Interactive Lecture
The Second Law of Thermodynamics and Irreversibility
by Dr Arnaldo Delli Carri
Entropy is a measure of disorder, randomness, or — most usefully for engineers — the unavailability of energy to do work. The same quantity reaches well beyond thermodynamics, into information theory and cosmology, but it is the engineering meaning this lecture develops.
Entropy is a state function that quantifies the degree of disorder in a system. In any real (irreversible) process, the total entropy of an isolated system always increases — that asymmetry is what gives thermodynamics its direction in time.
Entropy can be understood from multiple perspectives:
The statistical definition, introduced by Ludwig Boltzmann, provides deep insight:
\[S = k_B \ln(\Omega)\]
Where:
This equation is so fundamental that it's engraved on Boltzmann's tombstone!
The Second Law can be stated in several equivalent ways:
Clausius Statement: Heat cannot spontaneously flow from a colder body to a hotter body.
Kelvin-Planck Statement: It is impossible to construct a heat engine that operates in a cycle and produces no effect other than the absorption of heat from a reservoir and the performance of an equal amount of work.
Entropy Statement: The entropy of an isolated system never decreases; it either increases (irreversible process) or remains constant (reversible process).
\[\Delta S_{universe} \geq 0\]
For a reversible process, the entropy change is defined as:
\[dS = \frac{\delta Q_{rev}}{T}\]
For a finite reversible process:
\[\Delta S = \int_1^2 \frac{\delta Q_{rev}}{T}\]
Where:
For specific processes:
Isothermal Process (constant T):
\[\Delta S = \frac{Q}{T}\]
Isobaric Process (ideal gas):
\[\Delta S = m c_p \ln\left(\frac{T_2}{T_1}\right)\]
Isochoric Process (ideal gas):
\[\Delta S = m c_v \ln\left(\frac{T_2}{T_1}\right)\]
General Process (ideal gas):
\[\Delta S = m c_v \ln\left(\frac{T_2}{T_1}\right) + m R_s \ln\left(\frac{V_2}{V_1}\right)\]
Here \(R_s\) is the specific gas constant (J/(kg·K)), not the universal constant used elsewhere on this page: \(R_s = R/M\), where \(R = 8.314\) J/(mol·K) and \(M\) is the molar mass (kg/mol).
Explore the ideal reversible heat engine operating between two thermal reservoirs. The Carnot cycle represents the maximum possible efficiency for any heat engine and demonstrates fundamental principles of entropy.
The Carnot efficiency depends only on the temperatures of the hot and cold reservoirs: \(\eta_{Carnot} = 1 - \frac{T_C}{T_H}\). No real heat engine operating between these two reservoirs can exceed it.
Observe how a gas spontaneously expands into a vacuum, increasing entropy even though no heat is transferred. Watch the real-time entropy graph and density histogram to see the Second Law in action.
Entropy increases in free expansion even though \(Q = 0\) and \(W = 0\), because the number of accessible microstates grows: \(\Delta S = nR\ln(V_2/V_1)\). The gas will never spontaneously return to the left half — this irreversibility establishes the thermodynamic arrow of time.
The Second Law reaches into almost every branch of engineering — here's where entropy directly shapes design decisions:
Every real engineering process generates entropy. Minimising entropy generation is equivalent to minimising energy waste and improving efficiency — the principle behind design choices from power plants down to microprocessors.
| Concept | Equation | Description |
|---|---|---|
| Boltzmann Entropy | \(S = k_B \ln(\Omega)\) | Statistical definition of entropy |
| Entropy Change (Reversible) | \(dS = \frac{\delta Q_{rev}}{T}\) | Differential entropy change |
| Second Law | \(\Delta S_{universe} \geq 0\) | Entropy never decreases in isolated system |
| Carnot Efficiency | \(\eta_{Carnot} = 1 - \frac{T_C}{T_H}\) | Maximum heat engine efficiency |
| Isothermal Process | \(\Delta S = \frac{Q}{T}\) | Entropy change at constant temperature |
| Isobaric (Ideal Gas) | \(\Delta S = mc_p\ln(T_2/T_1)\) | Constant pressure process |
| Isochoric (Ideal Gas) | \(\Delta S = mc_v\ln(T_2/T_1)\) | Constant volume process |
| Free Expansion | \(\Delta S = nR\ln(V_2/V_1)\) | Entropy increase in free expansion |
| Gibbs Free Energy | \(G = H - TS\) | Determines spontaneity at constant T, P |
| Heat Engine Efficiency | \(\eta = \frac{W}{Q_H} = 1 - \frac{Q_C}{Q_H}\) | Actual heat engine efficiency |
| COP (Refrigerator) | \(COP_R = \frac{Q_C}{W} = \frac{Q_C}{Q_H - Q_C}\) | Coefficient of Performance |
| Clausius Inequality | \(\oint \frac{\delta Q}{T} \leq 0\) | For any cyclic process |
A Carnot heat engine operates between a hot reservoir at 800 K and a cold reservoir at 300 K. Calculate:
a) The thermal efficiency of the engine
b) If the engine absorbs 2000 J of heat from the hot reservoir per cycle, how much work does it produce?
c) How much heat is rejected to the cold reservoir?
