चित्र में दर्शाए अनुसार 1.0 मोल आदर्श गैस के लिए उत्क्रमणीय प्रक्रमों पर विचार कीजिए। w1,w2,w3 और w4 क्रमशः प्रक्रमों 1, 2, 3 और 4 में किए गए कार्य (कैलोरी में) को दर्शाते हैं; ΔU2 और ΔU4 क्रमशः प्रक्रमों 2 और 4 के लिए आंतरिक ऊर्जा में परिवर्तन हैं। [R=2 cal K−1 mol−1 का उपयोग करें] सही विकल्प है

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.