Here's what surprises most Physics students: Thermodynamics tests understanding, not memorisation. Yet 70% of aspirants score poorly because they memorise formulas without grasping concepts.

After analyzing 26 years of NEET papers (2000–2025), we found:

67 questions dedicated to Thermodynamics
4.68% of all Physics marks
Appears in 18 out of 26 years
Heavily conceptual (not calculation-heavy)

Yet most students:

Memorise ΔU = Q - W without understanding sign convention
Skip entropy (2nd law) thinking it's "too complex"
Confuse heat capacity, specific heat, latent heat
Don't understand why heat engines have efficiency limits

This post is your complete Thermodynamics roadmap: why the laws work, which PV diagrams repeat, which problems trick students, and how to master all concepts in 8 days.

1. Why Thermodynamics Matters

The Numbers:

Metric	Value
Questions (2000–2025)	67
% of Physics	4.68%
Years Appeared	18/26
Questions per paper	~1–2
Nature	Highly conceptual

2. Thermodynamics Core Concepts

Topic 1: First Law of Thermodynamics (~15–20 questions)

The Law:

ΔU = Q - W

Where:

ΔU = Change in internal energy
Q = Heat added to system (positive if added, negative if removed)
W = Work done BY system (positive if system expands, negative if compressed)

Physical Meaning: Energy is conserved. Heat added = increase in internal energy + work done by system.

Sign Convention (Critical):

Q positive: Heat enters system
Q negative: Heat leaves system
W positive: System expands (does work on surroundings)
W negative: System compressed (surroundings do work on system)

Internal Energy (U):

Depends ONLY on temperature
For ideal gas: U = nCᵥT (where Cᵥ = molar heat capacity at constant volume)
ΔU = nCᵥΔT (independent of path)

Common Question Types:

Type A: "In an expansion, gas does work on surroundings. Is W positive or negative?"
→ W is positive (system does work, energy leaves system)

Type B: "Calculate ΔU if Q = 100 J and W = 40 J (system expands)."
→ ΔU = Q - W = 100 - 40 = 60 J

Topic 2: Second Law of Thermodynamics & Entropy (~15–20 questions)

The Law:
Entropy of isolated system always increases (or stays same for reversible process).

Entropy (S):

Measure of disorder/randomness
ΔS = Q_rev / T (change in entropy = reversible heat / temperature)
For ideal gas expansion: ΔS = nR ln(V_f / V_i)

Physical Meaning:

Natural processes are irreversible (increase entropy)
Heat flows from hot → cold (natural direction)
All processes have some friction/irreversibility

Consequence for Heat Engines:

No heat engine is 100% efficient
Maximum efficiency = Carnot efficiency = 1 - T_cold / T_hot
Real engines always less efficient

Common Question Types:

Type A: "Why does heat naturally flow from hot to cold, not reverse?"
→ Entropy increases: Hot object loses heat (disorder increases) more than cold object gains (order increases less)
→ Net entropy increases (favorable)

Type B: "What is maximum efficiency of heat engine between 300 K and 600 K?"
→ η_max = 1 - 300/600 = 0.5 = 50%

Topic 3: Heat Capacity vs. Specific Heat vs. Latent Heat (~10–15 questions)

Heat Capacity (C):

Q = CΔT

Amount of heat needed to raise temperature by 1°C.
Units: J/°C or J/K

Specific Heat (c):

Q = mcΔT

Heat capacity per unit mass.
Units: J/(g·°C) or J/(kg·K)

Molar Heat Capacity (Cₘ):

Q = nCₘΔT

Heat capacity per mole.
For ideal gas:

Cᵥ (constant volume) = (f/2)R (where f = degrees of freedom)
Cₚ (constant pressure) = Cᵥ + R
For diatomic gas: Cᵥ = 5R/2, Cₚ = 7R/2

Latent Heat (L):

Q = mL

Heat needed to change phase (solid → liquid or liquid → gas) at constant temperature.
Examples:

Latent heat of fusion (solid → liquid)
Latent heat of vaporization (liquid → gas)

NEET Table:

Quantity	Formula	Units	Meaning
Heat Capacity	C = Q/ΔT	J/K	Total heat per degree
Specific Heat	c = Q/(mΔT)	J/(kg·K)	Per unit mass
Molar Heat Capacity	Cₘ = Q/(nΔT)	J/(mol·K)	Per mole
Latent Heat	L = Q/m	J/kg	For phase change

Common Question Types:

Type A: "How much heat to raise 2 kg of water by 10°C? (c = 4200 J/(kg·K))"
→ Q = mcΔT = 2 × 4200 × 10 = 84,000 J

