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Thermal Physics
Second Law of Thermodynamics
The second law of thermodynamics establishes the direction of spontaneous thermal processes and dictates that the total entropy of an isolated system can never decrease over time. It sets fundamental limitations on the conversion of heat into work through statements by Kelvin-Planck and Clausius.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Kelvin-Planck and Clausius statements, equivalence of statements, Carnot theorem, Carnot engine efficiency, absolute scale of temperature.
PGT extension
Clausius inequality, mathematical formulation of entropy change, entropy of ideal gases, principle of increase of entropy, thermodynamic potentials.
Prerequisites
Concept of Temperature
Zeroth and First Law of Thermodynamics
Reversible and Irreversible Processes
Kelvin-Planck Statement
It is impossible to construct a device operating in a thermodynamic cycle that produces no effect other than the extraction of heat from a single thermal reservoir and the performance of an equivalent amount of work.
This implies that no heat engine can have an efficiency of 100% because some heat must always be rejected to a lower-temperature sink.
It highlights the impossibility of converting thermal energy completely into mechanical work on a continuous basis.
Clausius Statement
It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a cooler body to a warmer body.
Heat cannot spontaneously flow from a colder body to a warmer body without the input of external work, which forms the operational basis for refrigerators and heat pumps.
It establishes the natural unidirectionality of heat conduction.
Equivalence of Kelvin-Planck and Clausius Statements
The two statements are logically equivalent; the violation of one implies the violation of the other.
If the Kelvin-Planck statement is violated, a cyclic device can convert heat completely into work, which can then be completely converted into heat in a colder reservoir via friction, violating the Clausius statement.
Similarly, if the Clausius statement is violated, heat moves from cold to hot without work input; coupling this with a standard heat engine yields a perpetual motion machine of the second kind, violating the Kelvin-Planck statement.
Entropy and the Second Law
Entropy is a state function defined via the infinitesimal heat transfer divided by the absolute temperature, d$S = dQ_rev / T$.
For any irreversible process in an isolated system, the total entropy change is strictly positive (dS > 0), whereas for a reversible process, it is zero (dS = 0).
The principle of increase of entropy dictates that natural processes proceed in the direction of increasing total entropy.
Derivations
Efficiency of a Carnot Engine
Conditions: Operating reversibly between two thermal reservoirs at temperatures T_1 (source) and T_2 (sink)
Write the net work output of the engine as $W = Q_1 - Q_2$, where Q_1 is heat absorbed from the source and Q_2 is heat rejected to the sink.
Apply the theorem that the ratio of heat exchanges in a reversible cycle depends only on the temperatures of the reservoirs: Q_1 / $Q_2 = T_1 / T_2$.
Substitute $Q_2 = Q_1 (T_2 / T_1) into the work equation$ to get $W = Q_1 (1 - T_2 / T_1)$.
Define efficiency η as W / Q_1 to obtain the final expression.
Result: $\eta = 1 - \frac{T_2}{T_1}$
Entropy Change for an Ideal Gas
Conditions: Quasi-static process of an ideal gas
Start from the first law of thermodynamics: d$U = dQ - dW$, or d$Q_rev = dU + P dV$.
Substitute d$U = C_v dT$ and $P = \frac{nRT}{V} into the heat equation: dQ_rev = C_v dT + \frac{nRT}{V} dV$.
Divide both sides by T to get the differential entropy change: d$S = \frac{dQ_{rev}}{T} = C_v \frac{dT}{T} + nR \frac{dV}{V}$.
Integrate between initial state (T_1, V_1) and final state (T_2, V_2) assuming constant heat capacity.
Result: \Delta $S = C_v \ln\left(\frac{T_2}{T_1}\right) + nR \ln\left(\frac{V_2}{V_1}\right)$
\Delta $S_{universe} = total entropy change$ J K^{-1}
Isolated systems or complete cycles including surroundings
Carnot Theorem Upper Bound
$$\eta_{any} \le \eta_{Carnot}$$
\et$a_{any} = efficiency of any irreversible engine$, \et$a_{Carnot} = Carnot efficiency$ Dimensionless
Same operating temperatures
Graphs and experiments
Carnot Cycle on P-V Diagram
Draw: A closed cycle consisting of two isotherms (curves) and two adiabats (steeper curves) plotted with Pressure on y-axis and Volume on x-axis.
Interpret: The area enclosed by the curve represents the net mechanical work done by the Carnot engine in one complete cycle.
Exam use: Tested in matching processes (isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression) and identifying work sign.
