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Understanding Energy Cycles in Thermodynamics

This lecture covers thermodynamic energy cycles, heat energy calculations, the importance of pathways like Carnot cycle and Hess' Law, and terms like enthalpy and isobaric processes. Solve problems on sublimation, vaporization, fusion, and more with practical examples and calculations. Learn about engines, expansion strokes, work done in cycles, isothermal and adiabatic expansions, converting energy units, and gas behaviors in pistons.

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Understanding Energy Cycles in Thermodynamics

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  1. Please Sit with Your Group • Please be sure each member of your team has a copy of • Energy Cycles Lecture Notes • Ice, Water, Steam Problem Set • Today’s reporter is the most talkative person

  2. Energy Cycles E. A. Mottel Integrated, First-Year Curriculum in Science, Engineering and Mathematics

  3. Energy Cycles • This lecture involves the concept of thermodynamic energy cycles and calculation of the heat energy released by these cyclic processes. • The importance of different pathways is exemplified by the Carnot cycle and Hess' Law.

  4. Energy Cycles • Change in Enthalpy (isobaric) • Isothermal • Constant external pressure • Varying external pressure • Adiabatic

  5. Energy Cycle • A series of energy steps following a defined pathway in which the final step returns the system to the original state conditions.

  6. sublimation Sublimation of Water at 0 °C H2O (l, 100 °C) heat capacity vaporization H2O (l, 0 °C) H2O (g, 100 °C) fusion heat capacity H2O (s, 0 °C) H2O (g, 0 °C) What thermodynamic terms are associated with each step?

  7. H2O (l, 0 °C) H2O (g, 100 °C) Determine the Enthalpy ofSublimation of Water at 0 °C H2O (l, 100 °C) -100 cal/g -418 J/g -540 cal/g -2259 j/g -80 cal/g -335 J/g +50 cal/g +209 J/g sublimation H2O (s, 0 °C) H2O (g, 0 °C) +670 cal/g +2803 J/g Assume the heat capacity of water vapor is constant from 0 °C to 200 °C.

  8. Ice, Water, Steam Problem Set • Select various masses of ice, water and steam at temperatures consistent with the phases. • Determine the final temperature of the mixture and the number of grams of each phase present in the final mixture.

  9. ¬1.5 atm ¬1.5 atm ¬1.5 atm Isothermal ExpansionConstant External Pressure The initial and final temperature of the gas in the cylinder is the same. Sketch a graph of this process One liter of a compressed gas in a cylinder causes a piston to expand against a constant external pressure of 1.5 atm until the total volume of the gas in the cylinder is three liters.

  10. 2 Pext (atm) 1 0 0 1 2 3 4 Volume (L) Vf Vf ó ó work = - P dV = - P dV õ õ Vi Vi Isothermal ExpansionConstant External Pressure Work may be represented as the area under the curve. = - P (Vf - Vi) Because pressure is constant, it can be factored out of the integral.

  11. Energy Units • L×atm can be converted to more common energy units (e.g., J or cal) by using the value of R as a conversion factor. work = - (1.5 atm) (3.0 L - 1.0 L) = -3.0 L×atm Determine the work done in calories and joules

  12. 1.987 cal×mol-1×K -1 = -3.0 L×atm × = -73 cal 0.08206 L×atm×mol-1×K -1 8.314 J×mol-1×K -1 = -3.0 L×atm × = -304 J 0.08206 L×atm×mol-1×K -1 Energy Units work = - (1.5 atm) (3.0 L - 1.0 L) = -3.0 L×atm Determine the work done in calories and joules

  13. Engines • An engine is a machine which can perform work. • The expansion of a gas in a piston can do work. • Describe the activities of an expanding gas at constant external pressure.

  14. ¬1.5 atm ¬1.5 atm ¬1.5 atm ¬1.5 atm ¬1.5 atm EnginesConstant External Pressure Compression Stroke Expansion Stroke

  15. 2 Pext (atm) 1 compression cycle 0 0 1 2 3 4 Volume (L) Engines expansion cycle As described this does not represent a practical engine because the final state does not equal the initial state. How much work is done in this overall process?

  16. ¬4.5 atm ¬2.25 atm ¬1.5 atm Isothermal ExpansionDecreasing External Pressure The initial and final temperature of the gas in the cylinder is the same. Sketch a graph of this process One liter of a compressed gas in a cylinder causes a piston to expand against a decreasing external pressure until the total volume of the gas in the cylinder is three liters.

  17. Isothermal ExpansionDecreasing External Pressure 4 3 PV = nRT Work may be represented as the area under the curve. 2 Pext (atm) 1 0 0 1 2 3 4 Volume (L) Vf Vf Vf nRT ó ó æ ö work = - P dV = - dV = - nRT ln õ õ è ø V Vi Vi Vi Isothermal ExpansionDecreasing External Pressure The pressure term is rewritten in terms of volume.

  18. Isothermal ExpansionDecreasing External Pressure 4 3 2 Pext (atm) 1 0 0 1 2 3 4 DE = Cv× DT Volume (L) isothermal Is there any net heat flow in this process? Is heat flowing into or out of the system? DE = q + w = 0 the internal energy of a phase is a function of its temperature

  19. Isothermal ExpansionDecreasing External Pressure 4 3 2 Constant External Pressure Pext (atm) 1 0 0 1 2 3 4 Volume (L) The decreasing external pressure piston performs more work (greater efficiency) than a piston working against a constant external pressure equal to Pfinal.

  20. Isothermal ExpansionDecreasing External Pressure 4 3 2 Pext (atm) 1 0 0 1 2 3 4 Volume (L) If the process is reversed, how much work is done?

  21. Room temperature gas colder gas q = 0 Gas Expansion When a gas expands against a low restraining pressure why does it cool? adiabatic expansion DE = q + w

  22. Gas ExpansionAir Conditioner

  23. gaseous CO2 liquid CO2 Gas ExpansionCarbon Dioxide Fire Extinguisher Why is it dangerous to point a carbon dioxide fire extinguisher at a person?

  24. Adiabatic Expansion • The same process occurs, except there is no heat flow allowed between the system and the surroundings. • On expansion, the gas will cool and follows a non-isothermal PV curve. • PVg = constant • for an ideal diatomic gas, g =1.67

  25. Adiabatic Expansion • In each of the examples, a different pressure change pathway is followed by the gas. • How much work will be done if the process is reversed to complete the cycle?

  26. Carnot Cycle • Consists of two isothermal and two adiabatic steps, occurring alternatively. • One of each type of step is involved in compression and expansion.

  27. Hess' Law Elements DHf,products - DHf,reactants DHrx Reactants Products The enthalpy change of a chemical reaction is equal to the difference of the enthalpy of the products and the enthalpy of the reactants.

  28. Hess' Law • DHrx = S DHf,products - SDHf,reactants • The system is exothermic if the enthalpy change is negative (DHrx < 0) and endothermic if the enthalpy change is positive (DHrx > 0).

  29. Thermodynamic ApplicationsComparison of Liquid Fuels • isooctane • methanol • ethanol

  30. Please Sit with Your Group • Please be sure each member of your team has a copy of • Thermodynamics Applications Lecture Notes • Today’s reporter is the most talkative person • Reading Assignment: Oxidation-Reduction: Zumdahl Chap. 4.10-4.12

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