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

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

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

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

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

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 path 2 : raise T 298  500 K lower T from 500  300 K T = Tfinal - Tinitial = 300 - 298 = 2 K

Extensive v.s. Intensive 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

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 = P2 = 1.5 atm 1.6 L P1 V1 = P2 V2 PV = nRT (6.0 atm) (0.4 L) = (1.5 atm) (V2) w = -Pext V w = -(1.5 atm) (1.6 - 0.4)L = -1.8 L atm (-1.8 L atm) (101.3 J/L atm) = -182 J

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

Same initial and final conditions  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 Ideal gases : if T = 0, E = E = q + w = 0

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

Endothermic Reaction Tprod < Treact K.E.prod < K.E.react P.E.prod > P.E.react products less stable than reactants

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