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Shweta Thakur posted an Question
November 02, 2020 • 18:13 pm 30 points
  • IIT JAM
  • Chemistry (CY)

Plz explain the highlighted part e whether the process is carried out reversibly or irreversibly, hence 5) au= 4ey-

Further, as the internal cnergy (U) is state function, thie value of AU is same whether the process is carried out reversibly or irreversibly, Hence 5) AU= 4ey-revireyre Combining results (4) and (5), we conclude that Irev irev rey irey Ton 2raya9AM 9revire 0 8) or T T Combining this result with the result given in equatíon (3), we observe that AsystemT uroundings 0 Thus, it may be concluded that in an irreversible process, the entropy change for the combined system and the surroundings is greater than zero, 1e., an irreversible process is accompanied by a net increase of entropy Since all spontancous processes are thermodynamically irreversible (first definition of second law of thermodynamics) it may be stated that Al spontaneous processes are accompanied by a net increase of entropy. This i

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    Priyanshu kumar Best Answer

    Here it is suppose that loss of heat by the surrounding takes place infinitesimally slowly as surrounding are much bigger in size and magnitude compared to system. So entropy change of surrounding is given by delS= -qirrev/T

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