Basic Thermodynamics 2.3

Basic Thermodynamics 2.3 - Entropy production irreversible...

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MSE 3050, Phase Diagrams and Kinetics, Leonid Zhigilei Entropy production – irreversible process Entropy created/produced: Example : One mole of supercooled liquid lead spontaneously freezes at 590 ° K and 1 atm pressure to the solid state. Calculate the entropy production. C p (liquid) = 32.4 - 3.1 × 10 -3 T [Jmol -1 K -1 ] C p (solid) = 23.6 + 9.75 × 10 -3 T [Jmol -1 K -1 ] Δ H m = 4810 Jmol -1 – latent heat of melting ( ) . res . th Pb . res . th initial Pb liquid . res . th final Pb solid S S S S S S S Δ + Δ = + + = Δ First, let’s calculate change in entropy for Pb Liquid Pb III II I Pb S S S S Δ + Δ + Δ = Δ The entropy changes along the actual path is unknown - the process is irreversible. But since entropy is a state function, we can calculate the entropy change along a reversible path. 590 ° K 600 ° K Solid Pb Actual path I II III
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MSE 3050, Phase Diagrams and Kinetics, Leonid Zhigilei Entropy change in the piece of lead, Δ S Pb : (I) isobaric reversible heating = Δ + Δ + Δ = Δ III II I Pb S S S S = Δ I S (II) isothermal isobaric process of freezing = Δ II S (III) isobaric reversible cooling = Δ III S
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