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Discussion :: Chemical Engineering Thermodynamics

  1. For a spontaneous process, free energy

    2. A.

     Is zero
    B.

     Increases
    C.

     Decreases whereas the entropy increases
    D.

     And entropy both decrease

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    Answer : Option C

    Explanation :

    From second law of thermodynamics
    TdS⩾δQ⇒TdS⩾dU+δW
    For an irreversible process TdS−dU−δW>0
    And for a reversible process TdS−dU−δW=0
    For any spontaneous process there should be finite changes so, we can consider it as an irreversible process and we know for irreversible process from second law of thermodynamics by above discussion: TdS−dU−PdV>0
    Under constant temperature and volume process −dF>0⇒dF<0
    Similarly for an constant temperature and pressure process d(TS−U−PV)>0⇒dG<0

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