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THERMODYNAMICS

Thermodynamics is the study of energy relationships that involve heat, mechanical work, and other aspects of energy and heat transfer. Central Heating. THERMODYNAMICS. State and apply the first and second laws of thermodynamics.

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THERMODYNAMICS

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  1. Thermodynamics is the study of energy relationships that involve heat, mechanical work, and other aspects of energy and heat transfer. Central Heating THERMODYNAMICS

  2. State and apply thefirstand second lawsofthermodynamics. • Demonstrate your understanding ofadiabatic, isochoric, isothermal, and isobaric processes. • Write and apply a relationship for determining theideal efficiencyof a heat engine. Objectives: After finishing this unit, you should be able to: • Write and apply a relationship for determiningcoefficient of performancefor a refrigeratior.

  3. A system is a closed environment in which heat transfer can take place. (For example, the gas, walls, and cylinder of an automobile engine.) Work done on gas or work done by gas A THERMODYNAMIC SYSTEM

  4. Usually the internal energy consists of the sum of the potential and kinetic energies of the working gas molecules. • The internal energy U of a system is the total of all kinds of energy possessed by the particles that make up the system. INTERNAL ENERGY OF SYSTEM

  5. +U WORK DONE ON A GAS (Positive) HEAT PUT INTO A SYSTEM (Positive) TWO WAYS TO INCREASE THE INTERNAL ENERGY, U.

  6. Wout Qout hot hot HEAT LEAVES A SYSTEM Q is negative -U Decrease TWO WAYS TO DECREASE THE INTERNAL ENERGY, U. WORK DONE BY EXPANDING GAS: W is positive

  7. THERMODYNAMIC STATE The STATE of a thermodynamic system is determined by four factors: • Absolute Pressure P in Pascals • Temperature T in Kelvins • Volume V in cubic meters • Number of moles, n, of working gas

  8. Wout Qin Heat input Initial State: P1 V1 T1 n1 Final State: P2 V2 T2 n2 Increase in Internal Energy, U. Work by gas THERMODYNAMIC PROCESS

  9. Win Qout Initial State: P1 V1 T1 n1 Final State: P2 V2 T2 n2 Work on gas Decrease in Internal Energy, U. Loss of heat The Reverse Process

  10. The net heat put into a system is equal to the change in internal energy of the system plus the work done BY the system. Q = U + W final - initial) • Conversely, the work done ON a system is equal to the change in internal energy plus the heat lost in the process. THE FIRST LAW OF THERMODYAMICS:

  11. +Wout +Qin -Win -Qout Q = U + W final - initial) • Heat Q input is positive U • Work BY a gas is positive U • Work ON a gas is negative • Heat OUT is negative SIGN CONVENTIONS FOR FIRST LAW

  12. Wout =120 J Qin 400 J Apply First Law: Q = U + W Example 1:In the figure, the gas absorbs400 Jof heat and at the same time does120 Jof work on the piston. What is the change in internal energy of the system? APPLICATION OF FIRST LAW OF THERMODYNAMICS

  13. Wout =120 J Qin 400 J  U = +280 J DQ is positive: +400 J (Heat IN) DW is positive: +120 J (Work OUT) Q = U + W U = Q - W U= Q - W = (+400 J) - (+120 J) = +280 J Example 1 (Cont.): Apply First Law

  14. Wout =120 J Qin 400 J  U = +280 J Energy is conserved: The 400 J of input thermal energy is used to perform 120 J of external work, increasing the internal energy of the system by 280 J Example 1 (Cont.): Apply First Law The increase in internal energy is:

  15. Q = U + W • Isochoric Process: V = 0, W = 0 • Isobaric Process: P = 0 • Isothermal Process: T = 0, U = 0 • Adiabatic Process: Q = 0 FOUR THERMODYNAMIC PROCESSES:

  16. 0 QIN QOUT -U +U HEAT IN = INCREASE IN INTERNAL ENERGY HEAT OUT = DECREASE IN INTERNAL ENERGY Q = U + W so that Q = U No Work Done ISOCHORIC PROCESS: CONSTANT VOLUME, V = 0, W = 0

  17. No Change in volume: P2 B PA P B = TA T B P1 A V1= V2 400 J 400 Jheat input increases internal energy by 400 Jand zero work is done. Heat input increases P with const. V ISOCHORIC EXAMPLE:

