DS = \frac{{d{Q_{rev}}}}{T}\)If the process is reversible and adiabatic: dQrev = 0 ⇒ dS = 0\({Q_{rev}} = \smallint TdS\)Entropy analysis for a closed system:\(ds=(\frac{\delta Q}{T} )+ (\delta S)_{gen}\)Case 1: For an internally reversible process (δs)gen = 0When heat is added to the system ⇒ (δQ) = +ve ⇒ ds = +veWhen heat is rejected from the system ⇒ (δQ) = -ve ⇒ ds = -veFor adiabatic process ⇒ (δQ) = 0 ⇒ ds = 0 (ISENTROPIC) Case 2: For internally irreversible process (δs)gen ≠ 0 ⇒ (δs)gen = +veWhen heat is added to the system ⇒ (δQ) = +ve ⇒ ds = +veWhen heat is rejected from the system ⇒ (δQ) = -ve ⇒ ds = -ve or +ve or ZEROSince (δs)gen = +ve and (δQ) = -ve so change in entropy, ds can be zero is (δQ/T) = δs)gen. In this case, the process will be ISENTROPICFor adiabatic process ⇒ (δQ) = 0 ⇒ ds = +ve Thus for a process to be isentropic there are TWO cases:Reversible adiabatic processThe irreversible process where (δQ/T) = δs)gen NOTE: Irreversible adiabatic process is not an isentropic process.