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 |