  18. QIN QOUT Work Out Work In HEAT IN = Wout + INCREASE IN INTERNAL ENERGY Q = U + W But W = P V -U +U ISOBARIC PROCESS: CONSTANT PRESSURE, P = 0 HEAT OUT = Wout + DECREASE IN INTERNAL ENERGY

  19. B A VA VB P = TA T B V1 V2 400 J 400 Jheat does 120 J of work, increasing the internal energy by 280 J. Heat input increases Vwith const. P ISOBARIC EXAMPLE (Constant Pressure):

  20. A B P VA VB = TA T B V1 V2 400 J PA = PB Work = Area under PV curve ISOBARIC WORK

  21. Q = U + W AND Q = W QIN QOUT Work Out Work In NET HEAT INPUT = WORK OUTPUT U = 0 U = 0 ISOTHERMAL PROCESS: CONST. TEMPERATURE, T = 0, U = 0 WORK INPUT = NET HEAT OUT

  22. A PA B PB V2 V1 PAVA =PBVB U = T = 0 Slow compression at constant temperature: ----- No change in U. ISOTHERMAL EXAMPLE (Constant T):

  23. A PA B PAVA = PBVB TA = TB PB VA VB U = T = 0 400 J of energy is absorbed by gas as 400 J of work is done on gas. T = U = 0 Isothermal Work ISOTHERMAL EXPANSION (Constant T):

  24. U = -W W = -U Work Out Work In U +U Q = 0 Work done at EXPENSE of internal energyINPUT Work INCREASES internal energy Q = U + W ; W = -U or U = -W ADIABATIC PROCESS: NO HEAT EXCHANGE, Q = 0

  25. A PA B PB V1 V2 Expanding gas does work with zero heat loss. Work = -DU ADIABATIC EXAMPLE: Insulated Walls: Q = 0

  26. A PA PAVA PBVB B = PB TA T B Q = 0 VA VB 400 J of WORK is done, DECREASING the internal energy by 400 J: Net heat exchange is ZERO. Q = 0 ADIABATIC EXPANSION:

  27. PAVA PBVB PV = nRT = TA T B Q = U + W U = nCv T REMEMBER, FOR ANY PROCESS INVOLVING AN IDEAL GAS:

  28. AB: Heated at constant V to 400 K. A 2-L sample of Oxygen gas has an initial temp-erature and pressure of 200 K and 1 atm. The gas undergoes four processes: • BC: Heated at constant P to 800 K. • CD: Cooled at constant V back to 1 atm. • DA: Cooled at constant P back to 200 K. Example Problem:

  29. B 400 K 800 K PB A 200 K 1 atm 2 L How many moles of O2 are present? Consider point A: PV = nRT PV-DIAGRAM FOR PROBLEM

  30. B 400 K 800 K PB A 200 K 1 atm PA P B = 2 L TA T B 1 atmP B P B = 2 atm = 200 K400 K or 203 kPa What is the pressure at point B? PROCESS AB: ISOCHORIC

  31. B 400 K 800 K PB A 200 K 1 atm 2 L Q = +514 J U = +514 J W = 0 Analyze first law for ISOCHORIC process AB. W = 0 Q = U = nCv T U = (0.122 mol)(21.1 J/mol K)(400 K - 200 K) PROCESS AB: Q = U + W

  32. B 400 K 800 K PB C 200 K VB V C 1 atm D = TB T C 2 L 4 L 2 LV C V C = V D = 4 L = 400 K800 K What is the volume at point C (& D)? PROCESS BC: ISOBARIC

  33. B 400 K 800 K 2 atm C 200 K 1 atm 2 L 4 L U = +1028 J Process BC is ISOBARIC. P = 0 U = nCv T U= (0.122 mol)(21.1 J/mol K)(800 K - 400 K) FINDING U FOR PROCESS BC.

  34. B 400 K 800 K 2 atm C 200 K 1 atm 2 L 4 L W = +405 J Work depends on change in V. P = 0 Work = PV W = (2 atm)(4 L - 2 L) = 4 atm L = 405 J FINDING W FOR PROCESS BC.

