?
In an irreversible process
Answer & Solution
Correct Answer:
Option
C
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 $$TdS - dU - \delta W = 0$$
$$\eqalign{ & TdS \geqslant \delta Q \cr & \Rightarrow TdS \geqslant dU + \delta W \cr} $$
For an irreversible process $$TdS - dU - \delta W > 0$$
And for an reversible process $$TdS - dU - \delta W = 0$$
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