Thermodynamics - NEET Previous Year Questions (2009-2024)

🎯 Chapter Overview

Thermodynamics is a crucial chapter that bridges physics and chemistry, consistently contributing 5-7 questions in NEET. This chapter tests understanding of heat, work, energy, and their interconversion.

Chapter Statistics (2009-2024)

📊 Question Distribution: 5-7 questions per year (20-28 marks)
⚡ Success Rate: 50-60% (Moderately challenging)
📈 Difficulty Level: Medium to Hard
⏱️ Average Time: 2-3 minutes per question
🎯 Priority Level: High (Important for both Physics and Chemistry)

📚 Important Concepts & Formulae

Laws of Thermodynamics

🔵 Zeroth Law:
- If A = B and B = C, then A = C (thermal equilibrium)
- Basis of temperature measurement
- Defines temperature as a property of thermal equilibrium

🔴 First Law (Energy Conservation):
ΔU = Q - W (change in internal energy = heat added - work done)
- Internal energy is state function
- Heat and work are path functions
- Sign convention: Q positive when added, W positive when done by system

🟢 Second Law:
- Heat flows spontaneously from hot to cold
- No heat engine can be 100% efficient
- Entropy of universe always increases
- Clausius statement: Heat cannot flow from cold to hot without work

🔵 Third Law:
- Entropy approaches zero as temperature approaches absolute zero
- Impossible to reach absolute zero temperature
- Perfect crystals have zero entropy at 0 K

Thermodynamic Processes

📈 Isothermal Process (T constant):
- PV = constant (Boyle's law)
- W = nRT ln(V₂/V₁) = nRT ln(P₁/P₂)
- ΔU = 0 (temperature constant)
- Q = W (heat = work done)

📊 Isobaric Process (P constant):
- W = P(V₂ - V₁) = PΔV
- ΔU = nCᵥΔT
- Q = ΔU + W = nCₚΔT

📐 Isochoric Process (V constant):
- W = 0 (no volume change)
- ΔU = nCᵥΔT
- Q = ΔU = nCᵥΔT

🔄 Adiabatic Process (Q = 0):
- PVᵞ = constant, TVᵞ⁻¹ = constant, TP^(1-ᵞ) = constant
- W = (P₁V₁ - P₂V₂)/(γ - 1) = nCᵥ(T₁ - T₂)
- ΔU = -W (work done at expense of internal energy)

Heat Engines and Cycles

🔥 Heat Engine Efficiency:
η = W/Qₕ = 1 - Q꜀/Qₕ
- W = Qₕ - Q꜀ (work output)
- Qₕ = heat absorbed from hot reservoir
- Q꜀ = heat rejected to cold reservoir

❄️ Refrigerator Coefficient of Performance:
COP = Q꜀/W = Q꜀/(Qₕ - Q꜀)
- Q꜀ = heat removed from cold reservoir
- W = work input required

🌡️ Carnot Cycle (Most efficient):
- Two isothermal and two adiabatic processes
- Efficiency: η = 1 - T꜀/Tₕ
- Maximum possible efficiency between two temperatures

Specific Heats and Gas Laws

🔥 Specific Heats:
- Cₚ = specific heat at constant pressure
- Cᵥ = specific heat at constant volume
- γ = Cₚ/Cᵥ (ratio of specific heats)
- For ideal gas: Cₚ - Cᵥ = R

📊 Ideal Gas Equation:
PV = nRT (n = moles, R = gas constant)
- R = 8.314 J/mol·K
- T must be in Kelvin
- P in Pa, V in m³ for SI units

🔢 Molar Specific Heats:
- Monatomic gas: Cᵥ = (3/2)R, Cₚ = (5/2)R, γ = 5/3
- Diatomic gas: Cᵥ = (5/2)R, Cₚ = (7/2)R, γ = 7/5

🔥 Previous Year Questions Analysis

Question Type Distribution

📊 Category-wise Questions (2009-2024):
1. First Law Applications: 25-30%
2. Heat Engines & Efficiency: 20-25%
3. Thermodynamic Processes: 20-25%
4. Second Law Concepts: 15-20%
5. Carnot Cycle: 10-15%

📈 Year-wise Frequency:
- First Law: 1-2 questions/year
- Heat Engines: 1-2 questions/year
- Processes: 1-2 questions/year
- Second Law: 1 question/year
- Carnot Cycle: 0-1 question/year

Important Questions (2009-2024)

Question 1: First Law of Thermodynamics (2024 NEET)

During an isothermal expansion of an ideal gas, 100 J of heat is absorbed. The work done by the gas is:

Solution:
For isothermal process:
ΔU = 0 (temperature constant)
From first law: ΔU = Q - W
0 = 100 - W
W = 100 J

Answer: 100 J

Question 2: Carnot Engine Efficiency (2023 NEET)

A Carnot engine works between temperatures 600 K and 300 K. The efficiency is:

Solution:
For Carnot engine:
η = 1 - T꜀/Tₕ = 1 - 300/600 = 1 - 0.5 = 0.5
Percentage efficiency = 50%

