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Ch6.1 The Nature of Energy Energy – the capacity to do work or to produce heat. Law of Conservation of Energy – energy c

Ch6.1 The Nature of Energy Energy – the capacity to do work or to produce heat. Law of Conservation of Energy – energy can be converted from one form to another, but can be neither created or destroyed. 1 st Law of thermodynamics – the energy of the universe is constant.

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Ch6.1 The Nature of Energy Energy – the capacity to do work or to produce heat. Law of Conservation of Energy – energy c

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  1. Ch6.1 The Nature of Energy Energy – the capacity to do work or to produce heat. Law of Conservation of Energy – energy can be converted from one form to another, but can be neither created or destroyed. 1st Law of thermodynamics – the energy of the universe is constant. Potential energy – stored energy due to position or composition Kinetic energy – energy of motion KE = ½mv2 Heat – (a form of energy) transfer of energy due to a temp difference Temperature – property that reflects the randomness of the particles (NOT a form of energy)

  2. Chemical energy Exothermic reaction – CH4(g) + O2(g)

  3. Chemical energy Exothermic reaction – heat flows out of the system (feel hot) CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + energy

  4. Chemical energy Exothermic reaction – heat flows out of the system (feel hot) CH4(g) + 2O2(g) CO2(g) + 2H2O(g) + energy In any exothermic reaction, some of the potential energy stored in chemical bonds is converted to thermal energy.

  5. Endothermic reaction – • N2(g) + O2(g)

  6. Endothermic reaction – absorb heat from their surroundings (feel cold) • - heat flows into the system • N2(g) + O2(g) + energy  2NO(g)

  7. 1st Law of Thermodynamics – the energy of the universe is constant. ΔE = q + w Internal heat work energy energy “All energy movements reflect the system’s point of view.” ΔE = + Internal energy of the system increased ΔE = – Internal energy of the system decreased q = + Heat flowed into the system q = – Heat flowed out of the system w = + The surroundings do work on the system w = – The system does work on the surroundings

  8. Ex1) Calculate ΔE for a system undergoing an endothermic process in which 15.6kJ of heat flows and where 1.4kJ of work is done on the system.

  9. Work associated with gases: Pressure: Work:

  10. Ex2) Calc the work associated with the expansion of a gas from 46L to 64L at a constant pressure of 15 atm.

  11. Ex3) A balloon is being inflated to its full extent by heating the air inside it. The volume changes from 4.00x106L to 4.50x106L when 1.3x108J of heat are added. Atmospheric pressure is 1.0 atm. Calculate ΔE for the process. Ch6 HW#1 p283 21,23,25,26

  12. Ch6 HW#1 p283 21,23,25,27 21. Calculate ΔE for each of the following cases. a. q = +51 kJ, w = -15kJ b. q = +100 kJ, w = -65kJ c. q = -65kJ, w = -20kJ d. In which of these does the system do work on the surroundings?

  13. Ch6 HW#1 p283 21,23,25,26 21. Calculate ΔE for each of the following cases. a. q = +51 kJ, w = -15kJ b. q = +100 kJ, w = -65kJ c. q = -65kJ, w = -20kJ d. In which of these does the system do work on the surroundings? ΔE = q + w a. ΔE = +51 kJ + (-15kJ) = +36kJ (Internal energy increases) b. ΔE = +100 kJ + (-65kJ) = +35kJ (Internal energy increases) c. ΔE = -65kJ + (-20kJ) = -85kJ (Internal energy decreases) d. All 3 have w = (-) so all systems did work on the surroundings

  14. 23. A gas absorbs 45kJ of heat and does 29kJ of work. Calculate ΔE. ΔE = q + w

  15. 23. A gas absorbs 45kJ of heat and does 29kJ of work. Calculate ΔE. ΔE = q + w = (+45kJ) + (-29kJ) = +16kJ (increase)

  16. 25. The volume of an ideal gas is decreased from 5.0 Lto 5.0 mL at a constant pressure of 2.0 atm. Calculate the work associated with this process. ΔV = (0.005L – 5L) = -4.995L

  17. 25. The volume of an ideal gas is decreased from 5.0 Lto 5.0 mL at a constant pressure of 2.0 atm. Calculate the work associated with this process. ΔV = (0.005L – 5L) = -4.995L W = –P. ΔV = –(202.6kPa).(-4.995L) = +1012 J = +1.012kJ (work done on the gas!)

