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I am confused about the application of the 2nd Law for reversible and irreversible processes and cycles.

I want to know how the Clausius principle, the Kelvin-Planck statement, and the Clausius Inequality are contained in this equation: $$\mathrm{d}S \geq \frac{\delta Q}{T}.$$

Also, is this for the entropy of the universe or only the entropy of a system? How are irreversibilities contained in these equations? What are the differences between these equations for processes and cycles? How do the above equations relate to this one : $$\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}}?$$

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  • $\begingroup$ This answer might be of help to you. $\endgroup$
    – march
    Commented May 29 at 19:08
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    $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Commented May 29 at 20:20
  • $\begingroup$ Exactly what is your understanding of how the "Clausius principle" is worded? $\endgroup$
    – Bob D
    Commented May 30 at 15:37
  • $\begingroup$ Heat flows from high temperature to low temperature and if you want to reverse the proces you have to add heat or work $\endgroup$
    – Skaeler
    Commented May 30 at 16:51
  • $\begingroup$ As pointed out by Fermi, "T represents the temperature of the source which surrenders the quantity of heat dQ, and is not necessarily equal to the temperature T' of the system (or of part of the system) which receives the heat dQ." That means that the T in the inequality is the temperature of the reservoir (or reservoirs) in the surroundings from which the system receives heat. $\endgroup$
    – Chet Miller
    Commented Jun 3 at 14:56

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The Clausius inequality applies to a closed system, and does not include the surroundings. The more correct form of this inequality is $$\Delta S_{syst}\geq \int{\frac{dQ}{T_B}}$$where $T_B$ is the temperature at the boundary interface between the system and surroundings (through which the heat increment dQ passes). The equal sign applies to any reversible path between the initial and final thermodynamic equilibrium states of the system, and the > sign applies to any irreversible path between initial and final thermodynamic equilibrium states. The Clausius inequality can be taken as a fundamental statement of the 2nd law of thermodynamics. For a cycle,the change in entropy is zero (since the initial and final states are the same), in which case $$\Delta S_{syst}=0\geq \int{\frac{dQ}{T_B}}$$

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  • $\begingroup$ Actually both the temperature and the entropy are those of the surrounding environment. $\endgroup$
    – Roger V.
    Commented May 30 at 9:34
  • $\begingroup$ @RogerV. I disagree totally. See Moran et al, Fundamentals of Engineering Thermodynamics $\endgroup$
    – Chet Miller
    Commented May 30 at 10:27
  • $\begingroup$ @RogerV. Indeed, $T$ in the Clausius inequality is temperature of the heat reservoir the system interacts with, the system itself need not have a temperature. But in case temperature is a continuous function of position, $T$ is correctly given by temperature on the boundary surface between the system and the reservoir. $\endgroup$
    – Ján Lalinský
    Commented May 30 at 10:50
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    $\begingroup$ @RogerV. Entropy of the reservoir $S_r$, in case it interacts only with the single system, and $Q_r$ is heat accepted by the reservoir from the system, obeys the Clausius inequality $\Delta S_r \geq \int \frac{dQ_r}{T_B}$ which in terms of $Q$ is $ \Delta S_r \geq - \int \frac{dQ}{T_B}$. $\endgroup$
    – Ján Lalinský
    Commented May 30 at 10:56
  • $\begingroup$ @ChetMiller How would you work the inequality out for reversible and irreversible cycles? I have tried to do it but my signs are wrong. For reversible it should be $\frac{Q_H}{T_H} $ = - $\frac{Q_L}{T_L}$. I arrive at this but without the minus sign. $\endgroup$
    – Skaeler
    Commented May 30 at 14:23

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