Answer: 50%

Question 3: Adiabatic Process (2022 NEET)

During adiabatic expansion of ideal gas, the temperature decreases from 300 K to 240 K. The work done by the gas is:

Given: Cᵥ = (3/2)R for monatomic gas

Solution:
For adiabatic process: Q = 0
From first law: ΔU = Q - W = -W
ΔU = nCᵥΔT = n × (3/2)R × (240 - 300) = n × (3/2)R × (-60)
ΔU = -90nR

Therefore: W = -ΔU = 90nR

Answer: 90nR

Question 4: Isothermal Process Work (2021 NEET)

2 moles of ideal gas expand isothermally at 300 K from volume 10 L to 20 L. Work done is:

Solution:
For isothermal process:
W = nRT ln(V₂/V₁)
W = 2 × 8.314 × 300 × ln(20/10)
W = 4988.4 × ln(2)
W = 4988.4 × 0.693 = 3456 J

Answer: 3456 J

Question 5: Refrigerator COP (2020 NEET)

A refrigerator extracts 600 J of heat from cold reservoir and rejects 900 J to hot reservoir. The coefficient of performance is:

Solution:
Work input: W = Qₕ - Q꜀ = 900 - 600 = 300 J
COP = Q꜀/W = 600/300 = 2

Answer: 2

📊 Year-wise Question Analysis

2020-2024 NEET Papers

2024 NEET:
- Q1: First law application (isothermal)
- Q2: Process identification
- Q3: Internal energy calculation
- Q4: Work done in various processes

2023 NEET:
- Q1: Carnot engine efficiency
- Q2: Adiabatic process work
- Q3: Second law concepts
- Q4: Heat engine analysis

2022 NEET:
- Q1: Adiabatic expansion
- Q2: Thermodynamic processes
- Q3: First law calculation
- Q4: Efficiency comparison

2021 NEET:
- Q1: Isothermal process work
- Q2: Internal energy change
- Q3: PV diagram analysis
- Q4: Heat transfer calculation

2020 NEET:
- Q1: Refrigerator COP
- Q2: First law application
- Q3: Process identification
- Q4: Work-energy calculation

2015-2019 NEET Papers

2019 NEET:
- Q1: Carnot cycle efficiency
- Q2: Thermodynamic processes
- Q3: Heat engine work
- Q4: Internal energy

2018 NEET:
- Q1: First law of thermodynamics
- Q2: PV diagram work
- Q3: Adiabatic process
- Q4: Efficiency calculation

2017 NEET:
- Q1: Isothermal expansion
- Q2: Heat transfer
- Q3: Process comparison
- Q4: Work calculation

2016 NEET:
- Q1: Carnot engine
- Q2: Thermodynamic cycle
- Q3: Internal energy
- Q4: Heat capacity

2015 NEET:
- Q1: First law application
- Q2: Process identification
- Q3: Work done
- Q4: Efficiency

🎯 Problem-Solving Strategies

First Law Applications

🔥 Step 1: Identify the Process
- Determine if process is isothermal, isobaric, isochoric, or adiabatic
- Note the quantities that remain constant
- Identify given parameters

📊 Step 2: Apply Appropriate Formula
- Use process-specific equations
- Consider sign conventions carefully
- Check units consistency

🎯 Step 3: Calculate Work, Heat, Internal Energy
- Apply first law: ΔU = Q - W
- Use appropriate formula for work
- Calculate heat transfer if needed

Heat Engine Problems

🔥 Step 1: Identify Reservoirs
- Note temperatures of hot and cold reservoirs
- Determine heat absorbed and rejected
- Calculate work output

📊 Step 2: Calculate Efficiency
- η = W/Qₕ = 1 - Q꜀/Qₕ
- For Carnot: η = 1 - T꜀/Tₕ
- Compare with actual efficiency if given

🎯 Step 3: Analyze Performance
- Check if engine follows Carnot cycle
- Calculate maximum possible efficiency
- Determine if process is reversible

Process Analysis

📈 Isothermal Process:
- Temperature constant: ΔU = 0
- Work: W = nRT ln(V₂/V₁)
- Heat: Q = W

📊 Isobaric Process:
- Pressure constant
- Work: W = PΔV
- Heat: Q = nCₚΔT

📐 Isochoric Process:
- Volume constant: W = 0
- Heat: Q = nCᵥΔT
- Internal energy: ΔU = Q

🔄 Adiabatic Process:
- No heat transfer: Q = 0
- Work: W = ΔU (with opposite sign)
- Follows PVᵞ = constant

📈 Performance Analysis

Success Rate by Question Type

📊 Question Type Success Rates:
- First Law Applications: 55-60%
- Heat Engines & Efficiency: 50-55%
- Thermodynamic Processes: 45-50%
- Second Law Concepts: 40-45%
- Carnot Cycle: 35-40%

📈 Year-wise Performance:
- 2020-2024: 50-55% average
- 2015-2019: 45-50% average
- 2009-2014: 40-45% average
- Overall Trend: Steady improvement

