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The energy of an isolated system in a process
1. Increases
2. Decreases
3. Remains constant
4. Not predictable

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Correct Answer - Option 3 : Remains constant

Explanation:

Isolated system

In an Isolated system, there is no mass and energy interaction across the system boundary i.e. interaction between the system and the surroundings is absent. Therefore mass and the energy of the isolated system is constant.

For an isolated system dQ = 0 and dW = 0

First law of thermodynamics gives

dQ = ΔE + dW

∴ ΔE = 0

⇒ E = Constant

Therefore the energy of an isolated system is always constant.

For any infinitesimal process undergone by a system,

\(ds \ge \frac{{dQ}}{T}\)

For an isolated system which does not undergo any energy interaction with the surrounding dQ = 0.

Therefore for an isolated system

dSiso ≥  0

For reversible process

dS = 0, ∴ S = constant

For an irreversible process

diso > 0

Therefore the entropy of an isolated system can never decrease. It always increases for the irreversible process and remains constant if the process is reversible. This is known as the principle of increase of entropy, or simply the entropy principle.

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