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To add to Whelp's Answer. Even though the $$\bar{d}Q=T\,dS$$ does not hold in an irreversible process, it still gives us something general. Consider a thermodynamic system linked to a system of reservoirs and which can only exchange heat and nothing else with the reservoirs. Let's call the thermodynamic system an engine for our convenience. Then the engine ...


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This relation is not true for general processes. For a closed system, the general relation is $\delta Q \leq TdS$, as is illustrated by the Clausius Theorem (http://en.wikipedia.org/wiki/Clausius_theorem). Another way of writing it is $dS=\delta Q/T + dS_{irr}$ where $ dS_{irr}$ is the entropy change due to irreversibilitiy in the closed system ...


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Reversibility implies that the entropy change of the universe is 0. An isentropic process need not necessarily be reversible, provided that the entropy of the surroundings is increased. For an irreversible process, heat can be removed from the system in order to make it isentropic (since $dQ < TdS$. As heat is removed, you will have an isentropic ...


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Entropy and irreversibility are related to each other. Irreversibility is the one because of which process takes place and due to this energy degradation takes place that is exergy (availability) decreases. And basically irreversibility(I) is defined as I = To * ( del s ) That is when you multiply change in entropy of universe with the reference ...


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The difference is that one expansion is quasi-static (the reversible one) while the other is spontaneous because of a dramatic change of the external constraints (the irreversible one). In the quasi-static case, you start off indeed in the state where gas pressure equates external pressure. An external operator then slightly decreases the outside pressure ...



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