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Thermal Physics
Entropy and the Second Law of Thermodynamics
Entropy is a state function representing a measure of the unavailability of a system's thermal energy for conversion into mechanical work. Mathematically, it is defined through the macroscopic change dQ/T for a reversible process.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of entropy, Clausius inequality, entropy changes in ideal gas processes (isothermal, isobaric, isochoric, adiabatic), and principle of increase of entropy.
PGT extension
Statistical interpretation of entropy (Boltzmann relation), entropy of mixing, exact differential property of state functions, and T-S diagram analysis for thermodynamic cycles.
Prerequisites
Concept of Temperature
Laws of Thermodynamics
Kinetic theory of gases
Macroscopic Definition and State Function Nature
Entropy (S) is an extensive thermodynamic state function, meaning its value depends only on the current state of the system, independent of the path taken.
The infinitesimal change in entropy dS for a reversible process is given by the ratio of heat added reversibly (dQrev) to the absolute temperature (T).
The cyclic integral of dQ/T over any closed reversible cycle is identically zero, proving that entropy is a point function.
Clausius Inequality and the Second Law
For any arbitrary cycle (reversible or irreversible), the cyclic integral of dQ/T is less than or equal to zero: \oint \frac{dQ}{T} \le 0.
For an irreversible process between two equilibrium states, the entropy change dS > \frac{dQ}{T}.
In an isolated system (dQ = 0), entropy can never decrease; it remains constant for reversible processes (dS = 0) and increases for irreversible processes (dS > 0).
Statistical Interpretation of Entropy
Boltzmann related macroscopic entropy to microscopic probability through the relation $S = k_{B} \ln(\Omega)$, where kB is the Boltzmann constant and Ω is the thermodynamic probability (number of microstates).
Entropy is a measure of the degree of molecular disorder or randomness in a system.
Absolute entropy at T = 0 K is zero for a perfectly crystalline substance (Nernst Heat Theorem / Third Law of Thermodynamics).
T-S Diagrams (Temperature-Entropy)
A Temperature-Entropy (T-S) diagram plots absolute temperature on the ordinate and entropy on the abscissa.
The area under the T-S curve for a reversible process represents the total heat transferred: Q = \int T dS.
Isothermal processes appear as horizontal lines on a T-S diagram, whereas isentropic (reversible adiabatic) processes appear as vertical lines.
Derivations
Entropy Change of an Ideal Gas
Conditions: One mole of an ideal gas undergoing a state change from (T1, V1) to (T2, V2)
Start from the first law of thermodynamics for a closed system: d$Q = dU + dW$
Substitute d$U = C_{V} dT$ and d$W = P dV = \frac{RT}{V} dV into the expression: dQ = C_V dT + \frac{RT}{V} dV$
Divide by absolute temperature T to obtain the entropy change: d$S = \frac{dQ}{T} = C_V \frac{dT}{T} + R \frac{dV}{V}$
Integrate between initial state (T1, V1) and final state (T2, V2): \Delta $S = \int_{T_1}^{T_2} C_V \frac{dT}{T} + \int_{V_1}^{V_2} R \frac{dV}{V}$
Result: \Delta $S = C_V \ln\left(\frac{T_2}{T_1}\right) + R \ln\left(/frac{V_2}{V_1}\right)$
Entropy Change during Phase Transition
Conditions: Reversible phase change at constant temperature T and pressure P (e.g., melting or vaporization)
During a phase change, the temperature remains constant at the transition temperature Tt.
The heat supplied reversibly is equal to the latent heat of transition (L): Q = L.
Substitute into the definition of entropy change: \Delta $S = \int \frac{dQ_{rev}}{T} = \frac{1}{T_t} \int dQ$
Evaluate the integral for mass m with latent heat L.
