Q.For an isolated system, $\Delta U = 0$, what will be $\Delta S$?

The key idea is the Second Law of Thermodynamics: for an isolated system, entropy never decreases.

Reasoning:

1. An isolated system exchanges no energy or matter with its surroundings, so $\Delta U = 0$ is given.
2. The Second Law states that for any spontaneous process in an isolated system, the total entropy change is $\Delta S \ge 0$. …

The Second Law of Thermodynamics is the guiding principle here. It tells us that for any spontaneous process in an isolated system, entropy never decreases — it either increases (irreversible process) or stays the same (reversible process). An isolated system exchanges neither energy nor matter with its surroundings, so $\Delta U = 0$ is simply a consequence of the first law (energy conservation) when no heat or work crosses the boundary. But the Second Law gives us the direction: entropy must increase or remain constant.

The key insight: $\Delta U = 0$ doesn't mean nothing happens. Inside an isolated system, processes like mixing, chemical reactions, or heat flow between parts can still occur. The First Law says total internal energy is constant, but the Second Law says entropy must rise (or stay the same) as these processes move the system toward equilibrium.

Let's work through the reasoning step by step.

1. Start with the definition of entropy change for a reversible process.

   For any infinitesimal process, $dS = \frac{\delta Q_{\text{rev}}}{T}$, where $\delta Q_{\text{rev}}$ is the heat transferred reversibly. But for an isolated system, $\delta Q = 0$ for any actual process (no heat exchange with surroundings). However, to compute $\Delta S$, we must imagine a reversible path between the same initial and final states — entropy is a state function, so its change depends only on the endpoints.

2. Apply the Clausius inequality.

   The Second Law is captured by the Clausius inequality: for any cyclic process, $\oint \frac{\delta Q}{T} \leq 0$. For a non-cyclic process between two states, this leads to $dS \geq \frac{\delta Q}{T}$, where equality holds for a reversible process. For an isolated system, $\delta Q = 0$ always, so $dS \geq 0$. Integrating from initial to final state gives $\Delta S \geq 0$.

3. Interpret the two cases.

   • If the process inside the isolated system is reversible (an idealisation for an isolated system — it would require perfectly quasi-static changes with no dissipative effects), then $\Delta S = 0$.
   • If the process is irreversible (which is the real-world case — e.g., free expansion of a gas, mixing, heat flow between parts at different temperatures), then $\Delta S > 0$. …