Consider an engine in which the working substance is 1.23 mol of an ideal gas fo
ID: 1419830 • Letter: C
Question
Consider an engine in which the working substance is 1.23 mol of an ideal gas for which = 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:
Heat absorbed by the gas:
Internal energy change of the gas:
Ua
Ub
Uc
Ud
Utotal
Wa = kJ Wb = kJ Wc = kJ Wd = kJ Wtotal = kJExplanation / Answer
Work done by the gas:
Wa = 1.23 * 8.314 * (600 - 300)
= 3067.86 J = 3.067 kJ
Wb = 1.23 * 8.314 * 600 * ln(5)
= 9.875 kJ
Wc = - 1.23 * 8.314 * (600 - 300)
= - 3067.86 J = - 3.067 kJ
Wd = - 1.23 * 8.314 * 600 * ln(5)
= - 9.875 kJ
Wtotal = 0 kJ
Heat absorbed by the gas:
Qa = 3/2 * 8.314 * 300 * 1.23 = 4.601 kJ
Qb = 9.875 kJ
Qc = - 4.601 kJ
Qd = - 9.875 kJ
Qtotal = 0 kJ
Internal energy change of the gas:
Ua = 4.601 - 3.067 = 1.534 kJ
Ub = 0 kJ
Uc = - 1.534 kJ
Ud = 0 kJ
Utotal = 0 kJ