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SSC JE ME Previous Paper 3 (Held on: 27 Sep 2019 Evening)

Option 3 : remains the same

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**Explanation:**

**For any reversible process(1-2)**

\({\rm{Δ }}S = \;\mathop \smallint \limits_1^2 \frac{{dQ}}{T}\)

Where;

**ΔS** = change in entropy; **dQ** = heat transferred; **T** = temperature

If the system is isolated, there is no change in the entropy of the surroundings and ΔS ≥ 0, for an isolated system.

Therefore the entropy of an isolated system either increases or, in the limit, remains constant.

The equality sign holds good when the process undergone by the system is reversible, the inequality sign holds good if there is any irreversibility present in the process. This statement is usually called the **principle of entropy increase.**

Irreversible or spontaneous processes can occur only in that direction for which the entropy of the universe or that of an isolated system, increases. These processes cannot occur in the direction of decreasing entropy.

For an isolated system,

- ΔS < 0, for irreversible processes
- ΔS = 0, for reversible processes
- ΔS > 0, the process is impossible

But for reversible process, **ΔS **= 0 (i.e. **entropy remains constant.)**