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Explain change in entropy of a system during a reversible process Delta S = (q_(rev))/(T) |
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Answer» Solution :As entropy change is the difference in the ENTROPIES, when the system passes from initial state to the FINAL state, `Delta S = S_(F) - S_(1)` where, `S_(F)` = entropy of the final state and `S_(1)` = entropy of the initial state. The absolute value of entropy of a system cannot be measured, but the entropy change `(Delta S)` for a process can be calculated using thermodynamic DATA. Entropy change of a process is the ratio of quantity of heat absorbed to the absolute temperature when the process is carried out reversible and isothermally. If `q_("reversible")` is the heat absorbed at temperature T, then `Delta S = (q_("reversible"))/(T) = (DeltaH)/(T)` Eq. (1) is a mathematical form of stating the second law of THERMODYNAMICS. Entropy change is used to predict the spontaneity of a process. It is found in an isolated system, a process proceeds spontaneously in the DIRECTION where entropy increases. In other words, a process is spontaneous when `Delta S` is positive. More positive the `Delta S`, higher will be the spontaneity. |
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