One mole of an ideal gas is subjected to a cyclic process 1 2 3 1. The states of the gas are given by their Temperature (T) and Volume (V) as follows: State 1 is (600 K, 20 L), State 2 is (300 K, 20 L), and State 3 is (300 K, 10 L). The process 1 2 is isochoric, 2 3 is isothermal, and 3 1 is an isobaric process. The magnitude of the net work done by the gas in the cycle is _______ J. [Given : R = 8.3 J K ⁻¹ mol ⁻¹ ,

Step-by-step solution

For the process 1 2 (isochoric): Since volume is constant, the work done W_ 1 2 = 0 . For the process 2 3 (isothermal): The work done is given by W_ 2 3 = -nRT ( V₃ V₂ ) W_ 2 3 = -1 8.3 300 ( 10 20 ) W_ 2 3 = -2490 (- 2) = 2490 0.7 = 1743 J. For the process 3 1 (isobaric): The work done is given by W_ 3 1 = -P(V₁ - V₃) = -nR(T₁ - T₃) W_ 3 1 = -1 8.3 (600 - 300) = -2490 J. The net work done in the cycle is: W_ net = W_ 1 2 + W_ 2 3 + W_ 3 1 W_ net = 0 + 1743 - 2490 = -747 J. The magnitude of the net work done is 747