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Thermodynamic Processes

Thermodynamic Processes. Thermodynamic Processes.ppt. F. D x. D V. D V = A D x. Area. Work Done on a Gas. W = F D x. W= PA D x. W = P D V. Thermodynamic Processes.ppt. 1. Isotherms (lines of constant temperature). Pressure. T 4. T 3. T 2. T 1. Volume.

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Thermodynamic Processes

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  1. Thermodynamic Processes Thermodynamic Processes.ppt

  2. F Dx DV DV = A Dx Area Work Done on a Gas W = F Dx W= PA Dx W = P DV Thermodynamic Processes.ppt

  3. 1 Isotherms (lines of constant temperature) Pressure T4 T3 T2 T1 Volume Pressure - Volume Graph P Area under curve represents work Internal energy is proportional to temperature V Thermodynamic Processes.ppt

  4. W Q First Law of Thermodynamics System U Environment DU = ±Q ±W Thermodynamic Processes.ppt

  5. Thermodynamic Processes A. Isobaric B. Isovolumetric C. Isothermal D. Adiabatic Thermodynamic Processes.ppt

  6. Po DV Isobaric (Constant Pressure) Expansion P a) W = -PΔV  negative W b) DU increases T4 T3 c) Q = DU + W T2 T1 V DU = Q - W Thermodynamic Processes.ppt

  7. Po DV Isobaric (Constant Pressure) Compression a) W = -PDV  positive W P b) U decreases ΔU<0 T4 T3 c) Q = -DU - W T2 T1 V -DU = Q + W Thermodynamic Processes.ppt

  8. Po Pf Isovolumetric (Constant Volume) Decrease in Pressure P a) W = 0 (ΔV = 0) b) DU<0 (U decreases) T3 c) Q = -DU T2 T1 V DU = Q + W Thermodynamic Processes.ppt

  9. Pf Po Isovolumetric (Constant Volume) Increase in Pressure P a) W = 0 b) DU increases T3 c) Q = DU T2 T1 V DU = Q + W Thermodynamic Processes.ppt

  10. Po Pf Vo Vf Isothermal (Constant Temperature) Expansion P W< 0 b) DU = 0 T3 c) Q = W T2 T1 V DU = Q + W Thermodynamic Processes.ppt

  11. Pf Po Vo Vf Isothermal (Constant Temperature) Compression P W> 0 b) DU = 0 T3 c) Q = -W T2 T1 Note: ΔU=0 when: Isothermic process Cyclic process V DU = Q + W Thermodynamic Processes.ppt

  12. Po Pf Vo Vf Adiabatic (No Heat Exchange) Expansion P a) W = -PΔV negative W b) DU = decreases T2 c) Q = 0 W= -DU T1 V DU = Q + W Thermodynamic Processes.ppt

  13. Pf Po Vf Vo Adiabatic (No Heat Exchange) Compression P a) W = -PΔV  positive W b) DU = increases T2 c) Q = 0 W= DU T1 V DU = Q +W Thermodynamic Processes.ppt

  14. The internal energy of a gas increases during a thermodynamic process. Which of the following are possible processes? Single Concept I. adiabatic II. isothermal III. isobaric no Illustration (A) I and II only (B) I only (C) 1 and III only (D) III only (E) II and III only Multi- Response Type 2003 AP Physics Workshop, Ted Vittitoe

  15. An ideal gas expands to 10 times its original volume, maintaining a constant 440 K temperature. If the gas does 3.3 kJ of work on its surroundings, how much heat does it absorb? W = -3.3 kJ In an isothermal process DU = 0 Thermodynamic Processes.ppt

  16. Pa B 2.0 1.0 A C 50 25 One mole of a monatomic ideal gas, initially at point A is taken through a 3-process cycle as shown in the the PV diagram. Thermodynamic Processes.ppt

  17. Pa B 2.0 1.0 A C 50 25 The change in internal energy in each process DU = Q + W Thermodynamic Processes.ppt

  18. Pa B 2.0 1.0 A C 50 25 Heat added and exhausted in each process DU = Q + W Thermodynamic Processes.ppt

  19. Pa B 2.0 1.0 A C 50 25 Work done by the system in each process DU = Q + W Thermodynamic Processes.ppt

  20. Pa B 2.0 1.0 A C 50 25 Temperature Determine the temperature of 1 mol of gas at A, B, and C Thermodynamic Processes.ppt

  21. Pa B 2.0 1.0 A C 50 25 Internal Energy Determine the internal energy of the gas at A, B, and C Thermodynamic Processes.ppt

  22. Pa B 2.0 1.0 A C 50 25 Process AB Work: Heat: QAB = ΔUAB = UB – UA = 7480 J – 3740 J=3740J Internal Energy: Thermodynamic Processes.ppt

  23. Pa B 2.0 1.0 A C 50 25 Process BC Work: Internal Energy: Heat: Thermodynamic Processes.ppt

  24. Pa B 2.0 1.0 A C 50 25 Process CA Work: Internal Energy: Heat: QCA = ΔU - W QCA = -3740 J -2500 J Thermodynamic Processes.ppt

  25. Pa B 2.0 1.0 A C 50 25 Determine the net work done on the gas in one cycle Net work is represented by the enclosed area of the curve WAB = 0 J WBC = -3750 J WCA = 2500 J Thermodynamic Processes.ppt

  26. Pa B 2.0 1.0 A C 50 25 Determine the net heat absorbed by the gas in one cycle QAB = 3740 J QBC = 3750 J QCA = -6240 J Qnet = 1250 J Thermodynamic Processes.ppt

  27. 2P0 Po V0 2Vo 3Vo II An ideal gas initially has pressure Po, at volume Vo and absolute temperature To. It then undergoes the following series of processes: III I IV I. Heated, at constant volume to pressure 2Po II. Heated, at constant pressure to volume 3Vo III. Cooled, at constant volume to pressure Po IV. Cooled, at constant pressure to volume Vo Thermodynamic Processes.ppt

  28. 2P0 P0 V0 2Vo 3Vo II 2To 6To III I To IV 3To Find the temperature at each end point in terms of To Thermodynamic Processes.ppt

  29. II 2P0 III I Find the net work done by the gas in terms of Po and Vo P0 IV V0 2Vo 3Vo Net work equals net area under curve Thermodynamic Processes.ppt

  30. 2Po Po Vo 2Vo 3Vo II III I Find the net change in internal energy IV Thermodynamic Processes.ppt

  31. 2Po Po Vo 2Vo 3Vo II III I Find the net heat absorbed in terms of Po and Vo IV Thermodynamic Processes.ppt

  32. 2Po Po Vo 2Vo 3Vo II 2To 6To III I Find the heat transferred during process II in terms of Po and Vo IV Thermodynamic Processes.ppt

  33. Thermodynamic Processes END Thermodynamic Processes.ppt

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