Assignment. Chapter 5 Problems: ONE COLLECTED TODAY 1,2,7,9,11,14,17,19,21,24,26,28,30,33,37,42,43,44,48, 49,51,53,55,57,62,63.
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1 Assignment Chapter 5 Problems: ONE COLLECTED TODAY 1,2,7,9,11,14,17,19,21,24,26,28,30,33,37,42,43,44,48, 49,51,53,55,57,62,63. REVIEW & PROBLEM SOLVING Tonight C-109 5PM Exam 2----Friday, August Coverage from end of Chapter 3 to page 236. Skip pages no calculations involving nonideal gases just conceptual.
2 Only heat and work can change internal energy. Surroundings +q = heat added +w = work done on the system Internal Energy System Rotation, Vibrations, Electronic, translational energies Surroundings -q = heat lost from the system -w = work done by the system
3 First Law of Thermodynamics A system contains only internal energy. A system does not contain heat or work. These only occur during a change in the system. Heat Work!E = q + w 1st Law of Thermodynamics: Conservation of Energy In interactions between a system and its surroundings the total energy remains constant energy is neither created nor destroyed just transformed between the two.!e system +!E surroundings = 0
4 Subtle But Key Points There are only two ways that a system exchanges energy with its surroundings: (1) heat transfer (q) 2) by work (w). There is no absolute heat there must be a change in the system. System q w Surroundings!Esys = q + w
5 PV = Energy = Work PV has units of energy we call it WORK J = 1 L atm w = P V height w = F orce Distance F orce Area w = height Area w = P (Area height) w = P V
6 System Watch the Units of Work q w Surroundings Note that: J = 1 L atm!esys = q + w w = -P!V!Esys = q - P!V!Esys = q - P(Vf - Vi)
7 PV-Work Example Calculate the work (in kilojoules) done during a reaction in which the volume expands from 12.0 L to 14.5 L against an external pressure of 5.0 atm. Watch the conversion of PV units to joules. P = 5.0 atm!v = ( ) L = 2.5 L w = P!V w = -(5.0 atm)(2.5 L) = L atm (-12.5 L atm)(101 J/L atm) = -1.3 " 10 3 J = -1.3 kj
8 The Key Riff Any thermodynamic system in an equilibrium state posses a state function called internal energy. Between any two states the change in internal energy that takes place is equal to the difference in heat transfer into or from the system and the work done by or on the system (energy changes because of heat and work only!)
9 State Functions Don t Depend On Path A state function is a function that only depends on the initial and final states and not how the path taken to between the two states. PE = mgh All state functions!!e = E final - E initial!p = P final - P initial h!v = V final - V initial!t = T final - T initial Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.!h = H final - H initial
10 Temperature is a state function, it does not depend whether we start with boiling water at 100 C or frozen water at 0 C. Once we get to 50 the difference in temperature is 50 C.
11 Internal Energy Changes In Chemical Reactions
12 Determining the Change in Internal Energy of a System PROBLEM: When gasoline burns in a car engine, the heat released causes the products CO 2 and H 2 O to expand, which pushes the pistons outward. Excess heat is removed by the car s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (!E) in J, kj, and kcal.
13 PROBLEM: When gasoline burns in a car engine cylinder, heat released causes the products CO 2 and H 2 O to expand, which pushes the pistons outward. Excess heat is removed by the car s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in internal energy (!E) in J, kj, and kcal. PLAN: Define system and surroundings, assign signs to q and w and calculate!e. The answer should be converted from J to kj and then to kcal. SOLUTION: q = J w = J!E = q + w = -325 J + (-451 J) = -776 J kj -776J 103 J kcal = kJ kJ 4.18kJ = kcal
14 Chemists don t use internal energy as a yardstick in monitoring energy changes in chemical reactions. We use its cousin and give it a di!erent name: ENTHALPY
15 Let s Play With The Internal Energy Equation!E = q - P!V CASE 1 CASE 2 At constant pressure:!e = q P - P!V q P =!E + P!V At constant volume:!v=0!e = q V - P!V!E = q V Enthalpy =!H = q P =!E + P!V
16 Chemical Reactions will either give off heat or absorb heat from the surroundings. The energy changes are state functions and normally carried out at constant pressure. We use ENTHALPY as the yardstick or ruler to quantify and communicate the heat given off or the heat absorbed! H 2 O (s) "# H 2 O (l)!h = 6.01 kj/mol That s called a thermochemical equation--- with the!h it tells us something about heat that is absorbed or given off by the chemical reaction
17 Exothermic process is any process that gives off heat transfers thermal energy from the system to the surroundings. 2H 2 (g) + O 2 (g) H 2 O (g) 2H 2 O (l) + energy H 2 O (l) + energy An exothermic reaction gives off thermal energy = heat = q Endothermic process is any process in which heat has to be supplied to the system from the surroundings. energy + 2HgO (s) 2Hg (l) + O 2 (g) energy + H 2 O (s) H 2 O (l) An endothermic reaction feels cold.
18 Enthalpy is very close to internal energy Enthalpy =!H = q P =!E + P!V I. In solids and liquids there is no volume change (or it is small) then P!V = 0 and!h =!E II. Reactions in which the number of moles of gas does not change. N2 + O2 ----> 2NO III. Reactions in which the number of moles of gas does change but q is >>> P!V.
19 A Comparison of!h and!e 2Na (s) + 2H 2 O (l) 2NaOH (aq) + H 2 (g)!h = kj/mol!e =!H - P!V At 25 C, 1 mole H 2 = 24.5 L at 1 atm P!V = 1 atm x 24.5 L = 2.5 kj!e = kj/mol 2.5 kj/mol = kj/mol
20 !H and It s Applications In Chemistry Enthalpies That We Measure In the Lab!H = q P =!E + P!V
21 Heat Transfer continues to equilibrium Heat transferred = q $!T q = C!T C = heat capacity
22 Heat Capacity Is Intensive Property of Matter Heat transferred = q $!T (!T = T final - T initial ) q = C!T C is a material dependent constant called the Heat Capacity units of Energy per degree T (J/ C, J/K, cal/ C, cal/k) C = q /!T Units of s: J / C, J/ K Heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius or Kelvin.
23 Physics of Heat Transfer C = q /!T We also find that heat capacity, C, of a material is proportional to mass of a substance under question. We can write: C m Units of s: J/g C, J/g K C m s where m is mass and s is a constant called the specific heat capacity
24 Specific Heat (Capacity), s The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius. C m s q = C!T (1) (2) Units of s: J/g C, J/g K Substituting specific heat (1) for heat capacity C in equation 2 gives: Heat (q) absorbed or released: q = s m!t
25 Calculating Specific Heat Capacity How much heat is required to raise the temperature of 1.0 x 10 3 kg of iron from 25 C to 75 C if the specific heat capacity of iron is 0.45 J/g C? Heat (q) absorbed or released: q = s m!t
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