Entropy and Irreversibility

Real-Life Hook: Why Can’t You Unscramble an Egg?

Drop an egg, and it splatters. Mix cream into coffee, and it spreads. Break a glass, and shards scatter. We’ve all seen these happen—but have you ever seen them reverse spontaneously?

The egg never reassembles itself. The cream never unmixes. The glass never repairs itself. Why?

The answer is entropy—the measure of disorder. The Second Law of Thermodynamics states: Entropy of the universe always increases. This gives time its arrow—the reason we remember the past but not the future, and why all processes have a preferred direction.

Understanding entropy explains:

• Why heat flows from hot to cold (never reversed)
• Why perpetual motion machines are impossible
• Why mixing is easy, unmixing is impossible
• Why engines can’t be 100% efficient
• The ultimate fate of the universe (heat death)

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What is Entropy?

Definition

Entropy (S): A measure of the disorder or randomness of a system.

Mathematical Definition

For a reversible process:

Note: Temperature T must be in Kelvin!

Key Properties

1. State function: Depends only on initial and final states, not path
2. Extensive property: Depends on amount of substance (like mass, volume)
3. Never decreases for isolated system (Second Law!)
4. Increases for irreversible processes
5. Constant for reversible processes in isolated system

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Entropy and Disorder

Microscopic View

Entropy measures the number of microstates (arrangements) corresponding to a macrostate.

Boltzmann’s formula:

$$S = k_B \ln \Omega$$

where:

• $k_B$ = Boltzmann constant = $1.38 \times 10^{-23}$ J/K
• $\Omega$ = Number of microstates

Examples

1. Ice melting:

• Ice: Molecules in rigid crystal structure (low disorder, low S)
• Water: Molecules move freely (high disorder, high S)
• Melting: S increases ✓

2. Mixing gases:

• Separated: Each gas in own compartment (lower S)
• Mixed: Gases spread throughout (higher S)
• Mixing: S increases ✓

3. Breaking glass:

• Intact: Atoms in ordered structure (low S)
• Broken: Fragments scattered randomly (high S)
• Breaking: S increases ✓

Order vs Disorder

System	Order	Entropy
Crystal	High	Low
Liquid	Medium	Medium
Gas	Low	High
Dispersed	Very Low	Very High

General principle: Disorder increases → Entropy increases

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The Second Law (Entropy Statement)

Clausius Inequality

For any cyclic process:

Entropy and Second Law

Implications

1. Entropy of universe always increases (or stays constant)
2. Reversible processes: $\Delta S_{universe} = 0$
3. Irreversible processes: $\Delta S_{universe} > 0$
4. No process can decrease $S_{universe}$
5. Direction of time: Processes occur in direction of increasing entropy

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Calculating Entropy Changes

For Different Processes

1. Isothermal Process (T = constant)

2. Adiabatic Process (Q = 0)

For reversible adiabatic:

$$\Delta S = 0$$

(isentropic process)

For irreversible adiabatic:

$$\Delta S > 0$$

3. Isochoric Process (V = constant)

4. Isobaric Process (P = constant)

5. Phase Change (at constant T and P)

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Entropy in Different Processes

Reversible vs Irreversible

Reversible process:

• Quasi-static (infinitely slow)
• Always in equilibrium
• $\Delta S_{universe} = 0$
• Can be reversed with no net change

Irreversible process:

• Finite speed
• Not always in equilibrium
• $\Delta S_{universe} > 0$
• Cannot be reversed without leaving trace

Examples

Reversible:

• Carnot cycle (theoretical)
• Slow compression with heat bath

Irreversible:

• Free expansion (gas into vacuum)
• Heat flow across finite ΔT
• Friction
• Mixing of gases
• All real processes!

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Entropy Change for Ideal Gas

General Formula

For ideal gas going from $(P_1, V_1, T_1)$ to $(P_2, V_2, T_2)$:

Derivation

From First Law: $dU = dQ - dW$

For reversible process: $dQ = dU + PdV$

For ideal gas: $dU = nC_VdT$, $PV = nRT$

$$dQ = nC_VdT + \frac{nRT}{V}dV$$ $$dS = \frac{dQ}{T} = \frac{nC_VdT}{T} + \frac{nRdV}{V}$$

Integrate:

$$\Delta S = nC_V\ln\frac{T_2}{T_1} + nR\ln\frac{V_2}{V_1}$$

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Special Cases

1. Free Expansion

Setup: Gas expands into vacuum (no external pressure)

Observations:

• $Q = 0$ (isolated)
• $W = 0$ (no external pressure)
• $\Delta U = 0$ (from First Law)
• $\Delta T = 0$ (for ideal gas)

But volumes change: $V_1 \to V_2$

Entropy change:

$$\Delta S = nR\ln\frac{V_2}{V_1} > 0$$

Key point: Entropy increases even though Q = 0!