T-S (Temperature-Entropy) Diagram of Carnot Cycle
Draw: A rectangle on the T-S plane bounded by temperatures T_1 and T_2, and entropy limits S_A and S_B.
Interpret: The rectangular area on the T-S diagram directly represents the net heat absorbed (and net work done) by the cycle, $Q_net = W = (T_1 - T_2)(S_B - S_A)$.
Exam use: Used to compute thermal efficiency visually and compare real cycles with ideal Carnot cycles.
Draw: Schematic setup of falling weights turning a paddle wheel inside an insulated vessel filled with water, measuring temperature rise.
Interpret: Demonstrates the complete conversion of mechanical work into heat, but the reverse process (extracting equivalent work from heat to lift the weight without external changes) is impossible.
Exam use: Conceptual foundation for unidirectional energy degradation and the validity of the second law.
Solved numericals
TGT core
A Carnot engine operates between reservoirs at temperatures of 500 K and 300 K. If it absorbs 1200 J of heat from the source per cycle, calculate the work done by the engine per cycle.
Calculate Carnot efficiency using η = 1 - (T_2 / T_1): η = 1 - (300 / 500) = 1 - 0.6 = 0.4.
Relate efficiency to work and heat absorbed: η = W / Q_1.
Solve for W: $W = η \times Q_1 = 0.4 \times 1200 = 480 J$.
Answer: 480 J
PGT extension
Calculate the change in entropy when 2 kg of water at 0 °C is heated to 100 °C at constant pressure. Assume the specific heat capacity of water is constant at 4200 J kg^-1 K^-1.
Identify given values: mass m = 2 kg, $c = 4200 J kg^-1 K^-1$, initial temperature $T_1 = 273 K$, final temperature $T_2 = 373 K$.
Use the entropy change formula for condensed phases (constant heat capacity): dS = m c \frac{dT}{T}.
Integrate from T_1 to T_2: \Delta $S = m c \ln\left(\frac{T_2}{T_1}\right)$.
One mole of an ideal monatomic gas expands isothermally from an initial volume of 10 L to 20 L at a temperature of 300 K. Calculate the entropy change of the gas and the universe.
For an isothermal process of an ideal gas, temperature remains constant ($T_1 = T_2)$, so the first term in entropy change vanishes.
Apply the formula \Delta $S_{gas} = n R \ln(V_2 / V_1)$.
Since the isothermal expansion is performed reversibly against a matching external pressure, the entropy change of the surroundings is equal and opposite: \Delta $S_{surr} = -5.76 J K^-1$.
Calculate total entropy change of the universe: \Delta $S_{universe} = \Delta S_{gas} + \Delta S_{surr} = 5.76 - 5.76 = 0 J K^-1 (reversible process)$.
Reservoirs are assumed to possess infinite heat capacity, meaning their temperatures do not change regardless of heat exchange.
Macroscopic thermodynamic definitions apply to systems in internal thermal equilibrium; entropy is not defined for non-equilibrium states without extension.
Quantum statistical fluctuations can cause microscopic violations of the second law over extremely short time intervals or small spatial scales, but it remains strictly valid for macroscopic systems.
Common misconceptions
Common error: The second law states that energy is destroyed during work conversion.
Correct: Energy is always conserved (First Law); the second law dictates that energy is degraded in quality and becomes less available for useful work.
Total energy remains constant, but entropy increases, reducing the fraction of energy convertible to work.
Common error: A refrigerator can operate without any external work input if it transfers heat from a cold body to a hot body.
Correct: External work input is mandatory to pump heat from a lower temperature to a higher temperature.
Violating this would breach the Clausius statement of the second law.
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), but this is always accompanied by a greater increase in the entropy of the surroundings.
The restriction dS >= 0 applies strictly to the total isolated system (universe), not to open or closed subsystems.
Quick revision
The second law prohibits 100% efficient heat engines and unassisted heat flow from cold to hot bodies.
The Kelvin-Planck statement asserts the impossibility of a 100% efficient heat engine operating in a cycle.
The Clausius statement dictates that heat cannot spontaneously flow from a colder body to a warmer body without external work.
The Kelvin-Planck and Clausius statements are logically equivalent; violating one implies violating the other.
For any reversible process in an isolated system, the total entropy change is zero ($dS = 0$).
For any irreversible process, the total entropy change is strictly positive ($dS > 0$).
Carnot engine efficiency depends only on source and sink temperatures: $\eta = 1 - \frac{T_2}{T_1}$.
The Coefficient of Performance of a Carnot refrigerator is given by $COP = \frac{T_2}{T_1 - T_2}$.
