1. Please Pick Up Ice, Water, Steam Quiz Internal Energy, Heat & Work problem Set

2. Energy Cycles Edward A. Mottel Department of Chemistry Rose-Hulman Institute of Technology

3. 6/14/2015 Energy Cycles  Reading Assignment: Chang, Chapter 6.6  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. 6/14/2015 Energy Cycles  A series of energy steps following a defined pathway in which the final step returns the system to the original state conditions.  Pathways Isobaric Isothermal (constant external pressure) Isothermal (varying external pressure) Adiabatic

5. 6/14/2015 Sublimation of Water at 0  C sublimation H 2 O (s, 0  C)H 2 O (g, 0  C) H 2 O (g, 100  C) H 2 O ( l, 100  C) H 2 O ( l, 0  C) Which thermodynamic terms are associated with each step? heat capacity of steam enthalpy of vaporization heat capacity of water enthalpy of fusion

6. 6/14/2015 Determine the Enthalpy of Sublimation of Water at 0  C Assume the heat capacity of water vapor is constant from 0  C to 200  C. H 2 O (s, 0  C)H 2 O (g, 0  C) H 2 O ( l, 100  C) H 2 O ( l, 0  C)H 2 O (g, 100  C) sublimation +50 cal/g +209 J/g -540 cal/g -2259 j/g -100 cal/g -418 J/g -80 cal/g -335 J/g +670 cal/g +2803 J/g

7. 6/14/2015 Ice, Water, Steam Quiz  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.  Confirm your answer with the program ICEWATER in the class folder.

8. 6/14/2015 Isothermal Expansion Constant External Pressure  1.5 atm 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. The initial and final temperature of the gas in the cylinder is the same. Sketch a graph of this process

9. 6/14/2015 Isothermal Expansion Constant External Pressure 01234 0 1 2 Volume (L) P ext (atm)

10. 6/14/2015 Isothermal Expansion Constant External Pressure 01234 0 1 2 Volume (L) P ext (atm) Work may be represented as the area under the curve. work = -   VfVf ViVi P dV= - P   VfVf ViVi dV = - P (V f - V i )

11. 6/14/2015 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. Determine the work done in calories and joules work = - (1.5 atm) (3.0 L - 1.0 L) = -3.0 L  atm

12. 6/14/2015 Energy Units Determine the work done in calories and joules work = - (1.5 atm) (3.0 L - 1.0 L) = -3.0 L  atm = -3.0 L  atm × 1.987 cal  mol -1  K -1 0.08206 L  atm  mol -1  K -1 = -73 cal = -3.0 L  atm × 8.314 J  mol -1  K -1 0.08206 L  atm  mol -1  K -1 = -304 J

13. 6/14/2015 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. 6/14/2015 Engines Constant External Pressure  1.5 atm Expansion Stroke  1.5 atm Compression Stroke  1.5 atm

15. 6/14/2015 Engines 01234 0 1 2 Volume (L) P ext (atm) expansion cycle As described this does not represent a practical engine because the final state does not equal the initial state. compression cycle How much work is done in this overall process?

16. 6/14/2015  1.5 atm Isothermal Expansion Decreasing External Pressure 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. The initial and final temperature of the gas in the cylinder is the same. Sketch a graph of this process

17. 6/14/2015 Isothermal Expansion Decreasing External Pressure  4.5 atm 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.  2.25 atm  1.5 atm The initial and final temperature of the gas in the cylinder is the same. Sketch a graph of this process

18. 6/14/2015 01234 0 1 2 Volume (L) P ext (atm) 3 4 Isothermal Expansion Decreasing External Pressure

19. 6/14/2015 Isothermal Expansion Decreasing External Pressure PV = nRT Work may be represented as the area under the curve. work = -   VfVf ViVi P dV 01234 0 1 2 Volume (L) P ext (atm) 3 4 The pressure term is rewritten in terms of volume. = -   VfVf ViVi dV nRT V = - nRT ln VfVf ViVi     Isothermal Expansion Decreasing External Pressure

20. 6/14/2015 01234 0 1 2 Volume (L) P ext (atm) 3 4 Isothermal Expansion Decreasing External Pressure Is there any net heat flow in this process?  E = q + w the internal energy of a phase is a function of its temperature  E = q v = m  C v  T isothermal = 0 Is heat flowing into or out of the system?

21. 6/14/2015 01234 0 1 2 Volume (L) P ext (atm) 3 4 Isothermal Expansion Decreasing External Pressure The decreasing external pressure piston performs more work (greater efficiency) than a piston working against a constant external pressure equal to P final. Constant External Pressure

22. 6/14/2015 01234 0 1 2 Volume (L) P ext (atm) 3 4 Isothermal Expansion Decreasing External Pressure If the process is reversed, how much work is done? If the isothermal expansion process is reversed by isobaric compression, how much work is done?