• Process is irreversible
• $\Delta S_{universe} > 0$

2. Heat Flow Across Finite ΔT

Setup: Heat Q flows from hot body ($T_H$) to cold body ($T_C$)

Entropy changes:

• Hot body: $\Delta S_H = -\frac{Q}{T_H}$ (loses heat)
• Cold body: $\Delta S_C = +\frac{Q}{T_C}$ (gains heat)

Total:

$$\Delta S_{total} = \Delta S_C + \Delta S_H = \frac{Q}{T_C} - \frac{Q}{T_H}$$

Since $T_C < T_H$:

$$\Delta S_{total} = Q\left(\frac{1}{T_C} - \frac{1}{T_H}\right) > 0$$

Entropy increases! Process is irreversible.

3. Mixing of Ideal Gases

Setup: Two gases at same T, P in separate compartments, partition removed

Entropy of mixing:

$$\Delta S_{mix} = -nR(x_1\ln x_1 + x_2\ln x_2)$$

where $x_1$, $x_2$ are mole fractions.

Always positive! (since $\ln x < 0$ for $x < 1$)

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Carnot Cycle and Entropy

Entropy Changes in Carnot Cycle

Step 1: Isothermal expansion at $T_H$

$$\Delta S_1 = \frac{Q_H}{T_H}$$

Step 2: Adiabatic expansion

$$\Delta S_2 = 0$$

Step 3: Isothermal compression at $T_C$

$$\Delta S_3 = -\frac{Q_C}{T_C}$$

Step 4: Adiabatic compression

$$\Delta S_4 = 0$$

Net entropy change of gas:

$$\Delta S_{gas} = \frac{Q_H}{T_H} - \frac{Q_C}{T_C}$$

For Carnot cycle: $\frac{Q_H}{T_H} = \frac{Q_C}{T_C}$ (can prove from efficiency)

Therefore:

$$\Delta S_{gas} = 0$$

Gas returns to same state → Entropy unchanged ✓

Entropy of Surroundings

Hot reservoir: Loses $Q_H$ at $T_H$

$$\Delta S_H = -\frac{Q_H}{T_H}$$

Cold reservoir: Gains $Q_C$ at $T_C$

$$\Delta S_C = +\frac{Q_C}{T_C}$$

Total:

$$\Delta S_{universe} = \Delta S_{gas} + \Delta S_H + \Delta S_C$$ $$= 0 - \frac{Q_H}{T_H} + \frac{Q_C}{T_C} = 0$$

For Carnot (reversible) cycle: $\Delta S_{universe} = 0$ ✓

For real (irreversible) engine: $\Delta S_{universe} > 0$

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Temperature-Entropy (T-S) Diagram

Why T-S Diagrams?

Just like P-V diagrams show work, T-S diagrams are useful for heat!

On T-S diagram:

• Area under curve = Heat absorbed/rejected
• Reversible process: Smooth curve
• Irreversible process: Doesn’t have unique path

Carnot Cycle on T-S Diagram

Shape: Rectangle!

• Top: Isothermal at $T_H$ (horizontal line)
• Bottom: Isothermal at $T_C$ (horizontal line)
• Sides: Adiabatic (vertical lines, since $\Delta S = 0$)

Area = Net work = $W = (T_H - T_C)(S_2 - S_1)$

Beautiful symmetry with P-V diagram!

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Common Mistakes & Tricks

Mistake 1: Confusing Entropy with Energy

❌ High entropy = High energy ✅ High entropy = High disorder

Energy and entropy are different!

• Energy: How much (quantity)
• Entropy: How spread out (quality)

Mistake 2: Thinking ΔS Can Be Negative

❌ Entropy can never decrease ✅ Entropy of isolated system never decreases

Local entropy can decrease (like freezing water), but universe entropy increases!

Mistake 3: Using °C in Entropy Formulas

❌ $\Delta S = Q/T$ with T in °C ✅ Always use Kelvin!

Temperature must be absolute (Kelvin) in all thermodynamic formulas.