Clausius Inequality states that for any thermodynamic cycle, $\oint \frac{dQ}{T} \le 0$.
Entropy change of an ideal gas in T-V form is $\Delta S = C_v \ln\left(\frac{T_2}{T_1}\right) + nR \ln\left(\frac{V_2}{V_1}\right)$.
The principle of increase of entropy states that $\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} \ge 0$.
Glossary
Kelvin-Planck StatementA principle stating that no heat engine can convert all extracted heat into work without rejecting some heat to a sink.
Clausius StatementA principle stating that heat cannot spontaneously transfer from a cooler body to a warmer body without external work input.
Entropy ($S$)A state function defined via infinitesimal reversible heat transfer divided by absolute temperature ($dS = dQ_{rev} / T$).
Clausius InequalityThe mathematical relation $\oint \frac{dQ}{T} \le 0$ applicable to any thermodynamic cycle.
Coefficient of Performance (COP)The ratio of heat extracted from the cold reservoir to the work input in a refrigerator.
Carnot EfficiencyThe maximum theoretical efficiency possible for a heat engine operating between two constant-temperature reservoirs.
Kelvin-Planck vs. Clausius Statements
Feature
Kelvin-Planck Statement
Clausius Statement
Focus
Performance of Heat Engines
Operation of Refrigerators and Heat Pumps
Primary Constraint
Cannot convert 100% heat into work
Heat cannot flow cold-to-hot spontaneously
Implication
Requires a lower-temperature sink
Requires external work input
Reversible vs. Irreversible Entropy Changes
Process Type
System Entropy Change ($\Delta S_{sys}$)
Universe Entropy Change ($\Delta S_{univ}$)
Reversible Process
Can be positive, negative, or zero
Zero (\Delta S_{univ} = 0)
Irreversible Process
Can be positive, negative, or zero
Strictly positive (\Delta S_{univ} > 0)
Practice Quiz
Choose an option to reveal the explanation.
1. Which of the following statements is a direct consequence of the Kelvin-Planck statement?
Explanation: The Kelvin-Planck statement implies that no heat engine can have an efficiency of $100\%$ because some heat must always be rejected to a lower-temperature sink.
2. The Clausius statement of the second law of thermodynamics forms the operational basis for which devices?
Explanation: The Clausius statement establishes that heat cannot flow from cold to hot without external work, which is the operational basis for refrigerators and heat pumps.
3. What is the efficiency of a Carnot engine operating between a source at $500\text{ K}$ and a sink at $300\text{ K}$?
Explanation: Using $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{500} = 1 - 0.6 = 0.4$ or $40\%$.
4. If the Kelvin-Planck statement is violated, which of the following is also violated?
Explanation: The Kelvin-Planck and Clausius statements are logically equivalent; the violation of one implies the violation of the other.
5. What is the mathematical definition of entropy change for a reversible path between equilibrium states?
Explanation: Entropy change is defined as the infinitesimal reversible heat transfer divided by the absolute temperature, $dS = \frac{dQ_{rev}}{T}$.
6. For any irreversible process in an isolated system, what is the total entropy change ($\Delta S$)?
Explanation: For any irreversible process in an isolated system, the total entropy change is strictly positive ($\Delta S > 0$).
7. What does the Clausius inequality $\oint \frac{dQ}{T} \le 0$ signify for a reversible thermodynamic cycle?
Explanation: For a reversible thermodynamic cycle, the cyclic integral of $\frac{dQ}{T}$ is exactly zero, while for irreversible cycles, it is strictly less than zero.
8. What is the expression for the entropy change of an ideal gas in terms of temperature ($T$) and volume ($V$)?
Explanation: The T-V form for entropy change of an ideal gas is $\Delta S = C_v \ln\left(\frac{T_2}{T_1}\right) + nR \ln\left(\frac{V_2}{V_1}\right)$.
9. What is the Coefficient of Performance (COP) of a Carnot refrigerator operating between temperatures $T_1$ (source) and $T_2$ (sink)?
Explanation: The COP of a Carnot refrigerator is given by $COP = \frac{Q_2}{W} = \frac{T_2}{T_1 - T_2}$.
10. Which of the following units is appropriate for entropy?
Explanation: Entropy has units of heat divided by temperature, which is $\text{J K}^{-1}$.
11. According to Carnot's theorem, how does the efficiency of any irreversible engine compare to a Carnot engine operating between the same two temperatures?
Explanation: Carnot's theorem establishes the upper bound: $\eta_{any} \le \eta_{Carnot}$ for engines operating between the same temperatures.