  35. B 400 K 800 K 2 atm C 200 K 1 atm 2 L 4 L Q = 1433 J U = 1028 J W = +405 J Analyze first law for BC. Q = U + W Q = +1028 J + 405 J Q = +1433 J FINDING Q FOR PROCESS BC.

  36. B 400 K 800 K PB C A 200 K D 1 atm PC P D = 2 L TC T D 2 atm1 atm T D = 400 K = 800 KTD What is temperature at point D? PROCESS CD: ISOCHORIC

  37. 800 K 400 K C PB 200 K 400 K 1 atm D 2 L Q = -1028 J U = -1028 J W = 0 Analyze first law for ISOCHORIC process CD. W = 0 Q = U = nCv T U = (0.122 mol)(21.1 J/mol K)(400 K - 800 K) PROCESS CD: Q = U + W

  38. 400 K 800 K 2 atm A 200 K 400 K 1 atm D 2 L 4 L W = -203 J Work depends on change inV. P = 0 Work = PV W= (1 atm)(2 L - 4 L) = -2 atm L = -203 J FINDING W FOR PROCESS DA.

  39. 400 K 800 K 2 atm A 200 K 400 K 1 atm D 2 L 4 L Q = -717 J U = -514 J W = -203 J Analyze first law for DA. Q = U + W Q = -514 J - 203 J Q = -717 J FINDING Q FOR PROCESS DA.

  40. For all processes: DQ = DU + DW PROBLEM SUMMARY

  41. +404 J B C B C -202 J 2 atm 2 atm Neg 1 atm 1 atm 2 L 4 L 2 L 4 L B C Area = (1 atm)(2 L) Net Work = 2 atm L = 202 J 2 atm 1 atm 2 L 4 L NET WORK FOR COMPLETE CYCLE IS ENCLOSED AREA

  42. Example 2: A diatomic gas at 300 K and 1 atm is compressed adiabatically, decreasing its volume by 1/12. (VA = 12VB). What is the new pressure and temperature? ( = 1.4)   B PAVA = PBVB PB A PA PAVA PBVB = TA T B VB VA ADIABATIC EXAMPLE: Q = 0

  43.  B PAVA = PBVB PB Solve for PB: 300 K 1 atm A Q = 0 VB 12VB PB = 32.4 atm or 3284 kPa ADIABATIC (Cont.): FIND PB

  44. TB=? B 32.4 atm 300 K 1 atm A Solve for TB Q = 0 VB 12VB (1 atm)(12VB)(32.4 atm)(1 VB) = (300 K)T B TB = 810 K ADIABATIC (Cont.): FIND TB

  45. B 810 K Since Q = 0, W = - U 32.4 atm 300 K 1 atm A Q = 0 8 cm3 96 cm3 PV RT n = W = - U = - nCVT & CV= 21.1 j/mol K ADIABATIC (Cont.): If VA= 96 cm3and VA= 8 cm3, FIND W Find n from point A PV = nRT

  46. PV RT (101,300 Pa)(8 x10-6 m3) (8.314 J/mol K)(300 K) n = = B 810 K 32.4 atm 300 K 1 atm A W = - 3.50 J 8 cm3 96 cm3 n = 0.000325 mol & CV= 21.1 j/mol K T = 810 - 300 = 510 K W = - U = - nCVT ADIABATIC (Cont.): If VA= 96 cm3and VA= 8 cm3, FIND W

  47. Hot Res. TH Qhot Wout Engine Qcold Cold Res. TC • Absorbs heat Qhot • Performs work Wout • Rejects heat Qcold A heat engine is any device which through a cyclic process: HEAT ENGINES

  48. Hot Res. TH It is impossible to construct an engine that, operating in a cycle, produces no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work. Qhot Wout Engine Qcold Cold Res. TC THE SECOND LAW OF THERMODYNAMICS Not only can you not win (1st law); you can’t even break even (2nd law)!

  49. Hot Res. TH Hot Res. TH 400 J 400 J 100 J 400 J Engine Engine 300 J Cold Res. TC Cold Res. TC • An IMPOSSIBLE engine. • A possible engine. THE SECOND LAW OF THERMODYNAMICS

  50. Hot Res. TH QH W W QH QH- QC QH Engine e = = QC Cold Res. TC QC QH e = 1 - The efficiency of a heat engine is the ratio of the net work done W to the heat input QH. EFFICIENCY OF AN ENGINE

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