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***Work… Work is the area under the curve Work for an isobaric process Work in an isochoric process Work for an isometric process Work in an isothermal expansion or compression of an ideal gas Work for an isothermal process What is the work in an isothermal process... Work for an adiabatic process Work in an adiabatic expansion or compression of an ideal monatomic gas Work is greatest for idiabats Work for a cyclic process Wnet = Area ***Cycles… One mole of an ideal monatomic gas is taken through the cycle shown. Assume that p = 2p0, V = 2V0, p0 = 1.08x105 Pa, and V0 = 0.0244 m3. What is the efficiency of this cycle? What is the ideal efficiency of a heat engine operating between the highest and the lowest temperature? 2T0 , V0 ,2 P0 4T0 ,2V0 ,2 P0 W = p0∙V0 Qabc = nCv∙(Tb-Ta)+nCp∙(Tc-Tb) = Qabc = n∙(3R/2)∙(Tb-Ta)+n∙(5R/2)∙(Tc-Tb) = = nRTa∙[3/2∙(Tb/Ta-1)+5/2∙(Tc/Ta-Tb/Ta)] = = p0∙V0∙[3/2∙(2-1)+5/2∙(4-2)] = 13/2∙(p0∙V0) T0 , V0 , P0 2T0 ,2V0 , P0 ε = W/Qh = 2/13 = 15.4% εideal = 1-Ta/Tc = 1 – ¼ = 75% ***Suppose the working substance of this heat engine is 0.75 kg of an ideal monatomic gas. How would the efficiency of this cycle compare with that of a Carnot engine operating between the highest and lowest temperatures of this cycle? W = Area = (250-150)·(2.25-1.75)·10 = 500 J Q412 = nCv∙(T1-T4)+nCp∙(T2-T1) = Q412 = n∙(3R/2)∙(T1-T4)+n∙(5R/2)∙(T2-T1) = = nRT4∙[3/2∙(T1/T4-1)+5/2∙(T2/T4-T1/T4)] = = p4∙V4∙[3/2∙(250/150-1)+5/2∙((250/150)(2.25/1.75)-(250/150)] = 2.19∙(p4∙V4) = 5749 J ε = W/Qh = 8.7% εideal = 1-T4/T2 = 1 – (150/250)·(1.75/2.25) = 53.3% ***Suppose 5 moles of a monatomic ideal gas follows the three-part cycle shown. What is the efficiency of the cycle? What is the ideal efficiency of an engine operating between the highest and lowest temperature of this cycle? TA 3 TC TB 4 TC Wnet Wab Wbc Wca Qin QCA Q AB nCv ·(T - TC ) 15 nR TC A 2 1 P V 1 PA PC VB VC 1 2 PC 3VC 3PC VC 3 nR TC 2 2 2 n 3 ·R 2TC 15 nR TC 9 nR TC 2 2 U AB 3 nR TB TA 3 nR TC 3 nR TC 2 2 e real Wnet /Qin 33% U AB QAB WAB QAB 2 PC 3VC 9 nR TC QAB 15 nR TC 2 eideal 1 - TC /TB 1 - 1/4 75% Suppose 9 moles of a monatomic ideal gas follows the three-part cycle, the first part of which involves an isothermal expansion from point A, where V = 7.5x10-5 m3 and P=6x105 Pa, to point B, where the pressure is lower by a factor of 6. What is the efficiency of the cycle? What is the ideal efficiency of an engine operating between the highest and lowest temperature of this cycle? P V P V Qin Q ca Q ab nC v ·(T - Tc ) Wab n(3/2)·R a a c c – Wab a Wab nRTab ·ln(Vb /Va ) Pa Va ·ln(P /Pb ) Pa Va ·ln(6) a nR Wbc Pbc ·Vbc Pbc 5Vc 3/2 5/6·P Va Pa Va ·ln(6) Pa Va ·5/4 ln(6) a Wnet Wab Wbc Wca e real Wnet /Qin ln(6) - 5 / 6 31.5% Pa Va ·ln(6)- Pbc 5Vc 0 Pa Va ·ln(6) - 5 / 6 5/4 ln(6) eideal 1 - Tc /Ta 1 - 1/6 83.3% 9 moles of a monatomic ideal gas follows the 3-part cycle as shown. The first part of which involves an isothermal expansion from point A, where V = 3x10-5 m3 and the pressure is 6x105 Pa, to point B, where the pressure is lower by a factor of 6. Fill in the missing entries and determine the actual and ideal efficiencies. Q(J) W(J) ΔU(J) A B 32.3 32.3 0 e real Wnet /Qin 17.3/54.8 31.6% BC -37.5 -15 -22.5 eideal 1 - Tc /Ta 1 - 1/6 83.3% CA 22.5 0 22.5 Two moles of