12. What happens to the total entropy of the universe during any natural (spontaneous) process?
Explanation: The principle of increase of entropy dictates that natural (spontaneous) processes proceed in the direction of increasing total entropy ($\Delta S_{universe} \ge 0$).
13. Can the entropy of a specific physical system (such as water freezing into ice) decrease?
Explanation: The entropy of a specific system can decrease (e.g., when water freezes), but this is always accompanied by a greater increase in the entropy of the surroundings, keeping total entropy non-decreasing.
14. What is the expression for entropy change of an ideal gas in T-P form?
Explanation: The T-P form for an ideal gas entropy change is $\Delta S = C_p \ln\left(\frac{T_2}{T_1}\right) - nR \ln\left(\frac{P_2}{P_1}\right)$.
15. What does a perpetual motion machine of the second kind (PMM2) violate?
Explanation: A PMM2 attempts to extract heat from a single thermal reservoir and convert it completely into work, violating the Kelvin-Planck statement of the Second Law.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A Carnot engine absorbs $1000\text{ J}$ of heat from a reservoir at $600\text{ K}$ and rejects heat to a sink at $300\text{ K}$. What is the work done by the engine?
3. If $2\text{ moles}$ of an ideal monoatomic gas expand isothermally and reversibly to double its volume, what is the change in entropy of the gas?
Explanation: For an isothermal expansion ($\Delta T = 0$), $\Delta S = C_v \ln(1) + nR \ln\left(\frac{V_2}{V_1}\right) = 2 \cdot R \ln(2) = 2nR \ln 2$.
4. A refrigerator has a Coefficient of Performance (COP) of $4$. If it extracts $200\text{ J}$ of heat from the cold reservoir per cycle, how much work is done on the refrigerator per cycle?
5. Which of the following relations correctly represents the Clausius inequality for a thermodynamic cycle?
Explanation: The Clausius inequality states that for any cycle, $\oint \frac{dQ}{T} \le 0$, where equality holds for reversible cycles.
6. Two identical bodies of constant heat capacity $C$ and initial temperatures $T_1$ and $T_2$ ($T_1 > T_2$) are brought to thermal equilibrium using a reversible heat engine. What is the final equilibrium temperature?
Explanation: For a reversible engine operating between two finite bodies until they reach thermal equilibrium $T_f$, the total entropy change is zero: $\int \frac{dQ}{T} = 0 \implies \int_{T_1}^{T_f} \frac{C dT}{T} + \int_{T_2}^{T_f} \frac{C dT}{T} = 0 \implies \ln(T_f/T_1) + \ln(T_f/T_2) = 0 \implies T_f^2 = T_1 T_2 \implies T_f = \sqrt{T_1 T_2}$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. What is the physical meaning of the second law of thermodynamics regarding energy degradation?
Explanation: Energy is conserved (First Law), but the second law dictates that energy is degraded in quality and becomes less available for useful work due to entropy increase.
2. A Carnot engine works between $400\text{ K}$ and $500\text{ K}$. If it receives $1200\text{ J}$ of heat from the source, how much heat is rejected to the sink?
3. What is the change in entropy when $1\text{ kg}$ of ice at $0\text{}^{\circ}\text{C}$ melts completely into water at $0\text{}^{\circ}\text{C}$? (Latent heat of fusion of ice $= 3.34 \times 10^5\text{ J kg}^{-1}$)
6. Why is it impossible to construct a 100% efficient heat engine?
Explanation: A $100\%$ efficient heat engine would convert all extracted heat into work without rejecting heat to a sink, violating the Kelvin-Planck statement.
7. What is the entropy change of the universe when a system undergoes an irreversible adiabatic process?
Explanation: In any irreversible process (including adiabatic ones), the total entropy change of the universe is strictly positive ($\Delta S_{universe} > 0$).
8. If the source temperature of a Carnot engine is increased while keeping the sink temperature constant, what happens to its efficiency?
Explanation: Efficiency $\eta = 1 - \frac{T_2}{T_1}$. As $T_1$ increases, the ratio $\frac{T_2}{T_1}$ decreases, making $\eta$ increase.
9. Which of the following relations is true for any thermodynamic cycle?
Explanation: Entropy $S$ is a state function, so its cyclic integral over any thermodynamic cycle (reversible or irreversible) is zero: $\oint dS = 0$.
10. A heat engine absorbs $Q_1$ from a hot reservoir and rejects $Q_2$ to a cold reservoir. If it violates the Clausius statement, which condition holds?
Explanation: The Clausius statement prohibits heat from spontaneously flowing from a colder body to a warmer body without external work input.