23. 6/14/2015 Gas Expansion Room temperature gas colder gas When a gas expands against a low restraining pressure why does it cool?  E = q + w adiabatic expansion q = 0

24. 6/14/2015 Gas Expansion Carbon Dioxide Fire Extinguisher Why is it dangerous to point a carbon dioxide fire extinguisher at a person? liquid CO 2 gaseous CO 2

25. 6/14/2015 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. PV  = constant for an ideal diatomic gas,  =1.67

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27. 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?

28. 6/14/2015 Carnot Cycle  Consists of two isothermal and two adiabatic steps, occurring alternatively.  One of each type of step is involved in compression and expansion.

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32. Enthalpy H = E + PV Heat energy content of a molecule internal molecular motion electronic energy pressure-volume work required to maintain the volume of the molecule Only defined for a constant pressure process

33. Enthalpy Absolute enthalpy values cannot be measured, but changes in enthalpy can be measured relative to an arbitrary reference state

34. Enthalpy Heat of Formation Values H 2 (g) O 2 (g) C (gr) Reference State An element in its most stable phase at 25 ºC and 1 atm pressure  H f º = 0 H 2 O 2 (l) StandardState CH 3 OH (l) C 10 H 8 (s) StandardState An element, molecule or ion at 25 ºC and 1 atm pressure  H f º = -187.78 kJ·mol -1  H f º = +78.53 kJ·mol -1  H f º = -238.66 kJ·mol -1

35. Enthalpy Values An element in its most stable phase at 25 ºC and 1 atm pressure An element, molecule or ion at 25 ºC and 1 atm pressure Each element has only one reference state Enthalpy of formation values are listed in a  H f º table  H f º = 0 Reference State Standard State

36. HfºHfº change (final value - initial value) enthalpy standard state 25 ºC and 1 atm formation Is enthalpy a function of temperature? Enthalpy of Formation What do the initial and final values refer to?

37. Enthalpy of Formation Examples  H f º at 25 ºC and 1 atm for one mole of O 2 (g) oxygen:  H f º = 0 O 3 (g)3/2 O 2 (g) ozone:  H f º = +142.7 kJ·mol -1 CH 4 (g) C(gr) + 2 H 2 (g) methane:  H f º = -74.81 kJ·mol -1 Which of the reactions is exothermic?

38. Write the equation and determine the enthalpy of formation at 25 ºC and 1 atm for one mole of HCHO(g) formaldehyde: Hfº =Hfº = C(dia) diamond: Hfº =Hfº = KBr(s) potassium bromide: Hfº =Hfº = C(gr) + H 2 (g) + 1/2 O 2 (g) -108.57 kJ·mol -1 C(gr) +1.897 kJ·mol -1 K(s) + 1/2 Br 2 (l) -392.17 kJ·mol -1

39. 6/14/2015 Enthalpy of Reaction CO(g)C(gr) + 1/2 O 2 (g)  H f = -110.52 kJ·mol -1 CO 2 (g)C(gr) + O 2 (g)  H f = -393.51 kJ·mol -1 CO 2 (g)CO(g) + 1/2 O 2 (g)  H rx = ? What does a negative enthalpy change indicate? Can carbon monoxide burn exothermically?

40. 6/14/2015 Hess' Law The enthalpy change of a chemical reaction is equal to the difference of the enthalpy of the products and the enthalpy of the reactants. ReactantsProducts Elements  H rx  H f,products  H f,reactants  H rx =  H f,products -   H f,reactants

41. 6/14/2015 Hess' Law CO(g) + 1/2 O 2 (g)CO 2 (g) C(gr) + O 2 (g)  H rx  H rx =  H f,products -   H f,reactants  H rx = [  H f,CO2 ] - [  H f,CO + 1/2  H f,O2 ] = [-393.51 ] - [-110.52 + 1/2 · 0 ] = -282.99 kJ · mol -1

42. Enthalpy of Reaction Hess’ Law During the winter, a typical house will require about a half million kilojoules per day for heating. How many moles of natural gas (methane) would be needed? Is there an advantage to produce liquid water or steam as a by-product?

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44. Thermodynamic Applications Comparison of Liquid Fuels  methanol  ethanol  isooctane

45. Write the equation and determine the enthalpy of formation at 25 ºC and 1 atm for one mole of HCHO(g) formaldehyde: Hfº =Hfº = C(dia) diamond: Hfº =Hfº = KBr(s) potassium bromide: Hfº =Hfº = C(gr) + H 2 (g) + 1/2 O 2 (g) -108.57 kJ·mol -1 C(gr) +1.897 kJ·mol -1 K(s) + 1/2 Br 2 (l) -392.17 kJ·mol -1

46. 6/14/2015 Engines Constant External Pressure  1.5 atm Expansion Stroke Compression Stroke

47. 6/14/2015 Hess' Law   H rx =  H f,products -   H f,reactants  The system is exothermic if the enthalpy change is negative (  H rx 0).