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NEET 2021 Chemistry Question 90 Solution

Topic: Thermodynamics · Official NTA Booklet Code M1

Question

For irreversible expansion of an ideal gas under isothermal condition, the correct option is:

✓

Verified Against NTA Final Answer Key

Option (3) : ΔU = 0, ΔStotal ≠ 0

📚 Thermodynamics Easy

📋 Given Parameters & Problem Setup

Working Substance Ideal gas

Process Type Irreversible expansion

Thermal Condition Isothermal (T = constant)

⚡

The 30-Second Exam Hall Shortcut: ⚡ Isothermal Ideal Gas $\implies$ ΔU = 0; Irreversible Process $\implies$ ΔStotal > 0

⏱️ Save ~15 seconds

For an ideal gas, internal energy depends SOLELY on temperature: isothermal means $\Delta T = 0 \implies \Delta U = n C_v \Delta T = 0$ . By the Second Law of Thermodynamics, any irreversible (spontaneous) process strictly produces positive total entropy: $\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0 \implies \Delta S_{\text{total}} \neq 0$ . Therefore, $\Delta U = 0, \Delta S_{\text{total}} \neq 0$ ! Match Option (3)!

1 Isothermal condition: Temperature is constant (

\Delta T = 0

2 For an ideal gas: Internal energy is a function of temperature alone (

U = f(T)

)

\implies \Delta U = 0

3 Expansion is irreversible: According to the Second Law of Thermodynamics, all spontaneous/irreversible processes increase the entropy of the universe.

4 Therefore:

\Delta S_{\text{total}} > 0

, which means

\Delta S_{\text{total}} \neq 0

. (

\Delta S_{\text{total}} = 0

only for a reversible process).

5 Result:

\Delta U = 0, \Delta S_{\text{total}} \neq 0

6 Match Option (3).

🛡️

Why You Can 100% Trust This Shortcut: Canonical First and Second Laws of Thermodynamics from NCERT Chemical Thermodynamics.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Applying the First Law to an Isothermal Ideal Gas

For an ideal gas, intermolecular forces of attraction are zero. Consequently, internal energy $U$ depends solely on temperature (Joule's Law of ideal gases): $U = f(T) \implies dU = n C_V dT$ In any isothermal process, temperature remains strictly constant ( $\Delta T = 0$ ): $\Delta U = n C_V \Delta T = 0$

Step 2

Applying the Second Law to an Irreversible Process

The total entropy change of the universe (system plus surroundings) is defined by the Second Law of Thermodynamics: $\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}$ - For an idealized reversible process: $\Delta S_{\text{total}} = 0$ . - For any real, irreversible (spontaneous) process: entropy generation occurs due to dissipation, requiring: $\Delta S_{\text{total}} > 0 \implies \Delta S_{\text{total}} \neq 0$

Step 3

Combining Both Thermodynamic Conditions

Combining the isothermal criterion ( $\Delta U = 0$ ) with the irreversibility criterion ( $\Delta S_{\text{total}} \neq 0$ ):

\mathbf{\Delta U = 0, \quad \Delta S_{\text{total}} \neq 0}

⭐

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 1

✗ Incorrect

Verdict & Scientific Reason:

$\Delta S_{\text{total}} = 0$ is true ONLY for a REVERSIBLE process, whereas the prompt explicitly states IRREVERSIBLE.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming total entropy is conserved in an irreversible process.

2 2

✗ Incorrect

Verdict & Scientific Reason:

$\Delta U = 0$ for any isothermal process of an ideal gas; claiming $\Delta U \neq 0$ violates Joule's law.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming internal energy changes during expansion.

3 3

✓ 100% Correct

Verdict & Scientific Reason:

Exact thermodynamic conditions: $\Delta U = 0$ (isothermal ideal gas) and $\Delta S_{\text{total}} > 0$ (irreversible process).

💡

Why this is airtight: None; rigorous First and Second Law mastery.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Violates both the isothermal condition and the Second Law for irreversible processes.

⚠️

Student Fallacy (Why candidates pick this trap): Complete inversion of thermodynamic principles.

⚠️

NTA Trap Alert: Reversible vs. Irreversible Total Entropy Trap

Candidates frequently choose Option (1) ( $\Delta U = 0, \Delta S_{\text{total}} = 0$ ) because they recall $\Delta S_{\text{total}} = 0$ from reversible Carnot cycles. For any IRREVERSIBLE expansion, total entropy of the universe MUST increase ( $\Delta S_{\text{total}} > 0 \implies \Delta S_{\text{total}} \neq 0$ ).

🛡️ How to avoid the -1 mark: Irreversible

\implies \Delta S_{\text{total}} > 0 \implies \Delta S_{\text{total}} \neq 0

. Isothermal ideal gas

\implies \Delta U = 0

📖

Core NEET Syllabus Standard

NCERT Chemistry · Thermodynamics Standard Benchmark

Core Scientific Principle Tested:

"For an isothermal expansion of an ideal gas, $\Delta U = 0$ . For any irreversible process, $\Delta S_{\text{total}} > 0$ , hence $\Delta S_{\text{total}} \neq 0$ ."

Syllabus Focus: First and Second Laws of Thermodynamics (Spontaneity and Entropy)

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Chemistry Question 90 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Isothermal Ideal Gas ⟹ \implies⟹ ΔU = 0; Irreversible Process ⟹ \implies⟹ ΔStotal > 0

Step-by-Step Rigorous Workflow & Mathematical Proof

Applying the First Law to an Isothermal Ideal Gas

Applying the Second Law to an Irreversible Process

Combining Both Thermodynamic Conditions

360° Option Elimination Engine Key Feature

NTA Trap Alert: Reversible vs. Irreversible Total Entropy Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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The 30-Second Exam Hall Shortcut: ⚡ Isothermal Ideal Gas $\implies$ ΔU = 0; Irreversible Process $\implies$ ΔStotal > 0