Chemical Engineering Thermodynamics

In an irreversible process

  • A. Tds = dE – dW = 0
  • B. DE – dW – Tds = 0
  • C. Tds – dE + dW< 0
  • D. Tds – dT + dW< 0
Answer: Option C.
Explanation: 

From second law of thermodynamics :
TdS⩾δQ
⇒TdS⩾dU+δW

For an irreversible process TdS−dU−δW>0

And for an reversible process TdS−dU−δW=0

Leave a Reply

Your email address will not be published. Required fields are marked *

Back to top button
error: Alert: Content is protected !!