NCERT Solutions for Class 11 Physics Chapter 12 – Thermodynamics
Chapter 12 of Class 11 Physics, Thermodynamics, is a conceptually profound chapter that studies the relationship between heat, work, and internal energy in thermodynamic systems. The chapter begins with thermal equilibrium and the Zeroth Law of Thermodynamics, which defines temperature as a measurable quantity. The First Law of Thermodynamics — a statement of energy conservation — establishes that the change in internal energy of a system equals heat added minus work done by the system (ΔU = Q − W). This law is applied to specific thermodynamic processes: isothermal (constant T), adiabatic (no heat exchange), isochoric (constant volume), and isobaric (constant pressure). Each process has unique characteristics and practical applications in engines and refrigerators. The Second Law of Thermodynamics introduces the concept of the direction of natural processes — heat flows spontaneously from hot to cold — and is expressed through two equivalent statements by Kelvin-Planck and Clausius. The Carnot heat engine, the most efficient possible engine operating between two temperature reservoirs, provides the upper limit of engine efficiency. Students also learn about the refrigerator as a reverse heat engine. The concept of entropy as a measure of disorder is introduced, providing deep insight into irreversibility and natural processes.
NCERT Solutions PDF – Class 11 Physics Chapter 12 (All Exercises)
Important Formulas – Chapter 12: Thermodynamics
| Formula | Expression | Description |
|---|---|---|
| First Law of Thermodynamics | ΔU = Q − W | ΔU = change in internal energy; Q = heat added; W = work done by system |
| Work done by gas | W = ∫P dV = PΔV (isobaric) | Area under P-V curve = work done |
| Isothermal process (ideal gas) | W = nRT ln(V₂/V₁) | Temperature constant; ΔU = 0 for ideal gas |
| Adiabatic process | PV^γ = constant; W = (P₁V₁−P₂V₂)/(γ−1) | No heat exchange; Q = 0 |
| Isochoric process | W = 0; ΔU = Q | Constant volume; all heat changes internal energy |
| Isobaric process | W = PΔV; Q = nCpΔT | Constant pressure |
| Carnot Efficiency | η = 1 − T₂/T₁ = W_net/Q_H | T₁ = hot reservoir, T₂ = cold reservoir (in Kelvin) |
| Coefficient of Performance (Refrigerator) | COP = Q₂ / W = T₂ / (T₁ − T₂) | Q₂ = heat extracted from cold body |
| Entropy Change | ΔS = Q_rev / T | For reversible processes; ΔS ≥ 0 for universe (2nd law) |
| Molar specific heat (ideal gas) | Cp − Cv = R | R = 8.314 J/mol·K; Mayer's relation |
| Ratio of specific heats | γ = Cp / Cv | Monoatomic: 5/3; Diatomic: 7/5; Polyatomic: 4/3 |
Subtopics Explained – Chapter 12: Thermodynamics
Zeroth Law and Thermal Equilibrium
The Zeroth Law states: if body A is in thermal equilibrium with body C, and body B is also in thermal equilibrium with body C, then A and B are in equilibrium with each other. This seemingly obvious law is the logical basis for defining temperature and the working of thermometers.
First Law of Thermodynamics
The First Law (ΔU = Q − W) is energy conservation applied to thermodynamic systems. For an ideal gas, internal energy depends only on temperature. The sign convention is crucial: Q is positive when heat flows into the system; W is positive when the gas expands (does work on surroundings).
Thermodynamic Processes on P-V Diagrams
Isothermal: hyperbolic curve (T constant, ideal gas). Adiabatic: steeper than isothermal (no heat exchange). Isochoric: vertical line on P-V diagram (V constant). Isobaric: horizontal line (P constant). The area under any P-V curve equals the work done — a key concept for NCERT numerical problems.
Second Law of Thermodynamics
The Second Law has two equivalent statements: (1) Kelvin-Planck: no heat engine can convert all absorbed heat into work; (2) Clausius: heat cannot spontaneously flow from a cold to a hot body. Both statements imply the existence of irreversible processes and give the arrow of time its direction.
Carnot Engine and Maximum Efficiency
The Carnot cycle (isothermal expansion → adiabatic expansion → isothermal compression → adiabatic compression) represents the maximum possible efficiency for any heat engine: η = 1 − T₂/T₁. No real engine can exceed this; achieving it requires perfectly reversible processes. Most exam problems test calculation of Carnot efficiency and comparison with real engine efficiency.
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Quick Reference Table – Thermodynamic Processes Comparison
| Process | Condition | Work Done (W) | Heat (Q) | ΔU |
|---|---|---|---|---|
| Isothermal | T = const | nRT ln(V₂/V₁) | = W | 0 (ideal gas) |
| Adiabatic | Q = 0 | (P₁V₁−P₂V₂)/(γ−1) | 0 | −W |
| Isochoric | V = const | 0 | nCvΔT | Q |
| Isobaric | P = const | PΔV | nCpΔT | nCvΔT |
| Cyclic | Returns to start | Area of loop | = W | 0 |
NCERT Solutions for Class 11 Physics Chapter 12: Thermodynamics
Thermodynamics is the branch of physics that studies heat, work, energy, and their transformations. The chapter explains how energy moves between systems and surroundings. Students learn the fundamental laws governing thermal processes and energy conservation. Understanding thermodynamics is essential for solving numerical and conceptual problems related to engines, refrigerators, and natural processes.
The First Law of Thermodynamics states that energy cannot be created or destroyed but can only change from one form to another. It relates heat supplied to a system with the change in internal energy and work done. This law is fundamental for understanding energy conservation in thermodynamic processes. Students frequently solve numerical questions based on this principle.
The Second Law explains the direction of natural processes and introduces the concept of entropy. It states that heat naturally flows from a hotter body to a colder body. This law helps explain why some energy transformations are irreversible. Understanding the Second Law is crucial for analyzing heat engines and refrigeration systems.
An isothermal process occurs at constant temperature, while an adiabatic process occurs without heat exchange with the surroundings. The chapter compares these thermodynamic processes and their graphical representations. Students learn how pressure, volume, and temperature change under different conditions. Numerical questions frequently involve these processes and their associated equations.
Heat engines convert thermal energy into mechanical work. The chapter explains how engines operate through cyclic processes involving heat absorption, work output, and heat rejection. Understanding heat engine efficiency helps students analyze real-world machines such as automobile engines and power plants. Questions based on efficiency and thermodynamic cycles are common in examinations.