JEE Main · 2025 · Shift-IIeasyTHERMO-152

An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing…

Thermodynamics · Class 11 · JEE Main Previous Year Question

Question

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\text{A} \rightarrow \text{B} \rightarrow \text{C} \rightarrow \text{D} \rightarrow \text{A} as shown in three p-V diagram cases.

Choose the correct option regarding ΔU\Delta U:

(a) ΔU(Case-III)>ΔU(Case-II)>ΔU(Case-I)\Delta U(\text{Case-III}) > \Delta U(\text{Case-II}) > \Delta U(\text{Case-I})

(b) ΔU(Case-I)>ΔU(Case-II)>ΔU(Case-III)\Delta U(\text{Case-I}) > \Delta U(\text{Case-II}) > \Delta U(\text{Case-III})

(c) ΔU(Case-I)>ΔU(Case-III)>ΔU(Case-II)\Delta U(\text{Case-I}) > \Delta U(\text{Case-III}) > \Delta U(\text{Case-II})

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

Options
  1. a

    ΔU(Case-III)>ΔU(Case-II)>ΔU(Case-I)\Delta U(\text{Case-III}) > \Delta U(\text{Case-II}) > \Delta U(\text{Case-I})

  2. b

    ΔU(Case-I)>ΔU(Case-II)>ΔU(Case-III)\Delta U(\text{Case-I}) > \Delta U(\text{Case-II}) > \Delta U(\text{Case-III})

  3. c

    ΔU(Case-I)>ΔU(Case-III)>ΔU(Case-II)\Delta U(\text{Case-I}) > \Delta U(\text{Case-III}) > \Delta U(\text{Case-II})

  4. d

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

    ✓
Correct Answerd

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

Detailed Solution

🧠 For any cyclic process, ΔU=0\Delta U = 0 because internal energy is a state function A state function depends only on the current state, not the path taken to reach it. Starting and ending at the same point means zero net change — regardless of the shape or direction of the cycle.

All three cases (I, II, III) return the system to point A. Therefore: ΔUCase I=ΔUCase II=ΔUCase III=0\Delta U_{\text{Case I}} = \Delta U_{\text{Case II}} = \Delta U_{\text{Case III}} = 0

⚠️ Trap: Work and heat are path-dependent and will differ across the three cycles. Only state functions like UU, HH, SS, GG are guaranteed zero for any closed cycle.

Answer: (d)\boxed{\text{Answer: (d)}}

Practice this question with progress tracking

Want timed practice with adaptive difficulty? Solve this question (and hundreds more from Thermodynamics) inside The Crucible, our adaptive practice platform.

Practice in The Crucible →Browse Thermodynamics chapter →

More JEE Main Thermodynamics PYQs

2025medium

The correct statement amongst the following is: (A) The term 'standard state' implies that the temperature is 0°C (B) The standard state of pure gas is the pure gas at a pressure of 1 bar and…

2023medium

What happens when methane undergoes combustion in in systems A and B respectively? System A: Adiabatic container System B: Diathermic container

2019easy

Among the following, the set of parameters that represents path functions is: (i) q + w  &nbsp (ii) q  &nbsp (iii) w  &nbsp (iv) H - TS

2025medium

Two vessels A and B are connected via stopcock. Vessel A is filled with a gas at a certain pressure. The entire assembly is immersed in water and allowed to come to thermal equilibrium. After opening…

2025hard

Arrange the following in order of magnitude of work done by the system / on the system at constant temperature: (a) |Wreversible| for expansion in infinite stage. (b) |Wirreversible| for expansion in…

View all 58 Thermodynamics PYQs →