One mole of an ideal monatomic gas undergoes a cyclic process as shown in the figure. If the temperature of the gas at state 1 is 300 K and at state 4 is 500 K , then heat exchanged during process 2 → 3 , is

Step-by-step solution

Q 23 = nC P T 3 – T 2 The processes 1 → 2 and 3 → 4 are isochoric in nature. Hence, for these processes T ∝ P . As a result, we obtain, T 3 = 1000   K T 2 = 600   K Q 23 = 1 × 5 R 2 × 400 = 1000 R