1. Δ H = 0 and ΔS = 0
  2. Δ H ≠ 0 and ΔS = 0
  3. Δ H ≠ 0 and ΔS ≠ 0
  4. Δ H = 0 and ΔS ≠ 0

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Answer: Option 2

For a reversible adiabatic process the entropy change is zero since $$dS = \frac{{\delta Q}}{T}$$ But the enthalpy change need not be zero since the internal energy and pressure energy may vary during the process.

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