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  1. Entropy change for an irreversible isolated system is

    2. A.

     ∞
    B.

     0
    C.

     < 0
    D.

     > 0

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    Answer : Option D

    Explanation :

    From second law of thermodynamics
    TdS⩾δQ⇒TdS⩾dU+δW
    For an irreversible process S−dU−δW>0.    So, system undergoing irreversible change under constant internal energy and constant volume will have entropy change greater than zero dS>0.
    And for an reversible process TdS−dU−δW=0.

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