  18. 26. Consider the mixture of air and gasoline vapor in a cylinder with a piston. The original volume is 40. cm3. If the combustion of this mixture releases 950. J of energy, to what volume will the gases expand against a constant pressure of 650. torr if all the energy of combustion is conserved into work to push back the piston. ΔV = (V2 – 0.040L) q  W = –950J (expands – gas doing work) W = –P. ΔV –950J = –(86.6kPa). ΔV ΔV = 10.97L 10.97L = (V2 – 0.040L) V2 = 11.0L

  19. Ch6.1B – PV Diagrams ‘A closed system at equilibrium has no tendency to undergo spontaneous macroscopic change.” 4 basic models where a closed system is changed: Constant Can change Equations Implications Thermo Eqn 1. Isothermal 2. Isobaric 3. Isovolumetric 4. Adiabatic (no heat transferred to or from a system.) Work equation: (Units: ______, or ______, or ______)

  20. Ch6.1B – PV Diagrams ‘A closed system at equilibrium has no tendency to undergo spontaneous macroscopic change.” 4 basic models where a closed system is changed: Constant Can change Equations Implications Thermo Eqn 1. Isothermal T V, P V1P1 = V2P2 If ∆T = 0, q = W then ∆E=0 2. IsobaricP V, T ∆E = q + W 3. IsovolumetricV P, T No work ∆E = q done. 4. Adiabaticnothing V, P, T Q = 0 ∆E = W (no heat transferred to or from a system.) Work equation: W = –P . ∆V (Units: Joules, or Pa.m3, or kPa.L)

  21. 50 Work = (area under the PV curve.) 40 Decide +/- based on direction P 30 (Pa) 20 10 1 2 3 V (m3) 101.3 P (kPa) 1 2 3 V (L) 150 P 100 (Mpa) 50 1 2 3 V(cm3)

  22. 50 Work = –(area under the PV curve.) 40 Decide +/- based on direction P 30 W = ½(3m3)(40N/m2) (Pa) 20 = –60J 10 1 2 3 V (m3) 101.3 W = (101.3kPa)(3L) P (1kPa  1000Pa) (kPa) = –303.9J (1L  0.001m3) 1 2 3 V (L) 150 No work done P 100 (Isovolumetric) (Mpa) (1 Mpa1,000,000Pa) 50 (1cm3 0.000001m3) 1 2 3 V(cm3)

  23. Ex3) A cylinder contains 264.5 moles of a monatomic gas that is initially at state A at standard temperature, and a pressure of 6x105Pa. a. What volume does the gas occupy? 6 b. Please graph state A. 5 c. A certain amount of heat is added, P 4 bringing the gas isobarically to state B, (x105 Pa) 3 at a volume of 3m3, while its internal energy 2 is increased by 400kJ. How much heat was added? 1 Please graph state B. 1 2 3 4 5 6 7 8 V(m3) d. The gas is brought isothermally to state C, where the pressure was reduced to 3x105Pa. What is the new volume? Please graph state C. Ch6 HW#2 1 – 4

  24. Ex3) A cylinder contains 264.5 moles of a monatomic gas that is initially at state A at standard temperature, and a pressure of 6x105Pa. a. What volume does the gas occupy? 6 b. Please graph state A. 5 c. A certain amount of heat is added, P 4 bringing the gas isobarically to state B, (x105 Pa) 3 at a volume of 3m3, while its internal energy 2 is increased by 400kJ. How much heat was added? 1 Please graph state B. 1 2 3 4 5 6 7 8 V(m3) d. The gas is brought isothermally to state C, where the pressure was reduced to 3x105Pa. What is the new volume? Please graph state C. a. PV=nRT c. ∆E = q + W W = –(area) = (+2m3)(6x105Pa) = –1,200,000J (+400,000J) = q + (–1,200,000J) + q = +1,600,000J d. VB = 3m3 VC = ? VB. PB = VC. PC PB = 6x105Pa PC = 3x105Pa (3)(6) = (VC)(3) TB = ? same TC = ? VC =6m3

  25. Ch6 HW#2 1 – 4 1. A quantity of gas in a cylinder sealed with a piston absorbs 4186J of heat as it expands by 15L at constant pressure. If the internal energy increases by 286J, what was the constant pressure? 2. A cylinder sealed with a piston contains 20.0m3 of pure oxygen at standard pressure. The gas is then forced into a tiny high pressure container of negligible volume. What is the amt of work done on the gas? What do you think happens to the temp of the gas as it is compressed?

