Q.69 Chemistry ⭐ 360° Option Elimination Active

NEET 2026 Chemistry Question 69 Solution

Topic: Thermodynamics · Official NTA Booklet Code 50

Question

Consider the reversible processes for 1.0 mol of an ideal gas as shown in the figure (processes 2 and 4 are adiabatic). w₁, w₂, w₃ and w₄ represent work done (in calories) in the processes 1, 2, 3 and 4, respectively; ΔU₂ and ΔU₄ are changes in the internal energy for the processes 2 and 4, respectively. [use R = 2 cal K⁻¹ mol⁻¹] The correct option is

✓

Verified Against NTA Final Answer Key

Option (1) : w₁ + w₃ = -2 T₁ ln(V₂/V₁) - 2 T₂ ln(V₄/V₃)

📚 Thermodynamics Medium

📋 Given Parameters & Problem Setup

Amount of Ideal Gas (n) 1.0 mol

Process 1 Reversible isothermal expansion at T₁ from V₁ to V₂

Process 2 Reversible adiabatic expansion from T₁, V₂ to T₂, V₃

Process 3 Reversible isothermal compression at T₂ from V₃ to V₄

Process 4 Reversible adiabatic compression from T₂, V₄ to T₁, V₁

Gas Constant (R) 2 cal K⁻¹ mol⁻¹

⚡

The 30-Second Exam Hall Shortcut: ⚡ Isothermal Work Formula in Chemistry

⏱️ Save ~60 seconds

w_rev = -nRT ln(V_final / V_initial)! Substitute n = 1 and R = 2 cal/(K mol)!

1 Process 1 is isothermal at T₁ from V₁ to V₂: w₁ = -1 × 2 × T₁ ln(V₂/V₁) = -2 T₁ ln(V₂/V₁).

2 Process 3 is isothermal at T₂ from V₃ to V₄: w₃ = -1 × 2 × T₂ ln(V₄/V₃) = -2 T₂ ln(V₄/V₃).

3 Sum of isothermal works: w₁ + w₃ = -2 T₁ ln(V₂/V₁) - 2 T₂ ln(V₄/V₃).

4 Look at Option (1): Exact mathematical match! Pick Option (1) in 15 seconds!

🛡️

Why You Can 100% Trust This Shortcut: IUPAC chemistry sign convention: Work done BY system during expansion is negative: w = -∫ P dV = -nRT ln(V_f / V_i).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Review the IUPAC Thermodynamic Definition of Reversible Work

In chemistry, work done during a reversible isothermal expansion or compression of an ideal gas is defined by: w = - ∫ P_ext dV = - ∫ (nRT / V) dV = - nRT ln(V_f / V_i)

w_{\text{rev, iso}} = -nRT \ln\left(\frac{V_f}{V_i}\right)

• For n = 1.0 mol and R = 2 cal K⁻¹ mol⁻¹, the prefactor nR = 1.0 × 2 = 2 cal K⁻¹.

Step 2

Calculate Work for Process 1 (Isothermal Expansion at T₁)

Process 1 is an isothermal expansion at temperature T₁ from initial volume V₁ to final volume V₂.

w_1 = -nR T_1 \ln\left(\frac{V_2}{V_1}\right) = -2 T_1 \ln\left(\frac{V_2}{V_1}\right)

• Since V₂ > V₁, ln(V₂/V₁) > 0, so w₁ < 0 (work is done by the system).

Step 3

Calculate Work for Process 3 (Isothermal Compression at T₂)

Process 3 is an isothermal compression at temperature T₂ from initial volume V₃ to final volume V₄.

w_3 = -nR T_2 \ln\left(\frac{V_4}{V_3}\right) = -2 T_2 \ln\left(\frac{V_4}{V_3}\right)

• Since V₄ < V₃, ln(V₄/V₃) < 0, so w₃ > 0 (work is done on the system).

Step 4

Compute Total Isothermal Work (w₁ + w₃)

Add the two expressions together to obtain the sum of the work done in the two isothermal stages.

w_1 + w_3 = -2 T_1 \ln\left(\frac{V_2}{V_1}\right) - 2 T_2 \ln\left(\frac{V_4}{V_3}\right)

• Matches Option (1).

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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 A

✓ 100% Correct

Verdict & Scientific Reason:

Exact thermodynamic formula under IUPAC chemistry sign convention.

💡

Why this is airtight: None; perfect understanding of isothermal work integrals.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. For adiabatic processes (q = 0), w = ΔU, so w₂ + w₄ = ΔU₂ + ΔU₄, not ΔU₂ - ΔU₄.

⚠️

Student Fallacy (Why candidates pick this trap): Sign error in the First Law for adiabatic processes.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. w₂ is an adiabatic process involving temperature change (w₂ = C_v(T₂ - T₁)), not an isothermal term.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing adiabatic process 2 with an isothermal process.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Total work in a cyclic engine equals the enclosed area on the P-V diagram (net work output ≠ 0).

⚠️

Student Fallacy (Why candidates pick this trap): Confusing state function cyclic change (ΔU_cycle = 0) with path function work (w_net ≠ 0).

⚠️

NTA Trap Alert: Assuming Net Work in a Cycle is Zero (w_net = 0)

Students frequently memorize "cyclic process change is zero" (which is true for ΔU, ΔH, and ΔS) and erroneously apply it to work (Option 4). Work is a PATH function, and net cyclic work represents the non-zero mechanical output!

🛡️ How to avoid the -1 mark: State functions (U, H, S, G) sum to 0 in a cycle; path functions (q, w) do NOT sum to 0.

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Core NEET Syllabus Standard

NEET Chemistry Core Syllabus · Thermodynamics Standard Benchmark

Core Scientific Principle Tested:

"In chemistry, reversible isothermal work is w = -nRT ln(V_f/V_i)."

Syllabus Focus: Work done in reversible isothermal and adiabatic processes; cyclic processes

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

State functions cycle to zero; Work produces power like a hero!

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NEET 2026 Chemistry Question 69 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Isothermal Work Formula in Chemistry

Step-by-Step Rigorous Workflow & Mathematical Proof

Review the IUPAC Thermodynamic Definition of Reversible Work

Calculate Work for Process 1 (Isothermal Expansion at T₁)

Calculate Work for Process 3 (Isothermal Compression at T₂)

Compute Total Isothermal Work (w₁ + w₃)

360° Option Elimination Engine Key Feature

NTA Trap Alert: Assuming Net Work in a Cycle is Zero (w_net = 0)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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