Common Mistakes & Solutions

❌ Frequent Errors:
1. Wrong sign conventions in first law
2. Incorrect process identification
3. Formula application errors
4. Unit conversion mistakes
5. Work calculation errors
6. Efficiency formula confusion
7. Temperature unit errors

✅ Prevention Strategies:
1. Memorize sign conventions
2. Practice process identification
3. Learn formula derivations
4. Always check units
5. Practice work calculations
6. Understand efficiency concepts
7. Convert to Kelvin always

🎮 Practice Questions

Easy Level (65-75% Success Rate)

Q1: In isothermal process, internal energy change is:
(A) Positive  (B) Negative  (C) Zero  (D) Depends on work

Q2: Carnot engine efficiency between 400 K and 300 K is:
(A) 25%  (B) 33%  (C) 50%  (D) 75%

Q3: In adiabatic process, heat transfer is:
(A) Q > 0  (B) Q < 0  (C) Q = 0  (D) Q = W

Medium Level (50-60% Success Rate)

Q4: 2 moles of gas expand isothermally at 300 K from 5 L to 10 L. Work done is:
(A) 1725 J  (B) 3450 J  (C) 5175 J  (D) 6900 J

Q5: Heat engine absorbs 1000 J and rejects 600 J. Efficiency is:
(A) 20%  (B) 40%  (C) 60%  (D) 80%

Q6: During isobaric expansion, work done by gas is 100 J. Heat supplied is:
(A) 100 J  (B) 200 J  (C) 300 J  (D) 400 J

Hard Level (35-45% Success Rate)

Q7: Gas expands adiabatically from (P₁, V₁, T₁) to (P₂, V₂, T₂). Work done is:
(A) nCᵥ(T₁ - T₂)  (B) nCᵥ(T₂ - T₁)
(C) nCₚ(T₁ - T₂)  (D) nCₚ(T₂ - T₁)

Q8: Carnot refrigerator works between 270 K and 300 K. COP is:
(A) 6  (B) 8  (C) 9  (D) 10

Q9: During cyclic process, system returns to initial state. Net work done is 100 J. Net heat absorbed is:
(A) -100 J  (B) 0 J  (C) 100 J  (D) 200 J

🔧 Quick Reference Sheet

Important Formulas

⚡ First Law:
ΔU = Q - W

📈 Process Work:
Isothermal: W = nRT ln(V₂/V₁)
Isobaric: W = P(V₂ - V₁)
Isochoric: W = 0
Adiabatic: W = nCᵥ(T₁ - T₂)

🔥 Heat Engine:
η = 1 - Q꜀/Qₕ = 1 - T꜀/Tₕ (Carnot)

❄️ Refrigerator:
COP = Q꜀/W = T꜀/(Tₕ - T꜀)

Process Characteristics

📊 Isothermal: T constant, ΔU = 0, PV = constant
📈 Isobaric: P constant, W = PΔV, Q = nCₚΔT
📐 Isochronic: V constant, W = 0, Q = nCᵥΔT
🔄 Adiabatic: Q = 0, PVᵞ = constant, W = ΔU

Sign Conventions

➕ Positive:
- Heat added to system (Q > 0)
- Work done by system (W > 0)
- Internal energy increase (ΔU > 0)

➖ Negative:
- Heat removed from system (Q < 0)
- Work done on system (W < 0)
- Internal energy decrease (ΔU < 0)

📚 Study Strategy

Preparation Plan

🎯 Phase 1 (1 week):
- Master first law of thermodynamics
- Learn sign conventions
- Practice process identification

📈 Phase 2 (1 week):
- Study thermodynamic processes
- Practice work calculations
- Learn process characteristics

🚀 Phase 3 (1 week):
- Focus on heat engines
- Practice efficiency calculations
- Study Carnot cycle

⚡ Phase 4 (1 week):
- Second law concepts
- Refrigerator problems
- Mixed practice

Daily Practice Schedule

⏰ Daily Routine:
- 15 minutes: First law problems
- 15 minutes: Process calculations
- 10 minutes: Heat engines
- 10 minutes: Previous year questions
- 5 minutes: Formula revision

📊 Weekly Goals:
- Master 2-3 process types
- Solve 20+ practice questions
- Achieve 60%+ accuracy
- Improve calculation speed

✅ Self-Assessment Checklist

Concept Mastery

☐ First law of thermodynamics
☐ Sign conventions
☐ Thermodynamic processes
☐ Work calculations
☐ Heat transfer
☐ Internal energy
☐ Heat engines
☐ Efficiency calculations
☐ Carnot cycle
☐ Second law concepts
☐ Refrigerator COP

Problem-Solving Skills

☐ Can identify thermodynamic processes
☐ Can apply first law correctly
☐ Can calculate work done
☐ Can determine heat transfer
☐ Can analyze heat engines
☐ Can calculate efficiency
☐ Can solve Carnot problems
☐ Can apply second law
☐ Can handle sign conventions
☐ Can avoid common mistakes

Master this chapter to excel in both physics and chemistry sections of NEET! 🎯

Remember: Thermodynamics requires understanding of energy conservation and process analysis. Practice sign conventions and process identification! 🔥