e. interpretation of graphs 1. move away from origin  +T and +U 2. move toward y-axis (compression)  +Win 3. area under the curve = -Win f. Summary calculations for each process (T2 T – Process = Qin + Win U T1) Isometric 3/2PV 0 = PV/nR U (V = 0) 3/2nRT Isobaric 3/2PV PV/nR Win U – –PV (P = 0) 3/2nRT Isothermal 0 0 = -Win = -Qin (T = 0) Adiabatic = Win 0 = U (Q = 0) 2. heat cycle  a. (Carnot) system takes a body of gas through a complete cycle where = 0 U b. solving heat cycle problems determine T at each determine Qin, Win and for each process U isometric, Isobaric, use formulas Isothermal, adiabatic, use given values missing values, use: for any step: = Qin + Win U for = = 0, Qin Qout = Wout T U – – Win c. ideal (Carnot) efficiency: ec = (THigh – TLow)/THigh d. law efficiency: e = |Wout Win|/Qin – Waste = Qin(1 e) – e. heat engine vs. heat pump 1. heat engine takes in heat to perform work P Area  – Win = QL QH – V heat engine expands at TH, contracts at TL Win < 0 (work is done by the engine)  2. heat pump uses work to remove heat P + Area  Win = QH QL – V heat pump expands at TL, contracts at TH Win > 0 (work is done to the refrigerator) 
e. interpretation of graphs 1. move away from origin  +T and +U 2. move toward y-axis (compression)  +Win 3. area under the curve = -Win f. Summary calculations for each process Process T (T2 – T1) U = Qin + Win Isometric (V = 0) PV/nR 3/2PV 3/2nRT = U 0 Isobaric (P = 0) PV/nR 3/2PV 3/2nRT U – Win –PV Isothermal (T = 0) 0 0 = -Win = -Qin Adiabatic (Q = 0) = Win 0 = U 2. heat cycle  a. (Carnot) system takes a body of gas through a complete cycle where U = 0 b. solving heat cycle problems determine T at each determine Qin, Win and U for each process isometric, Isobaric, use formulas Isothermal, adiabatic, use given values missing values, use: for any step: U = Qin + Win for T = U = 0, Qin – Qout = Wout – Win c. ideal (Carnot) efficiency: ec = (THigh – TLow)/THigh d. law efficiency: e = |Wout – Win|/Qin Waste = Qin(1 – e) e. heat engine vs. heat pump 1. heat engine takes in heat to perform work P – Area  Win = QL – QH V heat engine expands at TH, contracts at TL  Win < 0 (work is done by the engine) 2. heat pump uses work to remove heat P + Area  Win = QH – QL V heat pump expands at TL, contracts at TH  Win > 0 (work is done to the refrigerator)
AP%20Physics%208%20Worksheet.docx