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.

Choose the correct option regarding \(\Delta U\):

• A. \(\Delta U({Case-III})>\Delta U({Case-II})>\Delta U({Case-I})\)
• B. \(\Delta U({Case-I}) = \Delta U({Case-II}) = \Delta U({Case-III})\)
• C. \(\Delta U({Case-I})>\Delta U({Case-II})>\Delta U({Case-III})\)
• D. \(\Delta U({Case-I})>\Delta U({Case-III})>\Delta U({Case-II})\)

Hint:

For any cyclic process, \(\Delta U = 0\) because the system returns to the initial thermodynamic state.

Answer 1

The Correct Option is B

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:

\(\Delta U({Case-I}) = \Delta U({Case-II}) = \Delta U({Case-III})\)