is the temperature of the reservoir. So the integral is zero basically because we can calculate variation of entropy of a reservoir using its temperature, while for gases we need to use the temperature of the reservoir, right? – abc Jan 02 '22 at 09:34
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I mean that we can't treat a reservoir like the gas. In the entropy for the reservoir we need to consider the $T$ of the reservoir, instead of the $T$ of the gas, while for the gas we cannot take the $T$ of the gas (like we similarly did for the reservoir) but we have to take the $T$ of the reservoir. And I think that this is the only reason for the Clausius integral to be zero for an isolated system (where system is always gas+reservoir). – abc Jan 02 '22 at 10:45
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The T of the reservoir is the same as the T of the gas when the gas is in thermal equilibrium with the reservoir, i.e., when the heat transfers are reversible, in which case the Clausius integral will be zero. It's when the gas is not in thermal equilibrium with the reservoir so that the T of the gas is only the same as the T of the reservoir at the boundary between the gas and the reservoir, which makes the heat transfers irreversible and the Clausius integral non zero. – Bob D Jan 08 '22 at 19:17