An ideal gas undergoes a cyclic transformation starting from point A and coming back to the same point by tracing the path A→B→C→D→A as shown in the three cases below.
Choose the correct option regarding \(\Delta U\):
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For any cyclic process, \(\Delta U = 0\) because the system returns to the initial thermodynamic state.
An ideal gas undergoes a cyclic transformation, meaning it starts and returns to the same point. In a cyclic process, the change in internal energy (\(\Delta U\)) of an ideal gas is determined by the first law of thermodynamics: \[ \Delta U = Q - W \] where \(Q\) is the heat added to the system and \(W\) is the work done by the system. Since the system returns to its initial state, the internal energy, a state function, remains unchanged at the end of the cycle. Thus for any cyclic transformation, \(\Delta U = 0\). Therefore, regardless of the path taken (whether it's Case I, II, or III), the change in internal energy (\(\Delta U\)) will be the same for all cases. This means:
Step 1: Understanding the problem.
An ideal gas undergoes a cyclic transformation through points A → B → C → D → A in three different cases. We are asked to determine the relation regarding the change in internal energy (ΔU) for each case.
Step 2: Concept used.
For an ideal gas, the change in internal energy (ΔU) depends only on the initial and final temperatures of the system and not on the path taken. This means that internal energy is a state function.
In a cyclic process, the system returns to its initial state after completing one cycle. Therefore, the initial and final temperatures are equal, implying that:
\[
\Delta U_{\text{cycle}} = 0.
\]
Step 3: Application to all three cases.
In each of the given PV diagrams (Case I, Case II, and Case III), the process is cyclic — the gas starts from point A and eventually returns to the same point A. Thus, for all cases:
\[
\Delta U_{\text{Case I}} = \Delta U_{\text{Case II}} = \Delta U_{\text{Case III}} = 0.
\]
