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Understanding Entropy: Disorder, Thermodynamics, and Its Implications

This article explores the concept of entropy, a measure of disorder or randomness in systems, according to the Second Law of Thermodynamics. It delves into spontaneous processes, the importance of positional and energetic disorder, and how temperature influences these factors. The document covers reactions, both endothermic and exothermic, along with their entropic contributions. It also discusses the Third Law of Thermodynamics, the relationship of entropy with phase changes, and how the entropy of various states of matter compares.

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Understanding Entropy: Disorder, Thermodynamics, and Its Implications

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  1. Entropy = S Entropy is disorder or randomness

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

  3. Energetic Disorder P.E. K.E. a)  b)  random ordered reactants a) endothermic reaction b) exothermic reaction P.E. Hsystem0 < products Hsurroundings 0 > Ssurroundings > 0

  4. Ssurr = - Hsystem (J/K) T Ssurr depends on T heat  surroundings high T small effect low T relatively larger effect

  5. Positional Disorder (1/2) (1/2) (1/2) (1/2) (1/2) (1/2) = 1/64 With 1 mole of gas, Not probable

  6. RIP S = k ln W Positional Disorder W = degrees of freedom Boltzman 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. (J/K) Entropy [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 V2 wr= - Pext dV = - nRT ln (V2/V1)  V1

  9. Ssystem = qr T 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) = - 1120 J + 1120 J 3.77 J/K 3.77 J/K 3.77 J/K Ssurr = - Hsystem T = 1120 J 298 K

  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 S = 0 Entropy is absolute At 0K, 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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