a monatomic ideal gas are taken through the reversible cycle shown. Process bc is an adiabatic expansion, with Pb = 11.9 atm and Vb = 10 L. Find the efficiency of the cycle and the maximum efficiency for the temperature extremes of the cycle. V 5/3 1 Pa Pc b Pb V 11.9 0.372 atm c 8 3 (P P ) Vb 3 Q in Q ab nC v (Tb Ta ) n RC v b a (Pb Pa ) Vb 17.5 kJ 2 nR 2 Q out Q ca nCp T Pa Va Pc Vc Pa Va Vb 6.6 kJ 5 5 2 2 W 17.5 6.6 completecycle: W Q e act 62% Q in 17.5 Pa eideal 1 Ta / Tb 1 96.9% Pb ***Five moles of an ideal monatomic gas expand adiabatically, and its temperature decreases from 362 to 292 K. Determine the work done by the gas and the change in its internal energy. 3 W Q U U nRT 4363 J 2 ***In the cycle shown, 1 mol of an ideal diatomic gas is initially at a pressure of 1 atm and a temperature of 0°C. The gas is heated at constant volume to T2 = 141°C and is then expanded adiabatically until its pressure is again 1 atm. It is then compressed at constant pressure back to its original state. Find the efficiency of this cycle and the efficiency of a Carnot cycle operating between the temperature extremes of this cycle. Two moles of a diatomic gas are taken through the cycle ABCA as shown on the PV diagram. At A, the pressure and temperature are 4 atm and 592 K, respectively. The volume at B is twice that at A. The segment BC is an adiabatic expansion and the segment CA is an isothermal compression. Find the efficiency of this cycle and the efficiency of a Carnot cycle operating between the temperature extremes of this cycle. Wab Pa 2Va Va Pa Va nRTa A: 4 atm, 592 K 5 Wbc U n R Tc Tb 2 B: 4 atm, 2VA, TB=2TA 5 5 n R Ta 2Ta n R Ta 2 2 Wca nRTa ln Va / Vc nRTa ln Vb / 2Vc PbVb PcVc or TbVb 1 TcVc 1 Vc = Vb (Tb /Tc )1/( -1) Vb (2)1/( -1) Qin Qab nC p Tb Ta nR 2Ta Ta nR Ta 7 7 2 2 5 1 nRTa nRTa nRTa ln 2 (2)1/( -1) ereal Wtot / Qin 2 30.7% 7 nR T 2 ***Two moles of a diatomic gas are carried through the cycle ABCDA shown in the PV diagram. The segment AB represents an isothermal expansion, the segment BC an adiabatic expansion. The pressure and temperature at A are 5 atm and 563 K. The volume at B is twice that at A. The pressure at D is 1 atm. What is the efficiency of this cycle? What is the efficiency of a Carnot cycle operating between the temperature extremes of this cycle? Wab nRTa ln Vb / Va nRTa ln 2 A: P=5 atm, TA=563 K 5 ??? Wbc U n R Tc Tb B: 4 atm, 5VA/4, TB=TA 2 5 5 n R Ta 2Ta n R Ta 2 2 Pb Vb Pc Vc or Tb Vb 1 Tc Vc 1 Vc = Vb (Tb /Tc )1/( -1) Vb (2)1/( -1) Q in Q ab nCp Tb Ta nR 2Ta Ta nR Ta 7 7 D: P=1 atm, TD=TA/5 2 2 5 1 C: P=1 atm, TC=???TA/5 nRTa nRTa nRTa ln 2 (2)1/( -1) e real Wtot / Q in 2 30.7% 7 nR T 2 ***How much work must be done to 22 grams of CO at standard temperature and pressure to compress it to 1/5th of its initial volume if the process is isothermal? How much work must be done if the process is adiabatic? What if the gas was CO2? *** An air pump has a cylinder 0.280 m long with a movable piston. The pump is used to compress air from the atmosphere (at absolute pressure 1.01 105 Pa) into a very large tank at 4.50x105 Pa gauge pressure. (For air , Cv = 20.8 J/mol · K.) The piston begins the compression stroke at the open end of the cylinder. How far down the length of the cylinder has the piston moved when air first begins to flow from the