1. Key thermodynamic fact

For an ideal gas, internal energy $U$ depends only on temperature:

$U = U(T)$

Hence, the change in internal energy is

$\Delta U = nC_V\Delta T$

So, $\Delta U$ depends only on the initial and final states, not on the path.

2. Nature of the given process

The gas undergoes a cyclic transformation:

$A \to B \to C \to D \to A$

That means the system starts at point $A$ and returns to the same point $A$.

So initial state = final state.

Therefore,

$\Delta T = 0$

and hence

$\Delta U = 0$

for each cyclic process.

3. Apply to all three cases

Since all three diagrams represent cyclic processes starting and ending at the same state,

$\Delta U(\text{Case-I}) = 0$ $\Delta U(\text{Case-II}) = 0$ $\Delta U(\text{Case-III}) = 0$

Therefore,

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

4. Evaluate options

• A: False
• B: False
• C: True
• D: False

5. Final answer

The correct option is:

$\boxed{\text{C}}$