Diagram shows the variation of internal energy ( U ) with the pressure ( P ) of 2 . 0 mole gas in cyclic process abcda . The temperature of gas at c and d are 300 K and 500 K , respectively. Calculate the heat absorbed by the gas during the process.

Step-by-step solution

Change in internal energy for cyclic process Δ U = 0 For process a  →  b ,   P - constant W a  →  b = P Δ V = n R Δ T = - 4 0 0  R For process b  →  c ,   T - constant W b  →  c = nRTln P i P f = - 2 R 300 ln   2 For process c  →  d ,   P - constant W c  →  d = P Δ V = nR Δ T= + 4 0 0 R For process d  →  a ,   T - constant W d &#16