Q.39 Physics ⭐ 360° Option Elimination Active

NEET 2026 Physics Question 39 Solution

Topic: Thermodynamics · Official NTA Booklet Code 50

Question

One mole of an ideal monatomic gas undergoes a cyclic process as shown in the figure. The total heat supplied to the gas is:

✓

Verified Against NTA Final Answer Key

Option (3) : Q_net = 600 J

📚 Thermodynamics Easy

📋 Given Parameters & Problem Setup

Working Substance 1 mole of ideal monatomic gas

Process Closed rectangular cycle a -> b -> c -> d -> a

Pressure Range 100 N/m^2 to 300 N/m^2 (Delta P = 200 N/m^2)

Volume Range 2 m^3 to 5 m^3 (Delta V = 3 m^3)

Target Quantity Total net heat supplied to the gas over the cycle

⚡

The 30-Second Exam Hall Shortcut: ⚡ Cyclic Area Theorem Shortcut

⏱️ Bypasses computing separate heats along all 4 individual legs.

For any complete closed cycle: Delta U_cycle = 0, so Q_net = Work Done = Area of Cycle!

1 By First Law of Thermodynamics for a complete thermodynamic cycle: Delta U = 0.

2 Therefore, total net heat supplied equals total net work done: Q_net = W_net.

3 The indicator diagram is a simple rectangle on the P-V plane.

4 Area = (P_high - P_low) * (V_high - V_low) = (300 - 100) N/m^2 * (5 - 2) m^3 = 200 * 3 = 600 J.

5 Direct result: Q_net = 600 J. Solved in 10 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Internal energy is a thermodynamic state function. Over any closed path returning to the initial state, Delta U = 0 identically, guaranteeing Q_net = W_net = enclosed loop area.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Apply the First Law of Thermodynamics to a Complete Thermodynamic Cycle

Because internal energy U is a state function that depends solely on the thermodynamic state variables (P, V, T):

\Delta U_{\text{cycle}} = U_{\text{final}} - U_{\text{initial}} = 0

• First Law of Thermodynamics: Q_{net} = Delta U_{cycle} + W_{net} = 0 + W_{net} = W_{net}.
• Total net heat supplied over the cycle equals the net work done by the gas.

Step 2

Calculate the Net Work Done from the PV Diagram Enclosed Area

For a clockwise cyclic process, net work done is the positive enclosed area of the rectangle on the P-V diagram:

W_{\text{net}} = (P_{\text{high}} - P_{\text{low}}) \times (V_{\text{high}} - V_{\text{low}})

• P_{high} = 300 N/m^2, P_{low} = 100 N/m^2 ==> Delta P = 300 - 100 = 200 N/m^2.
• V_{high} = 5 m^3, V_{low} = 2 m^3 ==> Delta V = 5 - 2 = 3 m^3.

Step 3

Compute Net Heat and Confirm with Options

Multiply the dimensions of the rectangular cycle:

Q_{\text{net}} = W_{\text{net}} = (200\text{ N/m}^2) \times (3\text{ m}^3) = 600\text{ J}

• Matches Option (3) exactly.

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 Option (1)

✗ Incorrect

Verdict & Scientific Reason:

Arithmetic error.

⚠️

Student Fallacy (Why candidates pick this trap): Uses Delta V = 2 instead of 3: 200 * 2 = 400 J.

Note: Volume span is 5 - 2 = 3 m^3.

2 Option (2)

✗ Incorrect

Verdict & Scientific Reason:

Arbitrary distractor.

⚠️

Student Fallacy (Why candidates pick this trap): Takes average pressure 200 + 300.

Note: Enclosed area is Delta P * Delta V = 600 J.

3 Option (3)

✓ 100% Correct

Verdict & Scientific Reason:

100% mathematically exact; confirmed by Official NTA Final Answer Key.

💡

Why this is airtight: Flawless application of cyclic area theorem: Q_net = W_net = 200 * 3 = 600 J.

Note: Exact thermodynamics.

4 Option (4)

✗ Incorrect

Verdict & Scientific Reason:

Uses perimeter or upper area.

⚠️

Student Fallacy (Why candidates pick this trap): Calculates area under leg a->b (300 * 3 = 900) and subtracts wrong offset.

Note: Net work is the enclosed area inside the cycle.

⚠️

NTA Trap Alert: Calculating Separate Leg Heats (Wasting 4 Minutes)

Many candidates calculate Q_ab + Q_bc + Q_cd + Q_da using n*C_p*Delta T and n*C_v*Delta T, which wastes several minutes and invites calculation errors.

🛡️ How to avoid the -1 mark: Whenever total heat over a complete cycle is asked, ALWAYS use Q_net = Area of loop directly!

📖

Core NEET Syllabus Standard

NEET UG Physics Core Standards · Thermodynamics Standard Benchmark

Core Scientific Principle Tested:

"In any cyclic process, the system returns to its initial state, so Delta U = 0. Net heat absorbed equals net work done: Q_net = W_net = Area enclosed by the PV curve."

Syllabus Focus: Cyclic Processes & PV Indicator Diagrams

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2026 Physics Question 39 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Cyclic Area Theorem Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Apply the First Law of Thermodynamics to a Complete Thermodynamic Cycle

Calculate the Net Work Done from the PV Diagram Enclosed Area

Compute Net Heat and Confirm with Options

360° Option Elimination Engine Key Feature

NTA Trap Alert: Calculating Separate Leg Heats (Wasting 4 Minutes)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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