Question:

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.
 ideal gas undergoes a cyclic transformation
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.
Updated On: Mar 17, 2025
  • \(\Delta U({Case-III})>\Delta U({Case-II})>\Delta U({Case-I})\)
  • \(\Delta U({Case-I}) = \Delta U({Case-II}) = \Delta U({Case-III})\)
  • \(\Delta U({Case-I})>\Delta U({Case-II})>\Delta U({Case-III})\)
  • \(\Delta U({Case-I})>\Delta U({Case-III})>\Delta U({Case-II})\)
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The Correct Option is B

Solution and Explanation

  • For a cyclic process, the internal energy change (\(\Delta U\)) is always zero as the system returns to its initial state.
  • Since \(\Delta U\) is a state function, it depends only on the initial and final states, which are the same for all three cases.
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