Consider the reversible processes for 1.0 mol of an ideal gas as shown in the figure. \( w_1, w_2, w_3 \) and \( w_4 \) represent work done (in calories) in the processes 1, 2, 3 and 4, respectively; \( \Delta U_2 \) and \( \Delta U_4 \) are changes in the internal energy for the processes 2 and 4, respectively. [use \( R = 2 \text{ cal K}^{-1} \text{ mol}^{-1} \)] The correct option is

Internal energy is a state function. For a complete cycle \( \Delta U_{\text{net}} = 0 \). By first law of thermodynamics, \( \Delta U = q + w = 0 \), but since it's a closed cyclic process, the sum of total work done plus net heat exchange over the cycle equals zero, and since internal energy change over a complete cycle is zero, \( \oint dU = 0 \implies w_1 + w_2 + w_3 + w_4 = \Delta Q_{\text{net}} \)... wait, for cyclic process \( \oint dU = 0 \implies q_{ ext{net}} + w_{ ext{net}} = 0 \). But more directly, internal energy depends only on state. For any closed cycle, the sum of all works and changes satisfies cyclic properties. Specifically, \( w_1 + w_2 + w_3 + w_4 = 0 \) represents the total work in a closed cycle if it forms a closed loop, or let's check standard thermodynamic relations.