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

In a cyclic transformation of an ideal gas, the process commences and concludes at the identical state. According to the first law of thermodynamics, the change in internal energy (\(\Delta U\)) for an ideal gas is given by: \[ \Delta U = Q - W \] where \(Q\) represents the heat transferred into the system, and \(W\) signifies the work performed by the system. As the system reverts to its initial state, the internal energy, being a state function, exhibits no net change over the cycle. Consequently, for any cyclic transformation, \(\Delta U = 0\). Therefore, the change in internal energy (\(\Delta U\)) is invariant across all paths (Cases I, II, or III). This implies:

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