InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A reversible heat engine carries 1 mole of an ideal monatomic gas aroung the cycle 1-2-3-1. Process 1-2 takes place at constant volume, process 2-3 take place at constant volume, process 2-3 is adiabatic, and process 3-1 takes place at constant pressure. Complete the values for the heat `Delta Q`, the change in internal energy `Delta U`, and the work done `Delta `, for each of the three processes and for the cycle as a whole. |
|
Answer» For the process 1-2, we have `Delta W = 0` (since volume ramains constant) `Delta U = C, Delta T = (R )/(gamma - 1) Delta T = (8.3)/(((3)/(3))-1) xx (600 - 300) = 3735 J` For the process 2-3 we have `Delta Q = 0` (since the process is adiabatic) `Delta W = (p_(1) V_(1) - p_(2) V_(2))/(gamma - 1) = (R (T_(1) - T_(2)))/((gamma - 1)) = (8.3 (600 - 455))/(((5)/(3))- 1) = 1895 J` `Delta U = Delta Q - Delta W = 0 - 1805 = - 1805 J` For the process 3-1, we have `Delta Q = C_(P) Delta T = (gamma R)/((gamma - 1)) xx Delta T` or `Delta Q = (((5)/(3)) xx 8.3)/(((5)/(3))) xx (300 - 455) = - 3216 J` `Delta U = C_(V) Delta T = (R )/((gamma - 1)) Delta T = (8.3)/(((5)/(3)) -1) (300 - 455) = - 1286 J` `Delta W = Delta Q - Delta U = - 3116 - (- 1930) = - 1286 J` |
|