Thermodynamics · Question 39

[figure] 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 U :

1. U(Case-III) > U(Case-II) > U(Case-I)
2. U(Case-I) > U(Case-II) > U(Case-III)
3. U(Case-I) = U(Case-II) = U(Case-III)
4. U(Case-I) > U(Case-III) > U(Case-II)

Answer

(3) U(Case-I) = U(Case-II) = U(Case-III)

Solution

Internal energy U is a state function: its change depends only on the initial and final states, not on the path.

In every case the gas starts at A and returns to the same point A (cyclic process).

U = UA - UA = 0 for each cycle.

(The paths differ only in the work and heat exchanged: q = -w for each cycle.)

U(Case-I) = U(Case-II) = U(Case-III) = 0

Answer: (3)