Mistake 4: ΔS = 0 Means Reversible

❌ If $\Delta S_{system} = 0$, process is reversible ✅ Need $\Delta S_{universe} = 0$ for reversible

Check system + surroundings!

Mistake 5: Adiabatic Always Means ΔS = 0

❌ Adiabatic → $\Delta S = 0$ ✅ Reversible adiabatic → $\Delta S = 0$

Irreversible adiabatic (like free expansion) → $\Delta S > 0$!

Memory Trick: “SUDS”

Spontaneous processes increase Universe’s Disorder (Sentropy)

Trick: “Heat Flow”

Heat flows from hot to cold because:

$$\Delta S = \frac{Q}{T_C} - \frac{Q}{T_H} > 0$$

Smaller T → Bigger $\Delta S$ when gaining heat!

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Practice Problems

Level 1: JEE Main Basics

Q1. 100 g of ice at 0°C melts to water at 0°C. Find the entropy change. (Latent heat of fusion = 336 J/g)

Solution

Temperature: $T = 0°C = 273$ K

Heat absorbed: $Q = mL = 100 \times 336 = 33600$ J

Entropy change:

$$\Delta S = \frac{Q}{T} = \frac{33600}{273} = 123.1 \text{ J/K}$$

Entropy increases (ice is more ordered than water).

Q2. One mole of ideal gas expands isothermally at 300 K from 1 L to 5 L. Find entropy change.

Solution

For isothermal process:

$$\Delta S = nR\ln\frac{V_2}{V_1}$$ $$\Delta S = 1 \times 8.314 \times \ln(5)$$ $$\Delta S = 8.314 \times 1.609 = 13.38 \text{ J/K}$$

Entropy increases (gas expands, more disorder).

Q3. In which process does entropy of the system remain constant? (a) Isothermal (b) Adiabatic reversible (c) Isobaric (d) Isochoric

Solution

(b) Adiabatic reversible

In reversible adiabatic process:

• $Q = 0$ (adiabatic)
• $\Delta S = \int dQ/T = 0$

Also called isentropic process.

Other processes change entropy.

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Level 2: JEE Main/Advanced

Q4. 1 kg of water at 50°C is mixed with 1 kg of water at 30°C in an insulated container. Find the total entropy change. (Specific heat of water = 4200 J/kg·K)

Solution

Find final temperature:

Heat lost by hot water = Heat gained by cold water:

$$mc(50 - T_f) = mc(T_f - 30)$$ $$50 - T_f = T_f - 30$$ $$T_f = 40°C = 313 \text{ K}$$

Entropy change of hot water:

$$\Delta S_1 = mc\ln\frac{T_f}{T_1} = 1 \times 4200 \times \ln\frac{313}{323}$$ $$= 4200 \times \ln(0.969) = 4200 \times (-0.0315)$$ $$= -132.3 \text{ J/K}$$

Entropy change of cold water:

$$\Delta S_2 = mc\ln\frac{T_f}{T_2} = 1 \times 4200 \times \ln\frac{313}{303}$$ $$= 4200 \times \ln(1.033) = 4200 \times 0.0325$$ $$= 136.5 \text{ J/K}$$

Total entropy change:

$$\Delta S_{total} = \Delta S_1 + \Delta S_2 = -132.3 + 136.5$$ $$= 4.2 \text{ J/K}$$

Positive! As expected for irreversible process (heat flow across finite ΔT).

Q5. One mole of ideal gas ($\gamma = 1.4$) expands adiabatically from (2 atm, 300 K) to 1 atm. Find: (a) Final temperature (b) Entropy change if process is reversible (c) Entropy change if process is free expansion

Solution

(a) Final temperature:

For adiabatic: $TP^{(\gamma-1)/\gamma} = \text{const}$

$$T_2 = T_1\left(\frac{P_2}{P_1}\right)^{(\gamma-1)/\gamma}$$ $$T_2 = 300 \times \left(\frac{1}{2}\right)^{0.4/1.4}$$ $$T_2 = 300 \times (0.5)^{2/7} = 300 \times 0.82 = 246 \text{ K}$$

(b) Reversible adiabatic:

$$\Delta S = 0$$

(isentropic)

(c) Free expansion:

Even though Q = 0, process is irreversible!

