Entropy
Concept, Formula, TS Diagrams & Solved Problems — Complete Guide for Engineering Students
Last Updated: March 2026
📌 Key Takeaways
- Entropy (S) is a thermodynamic property that measures the degree of molecular disorder and the unavailability of energy for useful work.
- Definition: dS = δQrev/T — entropy change equals reversible heat transfer divided by absolute temperature.
- Entropy is a state function — it depends only on the current state, not the path taken.
- Entropy increase principle: ΔSuniverse ≥ 0. Total entropy never decreases in any natural process.
- For reversible processes: ΔSuniverse = 0. For irreversible processes: ΔSuniverse > 0.
- SI unit: J/K (or kJ/K). Specific entropy: J/(kg·K).
1. What is Entropy?
Entropy is one of the most important — and most misunderstood — concepts in thermodynamics. At its core, entropy is a measure of disorder or randomness at the molecular level. A system with high entropy has its energy distributed across many possible molecular configurations; a system with low entropy has its energy concentrated in an orderly arrangement.
But entropy is more than just “disorder.” From an engineering perspective, entropy measures how much of a system’s energy is unavailable for conversion to useful work. When entropy increases, the quality of energy degrades — high-grade energy (concentrated, ordered, capable of doing work) becomes low-grade energy (dispersed, disordered, useless for work).
Every natural process — heat flowing from hot to cold, gas expanding into a vacuum, a ball bouncing to a stop — increases the total entropy of the universe. This is the essence of the Second Law of Thermodynamics. Entropy gives the Second Law its quantitative teeth.
For engineering students, entropy is practically essential because it appears in efficiency calculations for every thermodynamic cycle, in the analysis of irreversibilities, in the design of turbines, compressors, and heat exchangers, and throughout steam tables and property diagrams.
2. Mathematical Definition
Entropy Change — Reversible Process
dS = δQrev / T
For a finite process: ΔS = S&sub2; − S&sub1; = ∫&sub1;² δQrev / T
Where: dS = infinitesimal entropy change (J/K), δQrev = heat transferred in a reversible process (J), T = absolute temperature (K)
Why “reversible” heat? Entropy is a state function — its change between two states is fixed regardless of the path. However, the formula dS = δQ/T only gives the correct entropy change when the heat transfer is reversible. For irreversible processes, δQactual/T < ΔS (the actual heat transfer underestimates the entropy change).
To calculate ΔS for an irreversible process, you do not use the actual irreversible path. Instead, you construct any convenient reversible path between the same initial and final states, and calculate ΔS along that reversible path. Since S is a state function, the answer is the same regardless of which reversible path you choose.
3. Entropy Change for Ideal Gases
For an ideal gas with constant specific heats, entropy change can be expressed in two equivalent forms:
Form 1 — Using T and V
ΔS = nCv ln(T&sub2;/T&sub1;) + nR ln(V&sub2;/V&sub1;)
Or per unit mass: Δs = cv ln(T&sub2;/T&sub1;) + R ln(v&sub2;/v&sub1;)
Form 2 — Using T and P
ΔS = nCp ln(T&sub2;/T&sub1;) − nR ln(P&sub2;/P&sub1;)
Or per unit mass: Δs = cp ln(T&sub2;/T&sub1;) − R ln(P&sub2;/P&sub1;)
Form 3 — Using P and V
ΔS = nCv ln(P&sub2;/P&sub1;) + nCp ln(V&sub2;/V&sub1;)
Choose the form based on which properties are known. All three are equivalent for ideal gases.
4. Entropy Change for Common Processes
| Process | Entropy Change Formula | Notes |
|---|---|---|
| Isothermal (T = const) | ΔS = Q/T = nR ln(V&sub2;/V&sub1;) | Expansion: ΔS > 0. Compression: ΔS < 0. |
| Isentropic / Adiabatic reversible | ΔS = 0 | No heat transfer, reversible → entropy unchanged. TVγ−1 = constant. |
| Isochoric (V = const) | ΔS = nCv ln(T&sub2;/T&sub1;) | Heating: ΔS > 0. Cooling: ΔS < 0. |
| Isobaric (P = const) | ΔS = nCp ln(T&sub2;/T&sub1;) | Heating: ΔS > 0. Cooling: ΔS < 0. |
| Phase change (at constant T, P) | ΔS = mL/T | L = latent heat (J/kg), T = phase change temperature (K). Melting/vaporisation: ΔS > 0. |
| Free expansion (ideal gas into vacuum) | ΔS = nR ln(V&sub2;/V&sub1;) > 0 | Irreversible. Q = 0 but ΔS > 0 because the process is irreversible. |
| Heat transfer across ΔT | ΔSuniverse = Q(1/TL − 1/TH) > 0 | Always generates entropy. The larger the temperature difference, the more entropy generated. |
5. TS Diagrams — Reading Them
The Temperature-Entropy (TS) diagram is one of the most useful tools in thermodynamics. Key properties of TS diagrams:
- Area under a curve on a TS diagram represents heat transfer: Q = ∫T dS. This is analogous to how area under a PV curve represents work.
- Vertical lines represent isentropic (constant-entropy) processes — adiabatic reversible compression or expansion.
- Horizontal lines represent isothermal processes.
- The vapour dome (bell-shaped curve) separates the liquid, two-phase, and vapour regions. The top of the dome is the critical point.
- Inside the dome, horizontal lines represent phase changes (evaporation or condensation at constant T and P).
- For any cycle, the enclosed area equals the net work output (= QH − QL).
Why TS diagrams matter: The Carnot cycle is a rectangle on the TS diagram (two isothermals + two isentropics). The Rankine cycle shows the phase change and superheating clearly. TS diagrams make it easy to visualise where entropy is generated and where efficiency is lost.