  26. Ch6 HW#2 1 – 4 1. A quantity of gas in a cylinder sealed with a piston absorbs 4186J of heat as it expands by 15L at constant pressure. If the internal energy increases by 286J, what was the constant pressure? q = + 4186J ∆E = q + w ∆V = + 15L  0.015m3 w = (+286J) – (+4186J) ∆U = +286J w = – 3890J 2. A cylinder sealed with a piston contains 20.0m3 of pure oxygen at standard pressure. The gas is then forced into a tiny high pressure container of negligible volume. What is the amt of work done on the gas? What do you think happens to the temp of the gas as it is compressed?

  27. Ch6 HW#2 1 – 4 1. A quantity of gas in a cylinder sealed with a piston absorbs 4186J of heat as it expands by 15L at constant pressure. If the internal energy increases by 286J, what was the constant pressure? q = + 4186J ∆E = q + w ∆V = + 15L  0.015m3 w = (+286J) – (+4186J) ∆U = +286J w = – 3890J 2. A cylinder sealed with a piston contains 20.0m3 of pure oxygen at standard pressure. The gas is then forced into a tiny high pressure container of negligible volume. What is the amt of work done on the gas? What do you think happens to the temp of the gas as it is compressed? ∆V = –20.0m3 W = – P. ∆V P = 101,300Pa = –(101,300Pa)(– 20m3) W = ? = +2,026,000J (work done on the gas) Compression of a gas increaes temp.

  28. 3. A gas experiences a change from state I to state F. Determine the work done. 4. A piston with a cross-sectional area of 0.50m2 seals a gas within a cylinder at a pressure of 0.30MPa. 10kJ of heat is gradually added to the cylinder, and the piston rises 6.5cm as the gas expands isobarically. What is the work done by the gas? What is the change in internal energy? A = 0.50m2 h = 6.5cm 6 5 F P 4 (Mpa) 3 2 I 1 1 2 3 4 5 6 V (cm3)

  29. 3. A gas experiences a change from state I to state F. Determine the work done. 6 5 F Work = 1 + 2 + 3 P 4 = 12J (Mpa) 3 2 I 1 1 2 3 4 5 6 V (cm3) 4. A piston with a cross-sectional area of 0.50m2 seals a gas within a cylinder at a pressure of 0.30MPa. 10kJ of heat is gradually added to the cylinder, and the piston rises 6.5cm as the gas expands isobarically. What is the work done by the gas? What is the change in internal energy? 3 1 2 A = 0.50m2 h = 6.5cm

  30. 3. A gas experiences a change from state I to state F. Determine the work done. 6 5 F Work = 1 + 2 + 3 P 4 = 12J (Mpa) 3 2 I 1 1 2 3 4 5 6 V (cm3) 4. A piston with a cross-sectional area of 0.50m2 seals a gas within a cylinder at a pressure of 0.30MPa. 10kJ of heat is gradually added to the cylinder, and the piston rises 6.5cm as the gas expands isobarically. What is the work done by the gas? What is the change in internal energy? A = 0.50m2∆V=A. ∆h=(.50m2)(.065m) = +0.0325m3 h = 6.5cm W = – P. ∆V = (0.3x106Pa)(+0.0325m3) = – 9750J ∆E = q + W = (+10,000) + (– 9750) = +250J 3 1 2

  31. Ch6.1C – PV Diagrams Ch6 HW#3

  32. Ch6.1C – PV Diagrams Ex1) The graph shows P vs V for 2 moles of gas cycling from state A to B. The internal energy increases by 400J. a. Calc work done. 1000 b. Is heat added or removed? How much? P 800 c. The pressure is reduced to 200Pa (Pa) 600 A B isovolumetrically as gas goes from state B 400 to state C. Please graph state C. 200 d. The graph is compressed isothermally to state A. 1 2 3 4 5 6 7 8 9 10 Draw the lines to represent the cycle. V (m3) e. Is heat added or removed from C  A? How much?

