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Thermodynamics – Class 11 Physics notes, important questions and MCQs

NCERT Class 11 Physics, chapter 11. Notes, key terms, important questions with answers and MCQs.

Chapter overview

Thermodynamics studies heat, work and internal energy. Its laws decide which processes are possible and set the limits of heat engines and refrigerators.

Laws and processes

Zeroth law defines temperature (thermal equilibrium).
First law: ΔQ = ΔU + ΔW, with W = PΔV.
Processes: isothermal (ΔT = 0, ΔU = 0), adiabatic (ΔQ = 0, PVᵞ = constant), isobaric (constant P), isochoric (constant V, W = 0).

Second law and engines

Second law: heat cannot flow spontaneously from cold to hot; no engine is 100% efficient.
Carnot efficiency η = 1 − T₂/T₁ (temperatures in kelvin).
Refrigerator coefficient of performance β = Q₂/W = T₂/(T₁ − T₂).

Key terms

Adiabatic process: process with no heat exchange
Internal energy: total energy of molecules of a system, a state function
Reversible process: a process that can be retraced exactly in reverse, passing through equilibrium states, with no dissipative losses

Formulas

ΔQ = ΔU + ΔW
η = 1 − T₂/T₁
PVᵞ = constant (adiabatic)

Important questions with answers

Why is the specific heat of a gas at constant pressure greater than at constant volume?2 marks

At constant pressure the gas expands and does work, so extra heat is needed besides raising internal energy; at constant volume all heat goes into internal energy. Cₚ − Cᵥ = R.

Derive the work done in an isothermal expansion of an ideal gas from V₁ to V₂.3 marks

W = ∫P dV with P = nRT/V at constant T: W = nRT ∫(dV/V) = nRT ln(V₂/V₁) = 2.303 nRT log(V₂/V₁).

Describe the Carnot cycle and derive its efficiency. A Carnot engine works between 527 °C and 27 °C and absorbs 4000 J per cycle. Find its efficiency and work output.5 marks

The Carnot cycle has four reversible steps: isothermal expansion at T₁ (absorbs Q₁), adiabatic expansion (cools to T₂), isothermal compression at T₂ (rejects Q₂), adiabatic compression (back to T₁). For the isothermal steps Q₁ = nRT₁ ln(V₂/V₁) and Q₂ = nRT₂ ln(V₃/V₄); the adiabatic steps give V₂/V₁ = V₃/V₄, so Q₂/Q₁ = T₂/T₁ and η = 1 − T₂/T₁. Here T₁ = 800 K, T₂ = 300 K: η = 1 − 300/800 = 0.625 = 62.5%. W = 0.625 × 4000 = 2500 J.

Quick revision

Isothermal: ΔU = 0.
Isochoric: W = 0.
Carnot engine is the most efficient between two temperatures.

Exam tips

Work done by gas = area under the P–V curve.
Sign convention: ΔQ positive when heat is given to the system; ΔW positive when the system does work.

Practice MCQs

In an isothermal process, which stays constant
1. temperature
2. volume
3. pressure
4. heat
Show answer
Answer: (a) temperature. Iso-thermal.
In an adiabatic process
1. ΔQ = 0
2. ΔT = 0
3. ΔV = 0
4. ΔP = 0
Show answer
Answer: (a) ΔQ = 0. No heat exchange.
A gas absorbs 500 J and does 200 J of work. Change in internal energy is
1. 300 J
2. 700 J
3. −300 J
4. 200 J
Show answer
Answer: (a) 300 J. 500 − 200.
Work done in an isochoric process is
1. zero
2. maximum
3. negative
4. PV
Show answer
Answer: (a) zero. ΔV = 0.
In an isobaric expansion at 2 × 10⁵ Pa, volume increases by 0.01 m³. Work done by the gas is
1. 2000 J
2. 200 J
3. 20 J
4. 20,000 J
Show answer
Answer: (a) 2000 J. PΔV.
Efficiency of a Carnot engine between 500 K and 300 K is
1. 60%
2. 30%
3. 40%
4. 50%
Show answer
Answer: (c) 40%. 1 − 300/500.
COP of a refrigerator between 300 K and 270 K is
1. 1.1
2. 10
3. 0.1
4. 9
Show answer
Answer: (d) 9. 270/30.
A heat engine takes 1000 J from the source and rejects 600 J to the sink. Its efficiency is
1. 40%
2. 60%
3. 160%
4. 25%
Show answer
Answer: (a) 40%. (1000 − 600)/1000.

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Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.