Consider an engine ... (See picture) Consider an engine in which the working sub
ID: 1419791 • Letter: C
Question
Consider an engine ... (See picture) Consider an engine in which the working substance is 1.23 mol of an ideal gas for which gamma = 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, and the internal energy change of the gas, first for each part of the cycle and then for the complete cycle. Work done by the gas: W_a = kJ W_b = kJ W_c = kJ W_d = kJ W_total = kJ Heat absorbed by the gas: Q_a = kJ Q_b = kJ Q_c = kJ Q_d = kJ Q_total = kJ Internal energy change of the gas: delta U_a = kJ delta U_b = kJ delta U_c = kJ delta U_d = kJ delta U_total = kJExplanation / Answer
Work done by gas in process a= nR deltaT = 1.23*8.3*(600-300)
= 3063 J
= 3.06 kJ
B)Heat absorbed by gas in a = Cp*n* deltaT
= [gamma/(gamma-1)]R*n deltaT
= 1.41/0.41*8.3*1.23*(600-300)
= 10533 J
= 10.53 kJ
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
Qtotal for cyclic process = Wtotal = 4.95 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
Total internal energy change = 0 J because it is cyclic process