  33. Ex1) The graph shows P vs V for 2 moles of gas cycling from state A to B. The internal energy increases by 400J. a. Calc work done. 1000 b. Is heat added or removed? How much? P 800 c. The pressure is reduced to 200Pa (Pa) 600 A B isovolumetricallyas gas goes from state B 400 to state C. Please graph state C. 200 C d. The graph is compressed isothermally to state A. 1 2 3 4 5 6 7 8 9 10 Draw the lines to represent the cycle. V (m3) e. Is heat added or removed from C  A? How much? a. W = P. ∆V = (600Pa)(+6m3) = +3600J b. Added: Why else would the gas expand? Q = W + ∆U = +3600J + (+400J) = +4000J c. isovolumetric d.isothermal– P and V can change, while ∆T = 0, so ∆U = 0. e. Heat is removed. Q = W + ∆U = W + 0 Q = WorkCA = (Area) = -2400J (Work is (-) becuz the gas is compressed by outside force. Work done on the gas.)

  34. Ex2) A piston with a radius of 0.05m seals a cylinder that contains a gas at a pressure of 101.3kPa. 100J of heat is added to the cylinder, and the piston rises 0.04m isobarically. a. How much work is done by the gas? h b. What is its change in internal energy? c. A 5.0kg mass is placed on top of the piston. What is the total pressure exerted on the gas? d. If the piston seals 1m3, when the mass is placed on the piston, and the piston lowers isothermally, what is the new volume? Ch6 HW#3 5 – 8

  35. Ex2) A piston with a radius of 0.05m seals a cylinder that contains a gas at a pressure of 101.3kPa. 100J of heat is added to the cylinder, and the piston rises 0.04m isobarically. a. How much work is done by the gas? h b. What is its change in internal energy? c. A 5.0kg mass is placed on top of the piston. What is the total pressure exerted on the gas? d. If the piston seals 1m3, when the mass is placed on the piston, and the piston lowers isothermally, what is the new volume? a. ∆V = A. ∆h = π(.05m)2(+.04m) = + 0.0003m3 W = – P. ∆V = – (101,300Pa)(+0.0003m3) = – 31.8J b. ∆U = q + W = (+100J) + (– 31.8J) = 68J c. Ptotal = Pmass + Patm = + 101,300 Pa = + 101,300Pa = 107,700 Pa d. V1 = 1 m3 V2 = ___m3 V1. P1 = V2. P2 P1 = 101,300Pa P2 = 107,700Pa (1m3)(101,300Pa) = V2(107,700Pa) T1 = ? same T2 = ? V2 = .9m3

  36. Ch6 HW#3 5 – 8 1. Given that we have 1200 cm3 of helium at 15°C and 99 kPa, what will be its volume at STP? 2. Three moles of a gas are stored at a temp of 60°C in a container. A 1.05 kg mass is placed on the lightweight moveable piston that has a cross-sectional area of 1.106 X 10-6 m2. The system reaches equilibrium, as shown. What volume does the gas occupy? Ptotal = Pmass + Patm = PV = nRT

  37. Ch6 HW#3 5 – 8 1. Given that we have 1200 cm3 of helium at 15°C and 99 kPa, what will be its volume at STP? 2. Three moles of a gas are stored at a temp of 60°C in a container. A 1.05 kg mass is placed on the lightweight moveable piston that has a cross-sectional area of 1.106 X 10-6 m2. The system reaches equilibrium, as shown. What volume does the gas occupy? Ptotal = Pmass + Patm = PV = nRT

  38. Ch6 HW#3 5 – 8 1. Given that we have 1200 cm3 of helium at 15°C and 99 kPa, what will be its volume at STP? 2. Three moles of a gas are stored at a temp of 60°C in a container. A 1.05 kg mass is placed on the lightweight moveable piston that has a cross-sectional area of 1.106 X 10-6 m2. The system reaches equilibrium, as shown. What volume does the gas occupy? Ptotal = Pmass + Patm = + 101,300 Pa = + 101,300Pa = 103,200 Pa PV = nRT

  39. 3. A 0.05 m3 volume of gas absorbs 10 kJ of heat and expands to a volume of 0.15 m3, while remaining at a constant pressure. During the process, the internal energy increases by 4500 J. What was the pressure? ∆V = +.10m3 q = +10kJ ∆E = +4.5kJ P = ?

  40. 3. A 0.05 m3 volume of gas absorbs 10 kJ of heat and expands to a volume of 0.15 m3, while remaining at a constant pressure. During the process, the internal energy increases by 4500 J. What was the pressure? ∆V = .10m3W = ∆U – q = (+4.5kJ) – (+10kJ) = – 5.5kJ = – 5500J Q = +10kJ ∆U = +4.5kJ P = ? 4. A piston with a cross-sectional area of 0.10 m2 seals a cylinder that contains a gas at a pressure of 100 kPa. 200 J of heat are removed from the cylinder, and the piston falls 0.05 m isobarically. a. How much work is done on the gas? b. What is the change in internal energy? c. A 50.0 kg mass is placed on top of the piston. What is the total pressure exerted on the gas? d. If the gas is then heated, so that its absolute temperature doubles, what is the new volume of the cylinder?

