Chemical Engineering Thermodynamics

Entropy change for an irreversible isolated system is

  • A. ∞
  • B. 0
  • C. < 0
  • D. > 0
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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