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Review of Energy and Thermodynamics in the Universe

This review explores the relationship between energy and thermodynamics in the universe, including the laws of thermodynamics, heat flow, work, and state functions. It delves into the concepts of exothermic and endothermic reactions, as well as the stability of products compared to reactants. The review also discusses ideal gases and thermochemistry.

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Review of Energy and Thermodynamics in the Universe

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  1. Review • system + surroundings Universe = • into system = • + • - • out of system = internal energy E =  kinetic energy • +  potential energy E = • K.E. - energy of motion P.E. - energy of position

  2. Euniverse = 1st Law of Thermodynamics Energy can not be created or destroyed Euniverse= 0 Esystem + Esurroundings = 0 - Esurroundings Esystem = heat = q Energy is exchanged as : work = w E = q + w

  3. heat flow of energy along a T gradient If T1 > T2 System at T1 Surroundings at T2 a) > b) < c) = • q system • 0 Exothermic reaction • q surroundings • > 0

  4. heat flow of energy along a T gradient If T1 < T2 System at T1 Surroundings at T2 > 0 • q system Endothermic reaction • q surroundings • < 0

  5. work Electrical work Mechanical work = force x distance force = pressure x area = P x m2 distance = m work = = P x V P x m2 x m Pext V - Wsystem=

  6. State Functions Property that depends only on the initial and final states Temperature, T : raise T from 298  300 K path 1 T = Tfinal - Tinitial= 300 - 298 = 2 K : raise T 298  500 K path 2 lower T from 500  300 K T = Tfinal - Tinitial = 300 - 298 = 2 K

  7. Extensive v.s. Intensive Extensive: Proportional to the mass of the system Intensive: Independent of the mass of the system Temperature: Intensive or Extensive Volume : Intensive or Extensive Pressure: Intensive or Extensive Internal Energy: Intensive or Extensive

  8. System 1 How much work will be done as the gas expands against the piston? Pext = 1.5 atm, T = 298 K P1 = 6.0 atm V1 = 0.4 L V2 = 1.6 L P2 = 1.5 atm P1 V1 = P2 V2 PV = nRT (6.0 atm) (0.4 L) = (1.5 atm) (V2) w = -PextV -(1.5 atm) -1.8 L atm w = (1.6 - 0.4)L = (-1.8 L atm) (101.3 J/L atm) = -182 J

  9. System 2 How much work is done when the stopcock is opened? ideal gas vacuum 0.4 L 1.2 L 6.0 atm P1 = 6 atm P2 = 1.5 atm V1 = 0.4 L V2 = 1.2 + 0.4 L = 1.6 L T1 = 298 K = T2 P1 V1 = P2 V2 w = -PextV = -(0 atm) (1.6 L - 0.4 L) w = 0

  10. System 1 System 2 P1 = P2 = V1 = V2 = T1 = T2 = 6.0 atm 1.5 atm P1 = P2 = V1 = V2 = T1 = T2 = 6.0 atm 1.5 atm 0.4 L 1.6 L 0.4 L 1.6 L 298 K 298 K w = -182 J q = +182 J w = 0 q = 0 Same initial and final conditions  all State functions are the same E will be the same if T = 0, E = 0 Ideal gases : E = q + w = 0

  11. Thermochemistry State 1 = State 2 = reactants products E = Eproducts - Ereactants Endothermicreaction  2NH3(g) + Ba(SCN)2(l) + 10H2O(l) + 2NH4SCN (s) Ba(OH)2•8H2O (s) a) < b) > c) = a) < b) > c) = q 0 w 0 reaction lowered T of system

  12. EndothermicReaction Tprod< Treact K.E.prod < K.E.react P.E.prod> P.E.react productsless stable than reactants

  13. Exothermic Reaction Tprod> Treact K.E.prod > K.E.react P.E.prod< P.E.react  mix of gases C12H24O12 (s) + KClO3(s) q 0 < w 0 < products more stable than reactants

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