Result: \Delta $S = \frac{m L}{T_t}$
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Infinitesimal Entropy Change
$$dS = \frac{dQ_{rev}}{T}$$
dQrev = heat added reversibly, T = absolute temperature J K-1
Reversible process
Clausius Inequality
$$\oint \frac{dQ}{T} \le 0$$
dQ = infinitesimal heat transfer along an arbitrary cycle Dimensionless integral
Any thermodynamic cycle
Entropy of Ideal Gas in terms of (T, V)
$$S2 - S1 = C_V \ln\left(\frac{T_2}{T_1}\right) + R \ln\left(\frac{V_2}{V_1}\right)$$
CV = molar heat capacity at constant volume, R = universal gas constant J K-1 mol-1
Ideal gas
Entropy of Ideal Gas in terms of (T, P)
$$S2 - S1 = C_P \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right)$$
CP = molar heat capacity at constant pressure, P = pressure J K-1 mol-1
Ideal gas
Boltzmann Entropy Relation
$$S = k_B \ln(\Omega)$$
kB = Boltzmann constant, $\Omega = number of accessible microstates$ J K-1
Statistical mechanics formulation
Principle of Increase of Entropy
$$dS_{universe} \ge 0$$
dSuniverse = sum of entropy changes of system and surroundings J K-1
All natural (real) processes are irreversible
Entropy Change in Isobaric Process
$$\Delta S = C_P \ln\left(\frac{T_2}{T_1}\right)$$
CP = heat capacity at constant pressure J K-1
Constant pressure process
Entropy Change in Isochoric Process
$$\Delta S = C_V \ln\left(/frac{T_2}{T_1}\right)$$
CV = heat capacity at constant volume J K-1
Constant volume process
Graphs and experiments
Carnot Cycle on T-S Diagram
Draw: A rectangle on the T-S plane bounded by two isotherms (horizontal lines) and two isentropes (vertical lines).
Interpret: The enclosed rectangular area represents the net work done (or net heat absorbed) in one complete Carnot cycle.
Exam use: Frequently tested in PGT/TGT for visualizing cycle efficiency and work output directly from thermal coordinates.
Throttling Process (Joule-Thomson Expansion)
Draw: Enthalpy is constant (h1 = h2), but entropy increases as pressure drops across a porous plug.
Interpret: Throttling is a highly irreversible process; hence \Delta S > 0 even though no heat is exchanged with the surroundings (adiabatic).
Exam use: Tested conceptually to verify that adiabatic does not imply isentropic if the process is irreversible.
Entropy vs Temperature for a Solid
Draw: An increasing curve starting from S = 0 at T = 0 K, rising steeply and then leveling off according to Debye's T3 law at very low temperatures.
Interpret: Validates the Third Law of Thermodynamics and illustrates how vibrational degrees of freedom populate with thermal energy.
Exam use: Used in PGT questions linking specific heat capacity measurements at low temperatures to absolute entropy.
Solved numericals
TGT core
Calculate the change in entropy when 2 kg of water at 100 °C is converted into steam at the same temperature. Given the latent heat of vaporization of water is 2.26 × 106 J kg-1.
Identify given values: Mass m = 2 kg, Temperature T = 100 °C = 373.15 K, Latent heat $L = 2.26 × 10^{6} J kg^{-1}$.
Apply the formula for entropy change during phase transition: \Delta S = \frac{m L}{T}
Calculate the final numerical value: \Delta S \approx 12112.8 \text{ J K}^{-1}
Answer: 1.21 × 10^4 J K^-1
PGT extension
One mole of an ideal monatomic gas (CV = 3/2 R) is heated from 300 K to 600 K while its volume is doubled. Calculate the total entropy change of the gas. (Take $R = 8.314 J mol^{-1} K^{-1}$, ln 2 = 0.693)
Identify given values: n = 1, T1 = 300 K, T2 = 600 K, V2 / V1 = 2, CV = 1.5 R.
Use the entropy change formula for an ideal gas: \Delta $S = C_V \ln\left(\frac{T_2}{T_1}\right) + R \ln\left(\frac{V_2}{V_1}\right)$
Substitute CV = 1.5 R: \Delta $S = 1.5 R \ln(2) + R \ln(2) = 2.5 R \ln(2)$
Substitute R = 8.314 and ln(2) = 0.693: \Delta S = 2.5 \times 8.314 \times 0.693
Calculate the result: \Delta S \approx 14.41 \text{ J K}^{-1}
Answer: 14.41 J K^-1
TGT core
A block of copper of mass 1 kg and specific heat capacity 400 J kg-1 K-1 is heated from 27 °C to 127 °C. Find the entropy change of the copper block.