Show Solution
Solution:
Given:
Part a) Thermal Efficiency:
The Carnot efficiency is given by:
\[\eta_{Carnot} = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{800} = 1 - 0.375 = 0.625\]
Answer: \(\eta = 62.5\%\)
Part b) Work Output:
\[W = \eta \times Q_H = 0.625 \times 2000 = 1250 \text{ J}\]
Answer: W = 1250 J
Part c) Heat Rejected:
From energy conservation: \(Q_H = W + Q_C\)
\[Q_C = Q_H - W = 2000 - 1250 = 750 \text{ J}\]
Alternatively: \(Q_C = (1-\eta) \times Q_H = 0.375 \times 2000 = 750\) J
Answer: QC = 750 J
A 2 kg block of copper (cp = 385 J/(kg·K)) is heated from 20°C to 120°C at constant pressure. Calculate the entropy change of the copper block.
Show Solution
Solution:
Given:
For a constant-pressure process, entropy change is given by:
\[\Delta S = mc_p\ln\left(\frac{T_2}{T_1}\right)\]
Calculation:
\[\Delta S = 2 \times 385 \times \ln\left(\frac{393}{293}\right)\]
\[\Delta S = 770 \times \ln(1.341) = 770 \times 0.2939\]
\[\Delta S = 226.3 \text{ J/K}\]
Answer: The entropy change is approximately 226 J/K
Two identical blocks of aluminium (mass = 1 kg each, cp = 900 J/(kg·K)) are initially at 400 K and 300 K respectively. They are brought into thermal contact in an insulated container until they reach thermal equilibrium. Calculate:
a) The final equilibrium temperature
b) The total entropy change of the system
c) Is this process reversible or irreversible?
Show Solution
Solution:
Given:
Part a) Final Temperature:
Energy conservation (insulated system, no heat loss):
\[m c_p (T_f - T_1) + m c_p (T_f - T_2) = 0\]
\[2 T_f = T_1 + T_2\]
\[T_f = \frac{400 + 300}{2} = 350 \text{ K}\]
Answer: Tf = 350 K
Part b) Total Entropy Change:
For block 1 (cooling from 400 K to 350 K):
\[\Delta S_1 = mc_p\ln\left(\frac{350}{400}\right) = 1 \times 900 \times \ln(0.875) = -120.18 \text{ J/K}\]
For block 2 (heating from 300 K to 350 K):
\[\Delta S_2 = mc_p\ln\left(\frac{350}{300}\right) = 1 \times 900 \times \ln(1.167) = 900 \times 0.1542 = 138.74 \text{ J/K}\]
Total entropy change:
\[\Delta S_{total} = \Delta S_1 + \Delta S_2 = -120.18 + 138.74 = 18.56 \approx 18.6 \text{ J/K}\]
Answer: ΔStotal = 18.6 J/K
Part c) Reversible or Irreversible?
Since \(\Delta S_{total} > 0\), heat has spontaneously flowed from hot to cold while generating entropy — consistent with the Second Law.
Answer: The process is irreversible.
One mole of an ideal gas initially occupies 0.01 m³ at 300 K. It undergoes free expansion into an evacuated chamber, doubling its volume. The process is adiabatic. Calculate:
a) The final temperature
b) The work done by the gas
c) The entropy change
Show Solution
Solution:
Given:
Part a) Final Temperature:
In free expansion, no work is done against external pressure (W = 0).
Since the process is adiabatic (Q = 0) and W = 0:
From the First Law: \(\Delta U = Q - W = 0\)
For an ideal gas, internal energy depends only on temperature.
Therefore, temperature remains constant.
Answer: T₂ = 300 K
Part b) Work Done:
No work is done against external pressure.
Answer: W = 0 J
Part c) Entropy Change:
Even though Q = 0, entropy still increases because the process is irreversible:
\[\Delta S = nR\ln\left(\frac{V_2}{V_1}\right) = 1 \times 8.314 \times \ln(2)\]
\[\Delta S = 8.314 \times 0.693 = 5.76 \text{ J/K}\]
Answer: ΔS = 5.76 J/K
Free expansion is spontaneous and irreversible. Entropy increases even though no heat is transferred!