Q.62 Chemistry ⭐ 360° Option Elimination Active

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NEET 2021 Chemistry Question 62 Solution

Topic: Thermodynamics · Official NTA Booklet Code M1

Question

Which one among the following is the correct option for right relationship between CP and CV for one mole of ideal gas?

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Verified Against NTA Final Answer Key

Option (2) : CP - CV = R

📚 Thermodynamics Easy

📋 Given Parameters & Problem Setup

System 1 mole of an ideal gas

Heat Capacities C_P (constant pressure) and C_V (constant volume)

⚡

The 30-Second Exam Hall Shortcut: ⚡ Mayer's Classic Thermodynamic Relation

⏱️ Save ~15 seconds

For one mole of an ideal gas, heating at constant pressure requires extra energy to perform boundary expansion work ( $P\Delta V = R\Delta T$ ). Hence, molar heat capacity at constant pressure exceeds that at constant volume by the gas constant $R$ : $C_P - C_V = R$ ! Match Option (2)!

1 Recall definition of enthalpy:

H = U + PV

2 For one mole of an ideal gas:

PV = RT \implies H = U + RT

3 Differentiate with respect to temperature:

\frac{dH}{dT} = \frac{dU}{dT} + R

4 By definition:

C_P = \frac{dH}{dT}

and

C_V = \frac{dU}{dT}

5 Substitute:

C_P = C_V + R \implies C_P - C_V = R

6 Match Option (2).

🛡️

Why You Can 100% Trust This Shortcut: Canonical Mayer's relation from NCERT Chemical Thermodynamics.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

First Law of Thermodynamics and Heat Capacities

Heat capacity at constant volume is defined as the rate of change of internal energy with temperature: $C_V = \left(\frac{\partial U}{\partial T}\right)_V$ . Heat capacity at constant pressure is defined as the rate of change of enthalpy with temperature: $C_P = \left(\frac{\partial H}{\partial T}\right)_P$ .

Step 2

Derivation Using Enthalpy of an Ideal Gas

Enthalpy is defined as $H = U + PV$ . For one mole ( $n = 1$ ) of an ideal gas, applying the ideal gas equation of state $PV = RT$ :

H = U + RT

Step 3

Differentiating with Respect to Temperature

Differentiating both sides with respect to temperature $T$ :

\frac{\partial H}{\partial T} = \frac{\partial U}{\partial T} + \frac{\partial (RT)}{\partial T} \implies C_P = C_V + R \implies C_P - C_V = R

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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 1

✗ Incorrect

Verdict & Scientific Reason:

Addition sign instead of subtraction; mathematically impossible since both $C_P$ and $C_V$ are greater than $R$ .

⚠️

Student Fallacy (Why candidates pick this trap): Writing addition instead of difference.

2 2

✓ 100% Correct

Verdict & Scientific Reason:

Exact canonical Mayer's equation for 1 mole of an ideal gas.

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Why this is airtight: None; textbook thermodynamic relation.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Multiplicative relation (dimensionally incorrect).

⚠️

Student Fallacy (Why candidates pick this trap): Dimensionally flawed equation.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Dimensionally incorrect expression.

⚠️

Student Fallacy (Why candidates pick this trap): Dimensionally flawed equation.

⚠️

NTA Trap Alert: Molar vs. Specific Heat Capacity Trap

Mayer's relation for 1 mole is $C_P - C_V = R$ . If specific heat capacities per gram are used ( $c_p$ and $c_v$ ), the relation becomes $c_p - c_v = R/M$ . The prompt explicitly states "for one mole of ideal gas", requiring $C_P - C_V = R$ .

🛡️ How to avoid the -1 mark: For 1 MOLE:

C_P - C_V = R

📖

Core NEET Syllabus Standard

NCERT Chemistry · Thermodynamics Standard Benchmark

Core Scientific Principle Tested:

"At constant pressure, gas expands and does work against surroundings, leading to $C_P - C_V = R$ ."

Syllabus Focus: Relationship Between Cp and Cv for an Ideal Gas

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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← Previous (Q.61) All Chemistry Questions Grid Next (Q.63) →

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NEET 2021 Chemistry Question 62 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Mayer's Classic Thermodynamic Relation

Step-by-Step Rigorous Workflow & Mathematical Proof

First Law of Thermodynamics and Heat Capacities

Derivation Using Enthalpy of an Ideal Gas

Differentiating with Respect to Temperature

360° Option Elimination Engine Key Feature

NTA Trap Alert: Molar vs. Specific Heat Capacity Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Thermodynamics

Work Done by an Ideal Monatomic Gas in Adiabatic Expansion

Net Heat Supplied in a Cyclic PV Indicator Diagram

Equipartition of Energy: Determining Vibrational Modes f from Specific Heat Ratio C_P / C_V

Entropy Changes in Isothermal Free Expansion of an Ideal Gas

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