cylinder into the tank? If the air is taken into the pump at 27.0°C, what is the temperature of the compressed air? Assume that the compression is adiabatic. ***A sample of an ideal gas goes through the process shown in Figure P20.32. From A to B, the process is adiabatic; from B to C, it is isobaric with 98 kJ of energy entering the system by heat. From C to D, the process is isothermal; from D to A, it is isobaric with 160 kJ of energy leaving the system by heat. Determine the difference in internal energy, Eint, B - Eint, A. ***Argon enters a turbine at a rate of 82.0 kg/min, a temperature of 850°C, and a pressure of 1.40 MPa. It expands adiabatically as it pushes on the turbine blades and exits at pressure 350 kPa. Calculate the maximum efficiency of the turbine. 0.5 mole of a diatomic gas, γ = 1.4, is taken through the cycle shown. Given that V1 = 5 L, V2 = 15 L, Th = 570°C, and Tc = 290°C, calculate the efficiency of the cycle from the definition e=W/Qin and compare to the maximum possible. V1 5 L, T1 570 C V2 15 L, T2 570 C T4 V4 1 T1V1 1 V4 13.7 L, T4 290 C T2 V2 1 T3 V3 1 V3 41.2 L, T3 290 C Q h Q12 W12 nRTh ln( V2 / V1 ) 3848 J Q c Q34 W34 nRTc ln( V4 / V3 ) 2573 J W Q h Q c 1275 J e W / Qh 0.33 e max 1 Tc / Th 0.33 ***What is the efficiency of this Carnot engine? Nicholas Léonard Sadi Carnot (1796 - 1832) Qh nRTh ln(V2 V1 ) Qc nRTc ln(V3 V4 ) P2 V2 P3 V3 and P4 V4 P1V1 T2 V2 1 T3 V3 1 and T4 V4 1 T1V1 1 Th V2 1 Tc V3 1 and Tc V4 1 Th V1 1 (V2 / V1 ) 1 (V3 / V4 ) 1 ln( V2 V1 ) ln( V3 V4 ) or Q c / Q h Tc / Th W / Qh (Qh Qc ) Qh 1 Tc Th Carnot cycle in a P-V and T-S diagrams P-V relationship for an adiabatic path dU ncv dT PdV PdV VdP nRdT R PdV VdP PdV cv c p cv dV 1 dV dP dV c V V P v V dP dV 0 P V ln P ln V constant PV constant ***Legs 1 and 3 are adiabats; legs 2 and 4 are isomets. Can this cycle achieve ideal efficiency? What does this plunger have to do with a diesel engine? The fuel-air mixture in the cylinder of a diesel engine at 20ºC is compressed from an initial pressure of 1 atm and volume of 800 cm3 to a volume of 60 cm3. Assuming the mixture behaves as an ideal gas with γ=1.40 and that the compression is adiabatic, find the final pressure and temperature of the mixture. Pf = Pi·(Vi/Vf)γ = 1·(800/60)1.4 = 37.6 atm Tf = (Pf·Vf)/(Pi·Vi)·Ti = 826 K ***Equipartition of energy... Molecular model of ideal gas 2 2mvxi 2mvxi mvxi Fi t (2 L / vx ) L 1 N 2 F mvxi 2 N P mv 2 A i 1 AL 3 V 2 2 1 PV N mv 2 3 2 1 2 3 mv k B T 2 2 Equipartition and specific heat of a diatomic gas Specific heats of gases TABLE 21.2 Molar Specific Heats of Various Gases Molar Specific Heat [J/mol·K] Gas CP CV CP - CV γ = CP / CV Monatomic Gases He 20.8 12.5 8.33 1.67 Ar 20.8 12.5 8.33 1.67 Ne 20.8 12.7 8.12 1.64 Kr 20.8 12.3 8.49 1.69 Diatomic Gases H2 28.8 20.4 8.33 1.41 N2 29.1 20.8 8.33 1.40 O2 29.4 21.1 8.33 1.40 CO 29.3 21.0 8.33 1.40 Cl2 34.7 25.7 8.96 1.35 Polyatomic Gases CO2 37.0 28.5 8.50 1.30 SO2 40.4 31.4 9.00 1.29 H2 O 35.4 27.0 8.37 1.30 CH4 35.5 27.1 8.41 1.31 All specific heats except those for water were obtained at 300 K. ***Specific heat vs molecular structure The difference between Cv and Cp dU dQ PdV dU p nC p dT nRdT dU v nCv dT dU p dU v (U U (T )) nCv dT nC p dT nRdT C p Cv R What is the specific heat of solids? Actual heat capacities for some solids Otto cycle