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Class 11·Physics
Class 11 · Physics

Thermodynamics

Thermodynamics governs all energy transformations involving heat. It explains why perpetual motion machines are impossible and why heat engines have efficiency limits.

Feynman Lens

Start with the simplest version: this lesson is about Thermodynamics. If you can explain the core idea to a friend using everyday language, examples, and one clear reason why it matters, you have moved from memorising to understanding.

Thermodynamics governs all energy transformations involving heat. The four laws apply universally — from steam engines to living cells. They explain why perpetual motion is impossible, why heat engines have efficiency limits, and why some processes are irreversible.

Zeroth Law and Thermal Equilibrium

If A is in thermal equilibrium with C and B is in thermal equilibrium with C, then A and B are in thermal equilibrium with each other. This justifies temperature as a state variable and the use of thermometers.

First Law of Thermodynamics

For a system: ΔU = Q − W, where Q is heat added to the system and W is work done by the system. The first law is conservation of energy applied to thermodynamic processes. Internal energy U is a state function; Q and W are path-dependent.

Special Quasi-static Processes (ideal gas)

Isothermal: T constant; ΔU = 0; W = nRT ln(V₂/V₁); Q = W.
Adiabatic: Q = 0; PV^γ = constant; W = (P₁V₁ − P₂V₂)/(γ − 1).
Isobaric: P constant; W = PΔV; Q = nC_p ΔT.
Isochoric: V constant; W = 0; Q = ΔU = nC_v ΔT.
Mayer's relation: C_p − C_v = R; ratio γ = C_p/C_v.

Second Law of Thermodynamics

Kelvin–Planck: No process is possible whose sole result is the absorption of heat from a reservoir and its complete conversion into work.
Clausius: No process is possible whose sole result is the transfer of heat from a colder body to a hotter body.
Both forbid 100% efficient heat engines and self-acting refrigerators.

Reversible vs. Irreversible Processes

A reversible process is an idealised quasi-static process with no dissipation; the system and surroundings can be returned to their initial states. All real processes (with friction, finite temperature differences, etc.) are irreversible and increase the total entropy of the universe.

Carnot Engine

A Carnot cycle (two isotherms + two adiabats) between hot reservoir T_H and cold reservoir T_C has the maximum possible efficiency: η = 1 − T_C/T_H. No real engine can exceed this, regardless of working substance.

Refrigerators and Heat Pumps

Coefficient of performance for a refrigerator: COP = Q_C / W. For a Carnot refrigerator: COP = T_C / (T_H − T_C). It becomes harder to extract heat as T_H − T_C grows.

Socratic Questions

  1. Why does compressing a gas adiabatically raise its temperature even though no heat enters it?
  1. An isothermal expansion of an ideal gas absorbs heat equal to the work done. How is this consistent with ΔU = Q − W?
  1. Why is the Carnot efficiency independent of the working substance?
  1. Why do real engines never reach Carnot efficiency, and what design choices push them closer?
  1. Why is it incorrect to say that heat is "stored" in a body? What quantity is actually a state function?
🃏 Flashcards — Quick Recall
Law
Zeroth law of thermodynamics
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If A↔C and B↔C are in thermal equilibrium, then A↔B is too. Justifies temperature as a state variable.
Law
First law of thermodynamics
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ΔU = Q − W. Heat added to a system minus work done by it equals the change in internal energy.
Concept
State function vs. path function
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U, P, V, T are state functions. Q and W depend on the path between states.
Process
Isothermal process (ideal gas)
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T constant ⇒ ΔU = 0; W = nRT ln(V₂/V₁); Q = W. Follows PV = constant.
Process
Adiabatic process (ideal gas)
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Q = 0; PV^γ = constant; T V^(γ−1) = constant; W = (P₁V₁ − P₂V₂)/(γ − 1).
Relation
Mayer's relation
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C_p − C_v = R per mole. The ratio γ = C_p/C_v: 5/3 (monatomic), 7/5 (diatomic at room T).
Law
Second law (Kelvin–Planck)
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No process can have as its sole result the complete conversion of heat from a single reservoir into work.
Law
Second law (Clausius)
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No process can have as its sole result the transfer of heat from a colder to a hotter body.
Engine
Carnot efficiency
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η_Carnot = 1 − T_C/T_H (with temperatures in kelvin). The upper bound on the efficiency of any heat engine operating between those reservoirs.
Refrigerator
Coefficient of performance
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COP = Q_C / W. For a Carnot refrigerator: COP = T_C / (T_H − T_C); it falls as the temperature gap widens.
📝 Quick Quiz — Test Yourself
In a process a gas absorbs 200 J of heat and does 80 J of work on the surroundings. The change in internal energy is:
  • A 280 J
  • B 120 J
  • C −120 J
  • D 80 J
A Carnot engine operates between 600 K and 300 K. Its maximum efficiency is:
  • A 50%
  • B 25%
  • C 75%
  • D 100%
For an ideal gas in an isothermal process, which is true?
  • A ΔU > 0 and Q = 0
  • B Q = 0 and W > 0
  • C ΔU = 0 and Q = W
  • D ΔU < 0 and Q = 0
An ideal gas is compressed adiabatically. Which statement is correct?
  • A Temperature decreases
  • B Internal energy decreases
  • C Heat flows out of the gas
  • D Temperature and internal energy increase
For a monatomic ideal gas, the ratio γ = C_p/C_v is:
  • A 7/5
  • B 5/3
  • C 4/3
  • D 1