An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A→B→C→D→A as shown in the three cases above.
Choose the correct option regarding ΔU :

Solution

Chemistry ma’am · 4 steps

Every cycle ends back at A, so the gas returns to the same state and Δ U = 0 each time.

1. State function. Internal energy depends only on the state of the gas (p, V, T), not on the path taken to reach it. For an ideal gas, U depends on T alone, and T is fixed once p and V are fixed.

   Δ U_(cycle) = U_A - U_A = 0

2. Shape does not matter. Case I uses curved paths, Case II is a rectangle and Case III is a diamond. Each cycle starts at A and ends at A, so Δ U = 0 in all three cases.
3. What does differ. The area enclosed by each loop is different, so the net work W, and with it the net heat q, change from case to case. But q and W always balance so that Δ U = q + W = 0.
4. So. Δ U(Case-I) = Δ U(Case-II) = Δ U(Case-III) = 0, which is option (c).

The idea to keep. For any complete cycle, every state function (U, H, S) changes by zero. Only path functions like q and W depend on the shape of the loop.

Answer from the NTA final answer key. Solution by PrepWiser.