1 / 13

Entropy = S

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

Download Presentation

Entropy = S

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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

More Related