Convert temperatures to Kelvin: T1 = 27 + 273 = 300 K, T2 = 127 + 273 = 400 K.
Use the entropy change formula for a solid/liquid: \Delta $S = \int \frac{m c dT}{T} = m c \ln\left(\frac{T_2}{T_1}\right)$
Substitute values: \Delta S = 1 \times 400 \times \ln\left(\frac{400}{300}\right)
Simplify the logarithmic term: \ln(4/3) = \ln(1.333) \approx 0.2877
Macroscopic entropy calculations via dQ/T assume the process can be approximated or executed as a reversible path between equilibrium end-states.
The ideal gas formulas assume constant specific heats CV and CP over the temperature range considered.
Common misconceptions
Common error: Adiabatic processes are always isentropic (constant entropy).
Correct: Only reversible adiabatic processes are isentropic. Irreversible adiabatic processes (like free expansion or throttling) involve an increase in entropy.
Irreversibility generates internal entropy even when no external heat is exchanged.
Common error: Entropy of a system can never decrease under any circumstance.
Correct: The entropy of a specific system can decrease (e.g., when water freezes into ice), but the entropy of the universe (system + surroundings) must always increase or remain constant.
Heat released to the surroundings during freezing increases the surroundings' entropy by a greater or equal amount.
Quick revision
Entropy is an extensive state function defined by d$S = dQ_{rev} / T with units J K^{-1}$.
For any natural (irreversible) process in an isolated system, dS > 0.
On a T-S diagram, area under the curve represents reversible heat transfer.
Must remember
Entropy ($S$) is an extensive thermodynamic state function and a point function.
For a reversible process, the infinitesimal entropy change is given by $dS = \frac{dQ_{rev}}{T}$.
Clausius inequality states that $\oint \frac{dQ}{T} \le 0$ for any thermodynamic cycle.
For an isolated system, $dS \ge 0$, where the equality holds for reversible processes and the inequality for irreversible processes.
Boltzmann's statistical relation connects macroscopic entropy to microscopic probability via $S = k_B \ln(\Omega)$.
According to the Third Law of Thermodynamics, the absolute entropy of a perfectly crystalline substance is zero at $T = 0\text{ K}$.
On a T-S diagram, the area under the curve represents the total heat transferred, $Q = \int T \, dS$.
Reversible adiabatic processes are isentropic, whereas irreversible adiabatic processes involve an increase in entropy.
Glossary
Entropy ($S$)An extensive property and state function representing the degree of molecular disorder or unavailability of thermal energy for work.
Clausius InequalityThe relation $\oint \frac{dQ}{T} \le 0$, which serves as a quantitative criterion for the reversibility or irreversibility of cycles.
Thermodynamic Probability ($\Omega$)The number of microstates corresponding to a given macroscopic state.
Boltzmann Constant ($k_B$)The gas constant per molecule, equal to $R/N_A$.
Isentropic ProcessA constant-entropy process, which is both reversible and adiabatic.
T-S DiagramA temperature-entropy coordinate plot used to visualize thermodynamic cycles and heat transfers.
Reversible vs Irreversible Adiabatic Processes
Feature
Reversible Adiabatic
Irreversible Adiabatic
Entropy Change ($dS$)
$dS = 0$ (Isentropic)
$dS > 0$
Heat Transfer ($dQ$)
Zero
Zero externally, but internal generation
Path Function Nature
Fully reversible path
Spontaneous / Free expansion / Throttling
State Function vs Path Function
Property
Entropy ($S$)
Heat ($Q$) and Work ($W$)
Mathematical Differential
Exact differential ($dS$)
Inexact differential ($dQ$, $dW$)
Cyclic Integral
$\oint dS = 0$
$\oint \frac{dQ}{T} \le 0$ or $\oint dQ \neq 0$
Dependence
Depends only on initial and final states
Depends on the specific path taken
Practice Quiz
Choose an option to reveal the explanation.
