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For an isolated system ∆U = 0, what will be ∆S ?

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\(\therefore\) ΔU = 0

\(\therefore\) T = 0

As we Know that ∆S = \(\frac{q_{rev}}{T}\) = \(\frac{q_{rev}}{0}\)

\(\therefore\) ΔS is more than zero,(ΔS> 0) i.e positive

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