Find volumes using ideal gas law:

$$V_1 = \frac{nRT_1}{P_1} = \frac{1 \times 8.314 \times 300}{2 \times 10^5}$$

For free expansion: $\Delta U = 0$ → $\Delta T = 0$ (ideal gas)

So $T_2 = T_1 = 300$ K (not 246 K!)

$$V_2 = \frac{nRT_2}{P_2} = \frac{1 \times 8.314 \times 300}{1 \times 10^5}$$ $$\frac{V_2}{V_1} = \frac{P_1}{P_2} = \frac{2}{1} = 2$$ $$\Delta S = nR\ln\frac{V_2}{V_1} = 1 \times 8.314 \times \ln(2)$$ $$= 8.314 \times 0.693 = 5.76 \text{ J/K}$$

Free expansion is irreversible → $\Delta S > 0$ ✓

Q6. A Carnot engine operates between 400 K and 300 K. It absorbs 1000 J from hot reservoir. Find: (a) Entropy change of hot reservoir (b) Entropy change of cold reservoir (c) Entropy change of universe

Solution

Carnot engine:

$$\eta = 1 - \frac{300}{400} = 0.25$$ $$W = 0.25 \times 1000 = 250 \text{ J}$$ $$Q_C = Q_H - W = 1000 - 250 = 750 \text{ J}$$

(a) Hot reservoir: Loses 1000 J at 400 K:

$$\Delta S_H = -\frac{1000}{400} = -2.5 \text{ J/K}$$

(b) Cold reservoir: Gains 750 J at 300 K:

$$\Delta S_C = \frac{750}{300} = 2.5 \text{ J/K}$$

(c) Universe:

$$\Delta S_{universe} = \Delta S_H + \Delta S_C + \Delta S_{gas}$$ $$= -2.5 + 2.5 + 0 = 0$$

Reversible process → $\Delta S_{universe} = 0$ ✓

If engine were irreversible: $\Delta S_{universe} > 0$

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Level 3: JEE Advanced

Q7. Two identical bodies at temperatures $T_1$ and $T_2$ are brought into thermal contact. Find the entropy change when they reach equilibrium. Show that $\Delta S > 0$ if $T_1 \neq T_2$.

Solution

Final temperature:

By energy conservation (equal heat capacities):

$$T_f = \frac{T_1 + T_2}{2}$$

Entropy change:

Each body: $\Delta S = C\ln(T_f/T_i)$

$$\Delta S_{total} = C\ln\frac{T_f}{T_1} + C\ln\frac{T_f}{T_2}$$ $$= C\ln\frac{T_f^2}{T_1T_2}$$

Substitute $T_f = (T_1 + T_2)/2$:

$$\Delta S = C\ln\frac{(T_1 + T_2)^2}{4T_1T_2}$$

Prove positive:

Need to show: $(T_1 + T_2)^2 > 4T_1T_2$

$$T_1^2 + 2T_1T_2 + T_2^2 > 4T_1T_2$$ $$T_1^2 - 2T_1T_2 + T_2^2 > 0$$ $$(T_1 - T_2)^2 > 0$$

This is always true if $T_1 \neq T_2$!

Therefore: $\Delta S > 0$ ✓

Special case: If $T_1 = T_2$ (already in equilibrium):

$$(T_1 - T_2)^2 = 0$$

→ $\Delta S = 0$ ✓

Conclusion: Entropy increases for irreversible process (heat flow across ΔT).

Q8. Show that for Carnot cycle, $\oint dQ/T = 0$.

Solution

Carnot cycle:

Process 1-2 (Isothermal at $T_H$):

$$\int_1^2 \frac{dQ}{T} = \frac{Q_H}{T_H}$$

Process 2-3 (Adiabatic):

$$\int_2^3 \frac{dQ}{T} = 0$$

(since Q = 0)

Process 3-4 (Isothermal at $T_C$):

$$\int_3^4 \frac{dQ}{T} = \frac{-Q_C}{T_C}$$

(Negative because heat is rejected)

Process 4-1 (Adiabatic):

$$\int_4^1 \frac{dQ}{T} = 0$$

Total:

$$\oint \frac{dQ}{T} = \frac{Q_H}{T_H} - \frac{Q_C}{T_C}$$

For Carnot engine:

From efficiency: $\eta = 1 - \frac{Q_C}{Q_H} = 1 - \frac{T_C}{T_H}$

Therefore: $\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$

Or: $\frac{Q_H}{T_H} = \frac{Q_C}{T_C}$

Substitute:

$$\oint \frac{dQ}{T} = \frac{Q_H}{T_H} - \frac{Q_C}{T_C} = 0$$

Proved!

This is Clausius theorem for reversible cycle.

