Q.29 Physics ⭐ 360° Option Elimination Active

NEET 2025 Physics Question 29 Solution

Topic: Thermodynamics · Official NTA Booklet Code 45

Question

A container has two chambers of volumes V₁ = 2 litres and V₂ = 3 litres separated by a partition made of a thermal insulator. The chambers contains n₁ = 5 and n₂ = 4 moles of ideal gas at pressures p₁ = 1 atm and p₂ = 2 atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of:

✓

Verified Against NTA Final Answer Key

Option (2) : 1.6 atm

📚 Thermodynamics Easy

📋 Given Parameters & Problem Setup

Chamber 1 V₁ = 2 litres, n₁ = 5 moles, p₁ = 1 atm

Chamber 2 V₂ = 3 litres, n₂ = 4 moles, p₂ = 2 atm

Boundary Condition Thermally insulated container (isolated system, ΔQ = 0, W = 0)

Operation Insulating partition is removed

⚡

The 30-Second Exam Hall Shortcut: ⚡ Internal Energy Conservation Shortcut

⏱️ Save ~40 seconds

Isolated container ⟹ U_final = U₁ + U₂ ⟹ P_eq (V₁ + V₂) = P₁V₁ + P₂V₂ ⟹ P_eq = (1×2 + 2×3) / (2 + 3) = 8 / 5 = 1.6 atm!

1 By First Law of Thermodynamics for rigid isolated container: Q = 0, W = 0 ⟹ ΔU = 0.

2 Internal energy of ideal gas: U = (f/2) nRT = (f/2) PV.

3 Conservation of internal energy: P_eq V_total = P₁V₁ + P₂V₂.

4 Total volume: V_total = V₁ + V₂ = 2 + 3 = 5 litres.

5 Compute products: P₁V₁ = 1 × 2 = 2 atm·L, and P₂V₂ = 2 × 3 = 6 atm·L.

6 Sum: 2 + 6 = 8 atm·L.

7 P_eq = 8 / 5 = 1.6 atm.

8 Select Option (2) in under 15 seconds!

🛡️

Why You Can 100% Trust This Shortcut: First Law of Thermodynamics & Kinetic Theory of Gas Mixtures (NCERT Chapters 11 & 12).

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Apply the First Law of Thermodynamics

The two chambers are enclosed in a container made of a thermal insulator, so no heat is exchanged with the surroundings (Q = 0). The walls of the container are rigid, so no external work is done (W = 0). According to the First Law of Thermodynamics:

\Delta U = Q - W = 0 \implies U_{final} = U_1 + U_2

• The total internal energy of the system remains conserved upon removing the partition.

Step 2

Relate Internal Energy to Pressure and Volume

For an ideal gas, internal energy is proportional to the product of pressure and volume: U = (f/2) nRT = (f/2) PV, where f is the degrees of freedom:

\frac{f}{2} P_{eq} (V_1 + V_2) = \frac{f}{2} P_1 V_1 + \frac{f}{2} P_2 V_2 \implies P_{eq} (V_1 + V_2) = P_1 V_1 + P_2 V_2

• Degrees of freedom f cancels out.

Step 3

Calculate Equilibrium Pressure

Substitute the given values P₁ = 1 atm, V₁ = 2 L, P₂ = 2 atm, V₂ = 3 L:

P_{eq} = \frac{P_1 V_1 + P_2 V_2}{V_1 + V_2} = \frac{(1\text{ atm} \times 2\text{ L}) + (2\text{ atm} \times 3\text{ L})}{2\text{ L} + 3\text{ L}} = \frac{2 + 6}{5} = \frac{8}{5} = 1.6\text{ atm}

• Note: The given mole numbers (n₁ = 5, n₂ = 4) are redundant data; equilibrium pressure is completely determined by P₁V₁ + P₂V₂.

⭐

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

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Calculation mistake.

⚠️

Student Fallacy (Why candidates pick this trap): Arithmetic misstep.

2 B

✓ 100% Correct

Verdict & Scientific Reason:

100% exact. P_eq = (P₁V₁ + P₂V₂) / (V₁ + V₂) = (2 + 6) / 5 = 1.6 atm.

💡

Why this is airtight: None; rigorous thermodynamics.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Simple average (1 + 2) / 2 = 1.5 rounded or miscalculated.

⚠️

Student Fallacy (Why candidates pick this trap): Taking an unweighted average of pressures.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Weighting by moles instead of volume.

⚠️

Student Fallacy (Why candidates pick this trap): Weighting pressures by moles: (1×5 + 2×4) / 9 = 13/9 ≈ 1.44 or miscalculation.

⚠️

NTA Trap Alert: The Redundant Moles Distractor Trap

The problem provides n₁ = 5 and n₂ = 4. Many students spend minutes trying to calculate initial temperatures T₁ = P₁V₁/n₁R and T₂ = P₂V₂/n₂R and final mixture temperature. Since U = (f/2)PV, you do NOT need temperature or moles at all: P_eq is directly (P₁V₁ + P₂V₂) / V_total!

🛡️ How to avoid the -1 mark: Remember: In isolated mixing, P_final = (P₁V₁ + P₂V₂) / (V₁ + V₂) directly!

📖

Core NEET Syllabus Standard

NEET Physics Core Syllabus · Thermodynamics Standard Benchmark

Core Scientific Principle Tested:

"In an isolated container with zero work, total internal energy is conserved: P_eq V_total = P₁V₁ + P₂V₂."

Syllabus Focus: First Law of Thermodynamics & Internal Energy

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Volume-Weighted Pressure: (1×2 + 2×3) / 5 = 8 / 5 = 1.6 atm!

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NEET 2025 Physics Question 29 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Internal Energy Conservation Shortcut

Step-by-Step Rigorous Workflow & Mathematical Proof

Apply the First Law of Thermodynamics

Relate Internal Energy to Pressure and Volume

Calculate Equilibrium Pressure

360° Option Elimination Engine Key Feature

NTA Trap Alert: The Redundant Moles Distractor Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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Entropy Changes in Isothermal Free Expansion of an Ideal Gas

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