Q.104 Chemistry ⭐ 360° Option Elimination Active

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NEET 2020 Chemistry Question 104 Solution

Topic: Thermodynamics · Official NTA Booklet Code E1

Question

The correct option for free expansion of an ideal gas under adiabatic condition is:

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Verified Against NTA Final Answer Key

Option (1) : q=0, ∆T=0 and w=0

📚 Thermodynamics Easy

📋 Given Parameters & Problem Setup

System Ideal gas

Process 1 Free expansion (expansion against vacuum, P_ext = 0)

Process 2 Adiabatic condition (no heat exchange with surroundings)

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The 30-Second Exam Hall Shortcut: ⚡ "Free Expansion -> w = 0; Adiabatic -> q = 0; Ideal Gas -> ΔT = 0!"

⏱️ 20 seconds

In free expansion, the gas expands against vacuum ( $P_{\text{ext}} = 0$ ), so work done $w = -P_{\text{ext}} \Delta V = 0$ . In an adiabatic process, there is no heat exchange, so $q = 0$ . By First Law, $\Delta U = q + w = 0$ . For an ideal gas, internal energy depends only on temperature ( $\Delta U = C_v \Delta T$ ), so $\Delta U = 0 \implies \Delta T = 0$ ! Mark Option (1) in 5 seconds.

1 Adiabatic condition means q = 0.

2 Free expansion into vacuum means P_ext = 0, hence w = -P_ext × ΔV = 0.

3 First law of thermodynamics: ΔU = q + w = 0 + 0 = 0.

4 For an ideal gas, ΔU = n Cv ΔT = 0, which requires ΔT = 0.

5 Result: q = 0, ΔT = 0, w = 0.

6 Select Option (1).

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Why You Can 100% Trust This Shortcut: NCERT Chemistry · Thermodynamics (First Law of Thermodynamics: Free Expansion)

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Define Free Expansion

Free expansion refers to the irreversible expansion of a gas against zero external opposing pressure ( $P_{\text{ext}} = 0$ , i.e. into a vacuum). The mechanical work done is: $w = -\int P_{\text{ext}} \, dV = -0 \times \Delta V = 0$

Step 2

Apply Adiabatic Condition

An adiabatic process is thermally insulated from its environment, meaning no heat enters or leaves the system: $q = 0$

Step 3

Apply First Law of Thermodynamics

According to the first law of thermodynamics: $\Delta U = q + w$ Substituting $q = 0$ and $w = 0$ : $\Delta U = 0 + 0 = 0$

Step 4

Deduce Temperature Change for Ideal Gas

For an ideal gas, internal energy is purely a function of temperature ( $U = f(T)$ , Joule's law of internal energy). Because there are no intermolecular attractive forces to overcome during expansion: $\Delta U = n C_v \Delta T = 0 \implies \Delta T = 0$ Therefore, the temperature of the ideal gas remains constant during adiabatic free expansion ( $q = 0, w = 0, \Delta T = 0$ ).

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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:

Factually and thermodynamically exact; vacuum expansion gives $w = 0$ , adiabatic gives $q = 0$ , and ideal gas yields $\Delta T = 0$ .

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Why this is airtight: None; core physical chemistry concept.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect; holds for reversible adiabatic expansion against non-zero external pressure where gas does work at the expense of internal energy ( $w < 0, \Delta T < 0$ ), not free expansion.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing free expansion into vacuum with expansion against a piston.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect; adiabatic condition strictly demands $q = 0$ , never $q < 0$ .

⚠️

Student Fallacy (Why candidates pick this trap): Violating the definition of an adiabatic system.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect; violates both adiabatic ( $q = 0$ ) and free expansion ( $w = 0$ ) definitions.

⚠️

Student Fallacy (Why candidates pick this trap): Random thermodynamic signs.

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NTA Trap Alert: Adiabatic Free Expansion vs Reversible Adiabatic Expansion

- In REVERSIBLE adiabatic expansion: Work is done against pressure ( $w \neq 0$ ), so $\Delta U < 0$ and the gas COOLS ( $\Delta T < 0$ ). - In FREE adiabatic expansion: $P_{\text{ext}} = 0 \implies w = 0$ , so $\Delta U = 0$ and $\Delta T = 0$ for an ideal gas!

🛡️ How to avoid the -1 mark: Look for "FREE EXPANSION": Work done is ALWAYS ZERO (

w = 0

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Core NEET Syllabus Standard

Chemistry · Thermodynamics Standard Benchmark

Core Scientific Principle Tested:

Syllabus Focus: Free Expansion of an Ideal Gas under Adiabatic Conditions: q = 0, w = 0, ΔT = 0

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NEET 2020 Chemistry Question 104 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ "Free Expansion -> w = 0; Adiabatic -> q = 0; Ideal Gas -> ΔT = 0!"

Step-by-Step Rigorous Workflow & Mathematical Proof

Define Free Expansion

Apply Adiabatic Condition

Apply First Law of Thermodynamics

Deduce Temperature Change for Ideal Gas

360° Option Elimination Engine Key Feature

NTA Trap Alert: Adiabatic Free Expansion vs Reversible Adiabatic Expansion

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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