1. Which of the following thermodynamic properties is an extensive state function?
Explanation: Entropy ($S$) is an extensive thermodynamic state function, meaning its value scales with system size and depends solely on the current state, not the path.
2. For a reversible process, the infinitesimal change in entropy is expressed as:
Explanation: By definition, the infinitesimal change in entropy for a reversible process is $dS = \frac{dQ_{rev}}{T}$.
3. The Clausius inequality for any arbitrary thermodynamic cycle is given by:
Explanation: The Clausius inequality states that for any cycle, reversible or irreversible, $\oint \frac{dQ}{T} \le 0$, where equality holds for a strictly reversible cycle.
4. In an isolated system, the entropy change for a natural (irreversible) process satisfies:
Explanation: For an isolated system ($dQ = 0$), natural irreversible processes always lead to an increase in entropy ($dS > 0$). State of equilibrium is reached when $dS = 0$.
5. Boltzmann's statistical relation linking macroscopic entropy to microscopic probability is:
Explanation: Ludwig Boltzmann expressed entropy as $S = k_B \ln(\Omega)$, where $k_B$ is the Boltzmann constant and $\Omega$ is the number of accessible microstates.
6. According to the Third Law of Thermodynamics (Nernst Heat Theorem), the absolute entropy of a perfectly crystalline substance at $T = 0\text{ K}$ is:
Explanation: At absolute zero ($T = 0\text{ K}$), a perfectly crystalline substance has only one microstate ($\Omega = 1$), leading to $S = k_B \ln(1) = 0$.
7. On a Temperature-Entropy (T-S) diagram, an isothermal process appears as a:
Explanation: Since absolute temperature remains constant during an isothermal process, it is represented as a horizontal line on a T-S diagram.
8. What does the area under the curve on a T-S diagram for a reversible process represent?
Explanation: Since $dQ_{rev} = T \, dS$, the area under the reversible T-S curve represents the total heat transferred: $Q = \int T \, dS$.
9. An ideal gas undergoes an isobaric process from temperature $T_1$ to $T_2$. The entropy change is given by:
Explanation: For a constant pressure (isobaric) process, the entropy change of an ideal gas is $\Delta S = C_P \ln\left(\frac{T_2}{T_1}\right)$.
10. Why is an adiabatic free expansion of a gas an irreversible process with an increase in entropy, despite $dQ = 0$?
Explanation: Free expansion into a vacuum is highly spontaneous and irreversible. While $dQ = 0$, entropy increases because $dS > \frac{dQ}{T}$ due to internal irreversibilities.
11. For an ideal gas, the entropy change in terms of temperature $T$ and volume $V$ is expressed as:
Explanation: Using the first law and ideal gas law, $dS = \frac{dU + P \, dV}{T} = C_V \frac{dT}{T} + R \frac{dV}{V}$, leading to $S_2 - S_1 = C_V \ln\left(\frac{T_2}{T_1}\right) + R \ln\left(\frac{V_2}{V_1}\right)$.
12. Which of the following processes represents an isentropic process?
Explanation: A reversible adiabatic process has $dQ_{rev} = 0$, which means $dS = 0$, making it an isentropic (constant entropy) process.
13. The cyclic integral of $\frac{dQ}{T}$ for any reversible cycle is:
Explanation: Since entropy is a state function, its net change over any closed cyclic path is zero, hence $\oint \frac{dQ_{rev}}{T} = 0$.
14. What happens to the total entropy of the universe during any real (spontaneous) process?
Explanation: According to the Second Law of Thermodynamics, all real processes are irreversible and result in an increase in the total entropy of the universe ($dS_{universe} > 0$).
15. If the thermodynamic probability $\Omega$ of a system increases, its entropy $S$:
Explanation: From $S = k_B \ln(\Omega)$, entropy is directly proportional to the natural logarithm of the number of microstates $\Omega$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. The cyclic integral $\oint \frac{dQ}{T} \le 0$ is known as:
Explanation: The mathematical expression $\oint \frac{dQ}{T} \le 0$ formulated by Rudolf Clausius establishes the criterion for reversibility and irreversibility of thermodynamic cycles.