6. Entropy Generation in Irreversible Processes
In any real (irreversible) process, entropy is generated — the total entropy of the universe increases. The entropy balance for a system is:
Entropy Balance
ΔSsystem = ∫ δQ/Tboundary + Sgen
Where: Sgen ≥ 0 always. Sgen = 0 for reversible processes, Sgen > 0 for irreversible processes.
Common sources of entropy generation (irreversibility):
| Source | Why It Generates Entropy |
|---|---|
| Heat transfer across finite ΔT | Energy moves from high-quality (hot) to low-quality (cold) form |
| Friction | Converts ordered mechanical energy into disordered thermal energy |
| Unresisted expansion (free expansion) | Gas fills available volume without doing useful work |
| Mixing of different substances | Increases randomness — mixed state is more disordered |
| Chemical reactions | Inherently irreversible rearrangement of molecular bonds |
| Electrical resistance (Joule heating) | Converts electrical energy to heat — work → thermal energy |
Engineering significance: Every unit of entropy generated represents lost potential for doing useful work. Minimising entropy generation is the fundamental goal of efficient engineering design. This is the basis of exergy (availability) analysis.
7. Worked Numerical Examples
Example 1: Entropy Change — Heating an Ideal Gas at Constant Volume
Problem: 2 kg of air (cv = 0.718 kJ/kg·K) is heated from 300 K to 600 K at constant volume. Calculate the entropy change.
Solution
ΔS = mcv ln(T&sub2;/T&sub1;) = 2 × 0.718 × ln(600/300)
= 1.436 × ln(2) = 1.436 × 0.6931
ΔS = 0.995 kJ/K
Entropy increases — the gas becomes more disordered as its temperature doubles.
Example 2: Entropy Change — Phase Change (Boiling Water)
Problem: 5 kg of water at 100°C is completely vaporised at atmospheric pressure. Latent heat of vaporisation = 2,257 kJ/kg. Find the entropy change.
Solution
T = 100 + 273 = 373 K
ΔS = mL/T = 5 × 2257 / 373 = 11,285 / 373
ΔS = 30.25 kJ/K
Large entropy increase — steam is far more disordered than liquid water.
Example 3: Entropy Generation — Heat Transfer Across Finite ΔT
Problem: 1,000 kJ of heat is transferred from a reservoir at 800 K to a reservoir at 300 K. Calculate the entropy generated.
Solution
ΔShot = −Q/TH = −1000/800 = −1.25 kJ/K (entropy decreases for hot reservoir)
ΔScold = +Q/TL = +1000/300 = +3.333 kJ/K (entropy increases for cold reservoir)
Sgen = ΔSuniverse = −1.25 + 3.333 = +2.083 kJ/K
2.083 kJ/K of entropy is generated — this represents permanently lost work potential.
Example 4: Isentropic Process — Verify Zero Entropy Change
Problem: Air (R = 0.287 kJ/kg·K, cv = 0.718 kJ/kg·K, γ = 1.4) is compressed isentropically from T&sub1; = 300 K, V&sub1; = 1 m³ to V&sub2; = 0.125 m³ (compression ratio = 8). Find T&sub2; and verify ΔS = 0.
Solution
T&sub2; = T&sub1; × (V&sub1;/V&sub2;)γ−1 = 300 × 80.4 = 300 × 2.297 = 689.1 K
Verify: ΔS = mcv ln(T&sub2;/T&sub1;) + mR ln(V&sub2;/V&sub1;)
Per unit mass: Δs = 0.718 × ln(689.1/300) + 0.287 × ln(0.125/1)
= 0.718 × 0.831 + 0.287 × (−2.079) = 0.597 − 0.597 = 0 ✓
8. Common Mistakes Students Make
- Using the actual irreversible Q to calculate ΔS: The formula ΔS = Q/T only works when Q is transferred reversibly. For irreversible processes, you must find a reversible path between the same states and calculate ΔS along that path.
- Thinking ΔS = 0 for all adiabatic processes: ΔS = 0 only for reversible adiabatic (isentropic) processes. An irreversible adiabatic process (e.g., free expansion, friction) has ΔS > 0 even though Q = 0.
- Claiming a system’s entropy cannot decrease: A system’s entropy CAN decrease (e.g., cooling, compression, freezing). The Second Law requires only that the total entropy (system + surroundings) does not decrease.
- Using Celsius in entropy formulas: Always use Kelvin. Using Celsius in ln(T&sub2;/T&sub1;) gives incorrect results because Celsius is not an absolute scale.
- Confusing entropy with enthalpy: Entropy (S, units J/K) measures disorder. Enthalpy (H, units J) measures total heat content at constant pressure. They are different properties with different physical meanings.
9. Frequently Asked Questions
What is entropy in simple terms?
Entropy measures how spread out or disordered the energy in a system is. A hot cup of tea in a cold room has low entropy — the energy is concentrated. After the tea cools to room temperature, the energy is spread evenly across the tea and room — high entropy. Every natural process moves in the direction of greater entropy (more spread-out energy), and this is what makes natural processes irreversible.
Can entropy of a system decrease?
Yes. When you freeze water into ice, the entropy of the water decreases — the ice molecules are more ordered than liquid molecules. But the freezer motor generates heat that increases the entropy of the room air by even more. The total entropy (ice + room) increases. The Second Law applies to the universe, not individual systems.
What is the formula for entropy change?
The fundamental definition is ΔS = ∫(δQrev/T). For ideal gases, the most useful forms are: ΔS = nCv ln(T&sub2;/T&sub1;) + nR ln(V&sub2;/V&sub1;), or ΔS = nCp ln(T&sub2;/T&sub1;) − nR ln(P&sub2;/P&sub1;). For phase changes at constant temperature: ΔS = mL/T. Choose the form based on which properties are given in the problem.