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
Solution (By Examveda Team)
From second law of thermodynamics :$$\eqalign{ & TdS \geqslant \delta Q \cr & \Rightarrow TdS \geqslant dU + \delta W \cr} $$
For an irreversible process $$TdS - dU - \delta W > 0$$
And for an reversible process $$TdS - dU - \delta W = 0$$
This Question Belongs to Chemical Engineering >> Chemical Engineering Thermodynamics
Join The Discussion