Type B: "What is Cₚ for a diatomic ideal gas?"
→ Cₚ = Cᵥ + R = 5R/2 + R = 7R/2

Topic 4: Processes (Isothermal, Adiabatic, Isobaric, Isochoric) (~15–20 questions)

Isothermal Process (T = constant):

Temperature doesn't change
For ideal gas: PV = constant (Boyle's law)
ΔU = 0 (internal energy unchanged)
Q = W (all heat becomes work)
On PV diagram: Hyperbola

Adiabatic Process (Q = 0, no heat exchange):

Insulated system, no heat in/out
ΔU = -W (work changes internal energy)
Temperature changes even though no heat exchange
On PV diagram: Steeper than isothermal

Isobaric Process (P = constant):

Pressure constant
W = PΔV
Q = nCₚΔT
On PV diagram: Horizontal line

Isochoric Process (V = constant):

Volume constant
W = 0 (no work)
Q = ΔU = nCᵥΔT
On PV diagram: Vertical line

NEET Process Table:

Process	Constant	ΔU	Q	W	PV Diagram
Isothermal	T	0	W	nRT ln(V_f/V_i)	Hyperbola
Adiabatic	None (but S const)	-W	0	ΔU	Steeper curve
Isobaric	P	nCᵥΔT	nCₚΔT	PΔV	Horizontal
Isochoric	V	nCᵥΔT	nCᵥΔT	0	Vertical

Common Question Types:

Type A: "In isothermal compression of ideal gas, is internal energy change zero?"
→ Yes, ΔU = 0 (T constant, internal energy depends only on T)

Type B: "In adiabatic expansion, does temperature increase or decrease?"
→ Decreases (Q = 0, W positive → ΔU negative → T decreases)

3. PV Diagrams (Most Common in NEET)

Reading PV Diagrams:

X-axis: Volume
Y-axis: Pressure
Area under curve: Work done by system

Common Path Types:

Isothermal (Hyperbola): PV = constant
Adiabatic (Steep curve): PV^γ = constant
Isobaric (Horizontal): P = constant
Isochoric (Vertical): V = constant

Cycle Processes:

Carnot Cycle: 2 isothermals + 2 adiabatics (maximum efficiency)
Otto Cycle: 2 isochoric + 2 adiabatic (car engine)

Common Question Types:

Type A: "On PV diagram, identify which process is isothermal."
→ The hyperbola-shaped curve

Type B: "Calculate work done in isobaric process: P = 100 Pa, V changes from 2 to 5 m³."
→ W = PΔV = 100 × (5-2) = 300 J

4. Most Common Mistakes

Mistake 1: Sign Convention for Work

Wrong: "W is always positive (work is always done)."
Reality: W positive = system expands (does work). W negative = system compressed.

Mistake 2: Confusing Q and W

Wrong: "Q - W means Q minus W numerically."
Reality: Must respect signs: ΔU = (+Q_in) - (+W_out)

Mistake 3: Thinking Entropy Only Decreases

Wrong: "Entropy of closed system can decrease."
Reality: Entropy of isolated system always increases (2nd law).

Mistake 4: Ignoring Heat Capacity Types

Wrong: Using specific heat when molar heat capacity is needed.
Reality: Check units and mass vs. moles in problem.

5. 8-Day Thermodynamics Mastery Plan

Days	Focus	Daily Work	Target
Days 1–2	First Law & Internal Energy	NCERT + sign convention	Master ΔU = Q - W
Days 3–4	Second Law & Entropy	Understand irreversibility	Know why efficiency < 100%
Days 5–6	Heat Capacity Types	Specific, molar, latent heat	Distinguish all types
Days 7–8	Processes & PV Diagrams	Isothermal, adiabatic, etc.	Solve 20 PV diagram problems

Daily Schedule (1.5 hours):

Concept read + diagrams (40 min)
Problem-solving: 10–12 questions (45 min)

6. Call to Action

Ready to Master Thermodynamics?

✅ Download the "First & Second Law Quick Reference" (ΔU formulas, sign conventions) → Study guide
✅ Get the "PV Diagram Process Identifier" (Isothermal, adiabatic, isobaric, isochoric) → Visual reference
✅ Access the "Heat Capacity Comparison Table" (C vs. c vs. Cₘ vs. L) → Quick reference
✅ Unlock All Thermodynamics PYQs (2000–2025) (67 questions, sorted by type) → Full question bank
✅ Join the Thermodynamics Study Group (PV diagram challenges, entropy discussions) → Community

Key Takeaways:

Thermodynamics is conceptual, not calculation-heavy
Master sign conventions for Q and W
Understand why heat engines have efficiency limits
Practice 20+ PV diagram problems
Focus on understanding, not memorization

NEET Master

NEET Master

NEET Master