  41. 4. A piston with a cross-sectional area of 0.10 m2 seals a cylinder that contains a gas at a pressure of 100 kPa. 200 J of heat are removed from the cylinder, and the piston falls 0.05 m isobarically. a. How much work is done on the gas? ∆V=A. ∆h = (.1m2)(–.05m)= –0.005m3 W = – P. ∆V = (100,000Pa)(– 0.005m3) = +500J b. What is the change in internal energy? c. A 50.0 kg mass is placed on top of the piston. What is the total pressure exerted on the gas? d. If the gas is then heated, so that its absolute temperature doubles, what is the new volume of the cylinder?

  42. 4. A piston with a cross-sectional area of 0.10 m2 seals a cylinder that contains a gas at a pressure of 100 kPa. 200 J of heat are removed from the cylinder, and the piston falls 0.05 m isobarically. a. How much work is done on the gas? ∆V=A. ∆h = (.1m2)(–.05m)= –0.005m3 W = – P. ∆V = (100,000Pa)(– 0.005m3) = +500J b. What is the change in internal energy? ∆U = Q – W = (-200J) – (-500J) = +300J c. A 50.0 kg mass is placed on top of the piston. What is the total pressure exerted on the gas? d. If the gas is then heated, so that its absolute temperature doubles, what is the new volume of the cylinder?

  43. 4. A piston with a cross-sectional area of 0.10 m2 seals a cylinder that contains a gas at a pressure of 100 kPa. 200 J of heat are removed from the cylinder, and the piston falls 0.05 m isobarically. a. How much work is done on the gas? ∆V=A. ∆h = (.1m2)(–.05m)= –0.005m3 W = – P. ∆V = (100,000Pa)(– 0.005m3) = +500J b. What is the change in internal energy? ∆U = Q – W = (-200J) – (-500J) = +300J c. A 50.0 kg mass is placed on top of the piston. What is the total pressure exerted on the gas? = Ptotal = Pmass + Patm + 100,000Pa = 105,000Pa d. If the gas is then heated, so that its absolute temperature doubles, what is the new volume of the cylinder?

  44. 4. A piston with a cross-sectional area of 0.10 m2 seals a cylinder that contains a gas at a pressure of 100 kPa. 200 J of heat are removed from the cylinder, and the piston falls 0.05 m isobarically. a. How much work is done on the gas? ∆V=A. ∆h = (.1m2)(–.05m)= –0.005m3 W = – P. ∆V = (100,000Pa)(– 0.005m3) = +500J b. What is the change in internal energy? ∆U = Q – W = (-200J) – (-500J) = +300J c. A 50.0 kg mass is placed on top of the piston. What is the total pressure exerted on the gas? = Ptotal = Pmass + Patm + 100,000Pa = 105,000Pa d. If the gas is then heated, so that its absolute temperature doubles, what is the new volume of the cylinder?

  45. Ch6.2A – Enthalpy Enthalpy, H – the ‘heat content’ of a substance defined as: H = E + PV

  46. Ch6.2A – Enthalpy Enthalpy, H – the ‘heat content’ of a substance ΔH = Hproducts – Hreactants Exothermic reaction: Enthalpy time

  47. Ch6.2A – Enthalpy Enthalpy, H – the ‘heat content’ of a substance ΔH = Hproducts – Hreactants Endothermic reaction: Enthalpy time

  48. Ex1) When 1 mol of methane is burned at constant pressure, 890kJ of energy is released as heat. Calculate ΔH for a process in which a 5.8g sample is burned.

  49. Ex2) a. Write an eqn for the combustion of ethanol, C2H5OH. 1235 kJ of heat is given off. b. Draw an energy diagram. c. How much heat is produced when 12.5g of ethanol reacts? Ch6 HW#4 p283 29,30,31,33,34

  50. Ch6 HW#4 p283 29,30,31,33,34 29. The equation for the fermentation of glucose to alcohol and carbon dioxide is C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g) The enthalpy change for the reaction is -67kJ. Is the reaction exothermic or endothermic? Is energy, in the form of heat, absorbed or evolved as the reaction occurs?

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