Saved Bookmarks
| 1. |
Figure shows the variation of internal energy (U) with the pressure (P) of 2.0 mole gas in cyclic process abeda. The temperature of gas at c and d are 300 and 500 K. How much will be the heat absorbed by the gas during the process? |
|
Answer» `400 R" In " 2` For process ` a to b, `(P - CONSTANT ) `W_(a to b) = P.dV = nRdT = - 400 R` For process `b to c `(T - constant ) `W_(b to c) = - 2R (300 ) ` In 2 For process` c to d` (T - constant) `W_(c to d) = + 400 R` For process `d to a ` (T - constant) `W_(d to a) = + 2R (500 )` In 2 Net work `(Delta W) = W_(a to b) + W_(b to c) + W_(c to d) + W_(d to a)` ` Delta W = 400 R " In " 2` `:. Delta Q = Delta U + Delta W` ` :. Delta Q = 400 R" In " 2` |
|