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THERMODYNAMICS

Thermodynamics involves heat and work transfer in a closed system, affecting internal energy. Learn the principles and applications of thermodynamic processes.

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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. A THERMODYNAMIC SYSTEM • 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

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

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

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

  6. 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

  7. THE FIRST LAW OF THERMODYAMICS: • 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.

  8. Wout =120 J Qin 400 J Apply First Law: Q = U + W APPLICATION OF FIRST LAW OF THERMODYNAMICS 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?

  9. Wout =120 J Qin 400 J  U = +280 J Example 1 (Cont.): Apply First Law 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

  10. Wout =120 J Qin 400 J  U = +280 J Example 1 (Cont.): Apply First Law 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 The increase in internal energy is:

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

  12. 0 QIN QOUT -U +U HEAT IN = INCREASE IN INTERNAL ENERGY HEAT OUT = DECREASE IN INTERNAL ENERGY ISOCHORIC PROCESS: CONSTANT VOLUME, V = 0, W = 0 Q = U + W so that Q = U No Work Done

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

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

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

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

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

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

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

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

  21. AB: Heated at constant V to 400 K. Example Problem: 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.

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

  23. 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 PROCESS AB: ISOCHORIC What is the pressure at point B?

  24. 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 PROCESS BC: ISOBARIC What is the volume at point C (& D)?

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

  26. 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 PROCESS CD: ISOCHORIC What is temperature at point D?

  27. 400 K 800 K 2 atm A 200 K 400 K 1 atm D 2 L 4 L W = -200 J FINDING W FOR PROCESS DA. Work depends on change inV. P = 0 Work = PV W= (1 atm)(2 L - 4 L) = -2 atm L = -200 J

  28. +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 2 atm Area = (1 atm)(2 L) Net Work = 2 atm L = 200 J 1 atm 2 L 4 L NET WORK FOR COMPLETE CYCLE IS ENCLOSED AREA

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

  30. 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

  31. 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

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

  33. Hot Res. TH 800 J W QC QH Engine e = 1 - 600 J 600 J 800 J Cold Res. TC e = 1 - e = 25% EFFICIENCY EXAMPLE An engine absorbs 800 J and wastes 600 J every cycle. What is the efficiency? Question: How many joules of work is done?

  34. Hot Res. TH QH W Engine TH- TC TH QC e = Cold Res. TC TC TH e = 1 - EFFICIENCY OF AN IDEAL ENGINE (Carnot Engine) For a perfect engine, the quantities Q of heat gained and lost are proportional to the absolute temperatures T.

  35. TC TH e = 1 - W QH e = 300 K 500 K e = 1 - Work = 120 J Example 3:A steam engine absorbs 600 J of heat at 500 K and the exhaust temperature is 300 K. If the actual efficiency is only half of the ideal efficiency, how much work is done during each cycle? Actual e = 0.5ei = 20% W = eQH = 0.20 (600 J) e = 40%

  36. Hot Res. TH Qhot Win Engine Qcold Cold Res. TC WIN = Qhot - Qcold REFRIGERATORS A refrigerator is an engine operating in reverse: Work is done on gas extracting heat from cold reservoir and depositing heat intohot reservoir. Win + Qcold = Qhot

  37. Hot Res. TH Qhot Engine Qcold Cold Res. TC THE SECOND LAW FOR REFRIGERATORS It is impossible to construct a refrigerator that absorbs heat from a cold reservoir and deposits equal heat to a hot reservoir with W = 0. If this were possible, we could establish perpetual motion!

  38. TheFirst Law of Thermodynamics:The net heat taken in by a system is equal to the sum of the change in internal energy and the work done by the system. Q = U + W final - initial) Summary • Isochoric Process: V = 0, W = 0 • Isobaric Process: P = 0 • Isothermal Process: T = 0, U = 0 • Adiabatic Process: Q = 0

  39. TheSecond Law of Thermo: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. Hot Res. TH Qhot Wout Engine Qcold Cold Res. TC Summary (Cont.)

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