Diagram shows the variation in the internal energy U with the volume V of 2.0 moles of an ideal gas in cyclic process abcda. The temperatures of the gas at b and c are 500 K and 300 K, respectively. Calculate the heat absorbed by the gas during the process.

Step-by-step solution

In the process a to b and c to d as Δ U = 0 , therefore Δ T = 0 or T = constant W = ∫ V i V f ⁡ PdV We have, PV = nRT ⇒ P = nRT V ∴              W = ∫ V i V f ⁡ nRT RV V = nRT ln V V i V f ⁡ = nRT ln V f ⁡ V i W ab = nRT b ln 2 V 0 V 0 = 2 R × 5 0 0 ln 2 = 1 0 0 0 R ln 2 and W cd = nRT c ln V 0 2 V 0 = 2 R × 3 0 0 ln 1 2 = - 6 0 0 R ln 2 There is no volume change from b to c and from