My work (optional) 11. 11/20 points I Previous Answers FRKestenCP 5 P.100. My No
ID: 2078231 • Letter: M
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My work (optional) 11. 11/20 points I Previous Answers FRKestenCP 5 P.100. My Notes Ask Your Teacher Consider an engine in which the working substance is 1.23 mol of an ideal gas for which Y 1.41. The engine runs reversibly in the cycle shown on the PV diagram (see figure below). The cycle consists of an isobaric (constant pressure) expansion a at a pressure of 15.0 atm, during which the temperature of the gas increases from 300 K to 600 K followed by an isothermal expansion b until its pressure becomes 3.00 atm. Next is an isobaric compression c at a pressure of 3.00 atm, during which the temperature decreases from 600 K to 300 K, followed by an isothermal compression d until its pressure returns to 15 atm. Find the work done by the gas, the heat absorbed by the gas, the internal energy change, and the entropy change of the gas, first for each part of the cycle and then for the complete cycle. 15 atm 3 atm 600 K I 300 K VIV4 V2 Work done by the gas: 3.0679 9.87 Wc 3.07 Wd -4.94 Wtotal 4.93 Heat absorbed by the gas: 0.74 9.87 5.4 Qc The heat absorbed by a gas in an isobaric process is QP nCPAT; since we only know y, we will need to represent cp n terms of Y alone. The heat absorbed by aExplanation / Answer
> Heat absorbed by gas in process c = Cp*n* deltaT
= [gamma/(gamma-1)]R*n deltaT
= 1.41/0.41*8.3*1.23*(300-600)
= -10533 J
= - 10.53 kJ
> internal energy change in process a = heat absorbed - work done
= 10.53 - 3.06 kJ
= 7.47 kJ
> internal energy change in process c = heat absorbed - work done
= -10.53 + 3.06 kJ
= - 7.47 kJ
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