Thermodynamics — Class 11 MCQs with Answers

Thermodynamics — important questions & MCQs with answers (Class 11 Physics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   Zeroth law of thermodynamics defines:

   ATemperatureBPressureCVolumeDEntropy
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   Answer: Temperature

   Explanation: Zeroth law: if A is in thermal equilibrium with C and B is also in equilibrium with C, then A and B share the same value of temperature — this is what the law defines, not pressure or entropy.

2. Q2 of 6Easy

   First law of thermodynamics: ΔU = ?

   AW − QBQ + WCQ − WDCannot determine
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   Answer: Q − W

   Explanation: First law of thermodynamics: ΔU = Q − W, heat added to the system minus work done BY the system — confusing this with W − Q reverses the sign of the work term, the usual slip.

3. Q3 of 6Medium

   Ideal gas in isothermal process. Work done in expansion is:

   APositive (gas does work)BNegativeCZeroDCannot say
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   Answer: Positive (gas does work)

   Explanation: Isothermal expansion: as volume increases the gas pushes outward on the surroundings, doing positive work (W > 0) even though ΔU = 0 — the heat absorbed exactly equals this work, Q = W.

4. Q4 of 6Medium

   Refrigerator (heat pump) takes heat from cold and rejects to hot. This requires:

   AVariableBNo workCHeat onlyDWork input
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   Answer: Work input

   Explanation: Second law: heat naturally flows hot to cold, so moving it from cold to hot in a refrigerator is non-spontaneous and needs external work input — assuming it runs on heat alone violates Clausius's statement.

5. Q5 of 6Hard

   Universe's entropy in irreversible process:

   AVariableBStays constantCCan decreaseDStrictly increases
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   Answer: Strictly increases

   Explanation: For any real, irreversible process, entropy generation is strictly positive, so the total entropy of the universe strictly increases — it can never stay constant or decrease, unlike a reversible one.

6. Q6 of 6Hard

   Engine of efficiency 30% absorbs 2000 J. Heat rejected:

   A2000 JB600 JC1400 JDCannot determine
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   Answer: 1400 J

   Explanation: Efficiency η = W/Q_h gives work output W = 0.3 × 2000 J = 600 J; by energy conservation heat rejected Q_c = Q_h − W = 2000 − 600 = 1400 J — forgetting to subtract W from Q_h gives 2000 J.

Thermodynamics notes: key concepts (Class 11 Physics)

The laws governing heat, work and internal energy: thermal equilibrium and the zeroth law, the first law as energy conservation, the four standard processes and the work each involves, the second law and its two statements, heat engines and refrigerators, and the Carnot cycle as the absolute limit on efficiency.

Thermodynamic system

A defined quantity of matter separated from its surroundings by a boundary, which may be open, closed or isolated depending on what can cross it.

Thermal equilibrium and zeroth law

If two systems are each in thermal equilibrium with a third, they are in equilibrium with each other — which is exactly what allows temperature to be defined and measured.

Internal energy

The total kinetic and potential energy of the molecules of a system. It is a STATE function, depending only on the current state and not on how it was reached.

Work done by a gas

Equal to the area under the process curve on a P-V diagram. Work is a PATH function, so different routes between the same two states do different amounts of work.

First law of thermodynamics

Heat supplied to a system either raises its internal energy or is spent doing work: ΔU = Q − W. It is conservation of energy including heat.

Isothermal process

Carried out at constant temperature, so for an ideal gas ΔU is zero and every joule of heat supplied appears as work done by the gas.

Adiabatic process

No heat enters or leaves, so Q is zero and the work done comes entirely out of internal energy. A rapid compression therefore heats the gas.

Isobaric process

Occurs at constant pressure, so work is simply PΔV and the heat supplied is nC_pΔT. Boiling water in an open vessel is isobaric.

Isochoric process

Occurs at constant volume, so no work is done at all and every joule of heat goes into internal energy, raising the temperature.

Cyclic process

The system returns to its initial state, so ΔU is zero over the cycle and the net heat absorbed equals the net work done — the area enclosed on the P-V diagram.

Reversible and irreversible

A reversible process passes through equilibrium states and can be exactly retraced. Every real process, involving friction or finite temperature differences, is irreversible.

Second law of thermodynamics

Clausius: heat cannot flow spontaneously from a colder to a hotter body. Kelvin-Planck: no engine can convert heat entirely into work in a cycle.

Heat engine

A device that takes heat from a hot reservoir, converts part of it into work, and rejects the rest to a cold reservoir. Its efficiency is always below one.

Refrigerator and heat pump

A heat engine run in reverse, using external work to move heat from cold to hot. Its performance is rated by a coefficient of performance, which can exceed one.

Carnot engine

The ideal reversible engine working between two temperatures. Its efficiency depends ONLY on those two temperatures and is the maximum any engine can achieve.

Entropy

A measure of a system's disorder and of energy unavailable for work. For any isolated system entropy never decreases, which sets the arrow of time.