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Entropy = S

Entropy = S. Entropy is. disorder. randomness. dispersal of energy. 2nd Law of Thermodynamics. > 0. for spontaneous processes. S universe. no external intervention. spontaneous =. S system. S surroundings. positional disorder. energetic disorder. Energetic Disorder.

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Entropy = S

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  1. Entropy= S Entropy is disorder randomness dispersal of energy

  2. 2nd Law of Thermodynamics > 0 for spontaneous processes Suniverse no external intervention spontaneous = Ssystem Ssurroundings positional disorder energetic disorder

  3. Energetic Disorder a)  b)  P.E. K.E. ordered random a) endothermic reaction b) exothermic reaction reactants P.E. Ssurr= -qsys (J/K) products T qsystem0 < > depends on T Ssurr qsurroundings 0 heat  surroundings Ssurroundings > 0 high T small effect low T relatively larger effect

  4. PositionalDisorder 2 dice microstates S = kB ln W kB = R/NA 3 4 5 6 7 8 9 10 11 12 2 distribution = state microstates = W = energy and position of atoms in state

  5. 23 ln 2 6.02 x 10 ∆SV → V = 2 1 S = kB ln W W 2 W kB = R/NA S2 – S1 = kB ln W2/W1 ∆S = = kB ln 2 x 2 x 2 for 1 mole gas kB ∆S = = 6.02 x 1023 kB ln 2 = Rln 2 Rln (V2/V1)

  6. S = kB ln W ∆S = Rln (V2/V1) Positional Disorder Boltzman W = microstates ordered states low probability low S high S disordered states high probability  Ssystem  Positional disorder Increases with number of possible positions (energy states) Ssolids Sliquids Sgases << <

  7. Entropy (J/K) [heat entering system at given T] convert q to S System 1 Pext = 1.5 atm E = 0 w = -182 J q = +182 J T = 298 K E = 0 System 2 Pext = 0 atm w = 0 q= 0

  8. = - nRT dV V  System 3 P1 = 6.0 atm P2 = 1.5 atm V1 = 0.4 L V2 = 1.6 L T1 = 298 K = T2 Pext = Pint + dP reversible process - infinitely slow n = .10 V2 = - nRT ln (V2/V1)  wr= - Pext dV V1 - nRT ln(1.6/4.0) = - 343.5 J wr=

  9. Ssystem = Ssystem System 1 Pext = 1.5 atm w = -182 J q = +182 J S = System 2 Pext = 0 atm w = 0 q = 0 S = System 3 Pext = Pint + dP wr = qr = S = -nRT ln (V2/V1) = - 343.5 J + 343.5 J 1.15 J/K 1.15 J/K 1.15 J/K ∆S = n Rln (V2/V1) qr = n R Tln (V2/V1) qr = 343.5 J 298 K T ∆S = n CP ln (T2/T1) ∆S = n CV ln (T2/T1)

  10. 3rd Law of Thermodynamics Entropy of a perfect crystalline substance at 0 K = 0

  11. Entropy curve gas liquid solid vaporization S qr T fusion 0 0 Temperature (K)

  12. Entropy At 0K, S = 0 Entropy is absolute S  0 for elements in standard states S is a State Function Sorxn = nSoproducts - nSoreactants S is extensive

  13. Increases in Entropy 1. Melting (fusion) Sliquid > Ssolid 2. Vaporization Sgas >> Sliquid 3. Increasing ngas in a reaction • Heating ST2 > ST1 if T2 > T1. • Dissolution Ssolution > (Ssolvent + Ssolute) ? 6. Molecular complexity number of bonds 7. Atomic complexity e-, protons and neutrons

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