For irreversible cycle: $\oint dQ/T < 0$

Q9. A gas undergoes free expansion from volume $V$ to $2V$. Calculate entropy change: (a) For the gas (b) For the surroundings (c) For the universe

Solution

Free expansion:

• Gas expands into vacuum
• $Q = 0$ (isolated)
• $W = 0$ (no external pressure)
• $\Delta U = 0$ → $\Delta T = 0$ (ideal gas)

(a) Entropy of gas:

Even though Q = 0, entropy increases (irreversible!):

$$\Delta S_{gas} = nR\ln\frac{V_2}{V_1} = nR\ln\frac{2V}{V}$$ $$= nR\ln 2 = 0.693nR$$

(b) Entropy of surroundings:

Since system is isolated:

$$\Delta S_{surr} = 0$$

(c) Entropy of universe:

$$\Delta S_{univ} = \Delta S_{gas} + \Delta S_{surr}$$ $$= 0.693nR + 0 = 0.693nR > 0$$

Key insight: Entropy increases even with Q = 0!

• Process is irreversible
• Second Law: $\Delta S_{univ} > 0$ ✓

Contrast with reversible expansion: For reversible isothermal expansion (same $\Delta V$):

• Gas: $\Delta S = nR\ln 2$ (same!)
• Surroundings: $\Delta S = -nR\ln 2$ (absorbs work as heat)
• Universe: $\Delta S = 0$ (reversible)

Q10. Prove that efficiency of Carnot engine can be written as:

$$\eta = 1 - \frac{T_C}{T_H}$$

using entropy considerations.

Solution

For Carnot cycle (reversible):

Net entropy change = 0 (returns to same state):

$$\Delta S_{cycle} = 0$$

Entropy changes:

From hot reservoir ($T_H$):

$$\Delta S_1 = \frac{Q_H}{T_H}$$

(Entropy gained from hot source)

To cold reservoir ($T_C$):

$$\Delta S_2 = -\frac{Q_C}{T_C}$$

(Entropy lost to cold sink, negative sign because rejected)

Total:

$$\Delta S_{cycle} = \frac{Q_H}{T_H} - \frac{Q_C}{T_C} = 0$$

Therefore:

$$\frac{Q_H}{T_H} = \frac{Q_C}{T_C}$$ $$\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$$

Efficiency:

$$\eta = 1 - \frac{Q_C}{Q_H} = 1 - \frac{T_C}{T_H}$$

Proved!

This shows: Carnot efficiency follows directly from entropy being state function (cycle returns to same S).

For irreversible engine:

$$\Delta S_{univ} > 0$$ $$\frac{Q_H}{T_H} - \frac{Q_C}{T_C} < 0$$ $$\frac{Q_C}{Q_H} > \frac{T_C}{T_H}$$ $$\eta = 1 - \frac{Q_C}{Q_H} < 1 - \frac{T_C}{T_H}$$

Irreversible efficiency is lower! ✓

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Connection to Other Topics

→ Second Law of Thermodynamics

Entropy is alternative statement of Second Law: Second Law →

→ Heat Engines

Carnot efficiency from entropy considerations: Heat Engines →

→ Thermodynamic Processes

Entropy changes in different processes: Processes →

→ Statistical Mechanics

Boltzmann’s formula connects entropy to microstates:

$$S = k_B\ln\Omega$$

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Philosophical Implications

The Arrow of Time

Why does time flow forward?

Entropy! The increase of entropy defines the direction of time:

• Past: Lower entropy (more ordered)
• Future: Higher entropy (more disordered)

We remember the past but not the future because:

• Our brain creates memories (ordered structures) from disordered surroundings
• This only works in direction of increasing entropy
• Reversing time would require decreasing universe entropy (forbidden!)

Heat Death of Universe

If entropy always increases, what’s the endgame?

Heat death: State of maximum entropy

• No temperature differences
• No energy gradients
• No work possible
• Universe reaches thermal equilibrium

Estimated: Trillions of years from now!

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Quick Revision Formula Sheet

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JEE Strategy Tips

1. Always use Kelvin in entropy calculations
2. State function: ΔS depends only on initial and final states
3. Isolated system: ΔS ≥ 0 (never negative!)
4. Reversible adiabatic: ΔS = 0 (isentropic)
5. Free expansion: ΔS > 0 even though Q = 0
6. For universe: Check system + surroundings
7. Carnot cycle: ΔS = 0 for gas, universe

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Back to: Thermodynamics Introduction

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Last updated: February 2025