2. When an ideal gas undergoes a reversible isothermal expansion, the entropy change of the gas is:
Explanation: In an isothermal expansion, heat is absorbed by the gas from the surroundings ($Q > 0$). Since $T$ is constant, $\Delta S = \frac{Q}{T} > 0$.
3. Which of the following functions is an exact differential?
Explanation: Heat ($dQ$) and work ($dW$) are path functions with inexact differentials, whereas entropy ($S$) is a state function possessing an exact differential ($dS$).
4. The entropy change of $2$ moles of an ideal gas when its volume doubles at constant temperature is:
Explanation: For an isothermal process, $\Delta S = nR \ln\left(\frac{V_2}{V_1}\right)$. Given $n = 2$ and $\frac{V_2}{V_1} = 2$, we get $\Delta S = 2R \ln(2)$.
5. Which statement correctly distinguishes a reversible adiabatic process from an irreversible adiabatic process?
6. Boltzmann's constant $k_B$ is related to the universal gas constant $R$ and Avogadro's number $N_A$ as:
Explanation: Boltzmann's constant represents the gas constant per molecule, defined as $k_B = \frac{R}{N_A}$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. If $100\text{ J}$ of heat is reversibly transferred to a system at a constant temperature of $250\text{ K}$, the entropy change of the system is:
Explanation: Using $\Delta S = \frac{Q_{rev}}{T} = \frac{100\text{ J}}{250\text{ K}} = 0.4\text{ J K}^{-1}$.
2. The entropy of an isolated system can never:
Explanation: According to the principle of increase of entropy, the entropy of an isolated system can never decrease ($dS \ge 0$). It increases for irreversible processes and remains constant for reversible ones.
3. On a T-S diagram, a reversible adiabatic process is represented by a:
Explanation: Since entropy remains constant ($dS = 0$) during a reversible adiabatic (isentropic) process, it appears as a vertical line on a T-S diagram.
4. What is the entropy change of the universe when a reversible engine operates between two heat reservoirs?
Explanation: For a completely reversible cycle, the total entropy change of the system and surroundings combined (universe) is zero ($dS_{universe} = 0$).
5. An ideal gas is expanded adiabatically and irreversibly. Which of the following is true?
Explanation: An irreversible adiabatic expansion (such as free expansion) involves internal dissipation, leading to an increase in the system's entropy ($\Delta S_{system} > 0$). Because it is isolated, $\Delta S_{universe} > 0$ as well.
6. The unit of entropy is:
Explanation: Entropy is defined as $dS = \frac{dQ}{T}$, making its SI unit joules per kelvin ($\text{J K}^{-1}$).
7. For a constant volume (isochoric) heating of an ideal gas, the entropy change is given by:
Explanation: For an isochoric process ($dV = 0$), $dS = \frac{C_V dT}{T}$, which integrates to $\Delta S = C_V \ln\left(\frac{T_2}{T_1}\right)$.
8. Which of the following laws of thermodynamics introduces the concept of entropy?
Explanation: The Second Law of Thermodynamics introduces entropy as a state function to measure the direction of spontaneous processes and irreversibility.
9. If two identical bodies of equal heat capacity $C$ and initial temperatures $T_1$ and $T_2$ are brought to thermal contact, what is the net entropy change of the system?
Explanation: The final equilibrium temperature is $T_f = \frac{T_1 + T_2}{2}$. The entropy change is $\Delta S = C \ln\left(\frac{T_f}{T_1}\right) + C \ln\left(\frac{T_f}{T_2}\right) = 2C \ln\left(\frac{T_1 + T_2}{2 \sqrt{T_1 T_2}}\right)$.
10. A system undergoes a cyclic process absorbing $500\text{ J}$ of heat and rejecting $300\text{ J}$ of heat to a single reservoir at $300\text{ K}$. What can be inferred about the process?
Explanation: The net cyclic integral of $\frac{dQ}{T}$ must be $\le 0$. If heat is absorbed from a single reservoir with net work output, it violates the Kelvin-Planck statement of the Second Law.