Thermodynamics — NEET Physics MCQs with answers

Thermodynamics is a Class 11 Physics chapter in the NEET (UG) syllabus. NEET720 has 795 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Refrigerators and heat pumps · Heat engines efficiency · Specific heats Cp and Cv relation · Isothermal and adiabatic processes · Free expansion · Entropy · Reversible and irreversible processes · Carnot engine and efficiency · Cyclic processes and area under P-V curve · Thermal equilibrium and zeroth law · First law and internal energy · Work and P-V diagrams · Specific heats of gases · Thermodynamic processes · Second law and reversibility · Zeroth law and thermal equilibrium · Thermometry · Thermodynamic processes · Cyclic processes · State functions vs path functions · Refrigerators · Kinetic Theory of Gases · Thermodynamics · Refrigerator · Isochoric process · Isobaric process · Carnot theorem · P-V diagrams · Isothermal-Adiabatic · Carnot Engine · Specific Heat Capacity · Heat Transfer · First Law · Isothermal vs Adiabatic · Internal Energy vs Heat · Second Law · Cyclic Process · Cp and Cv · Zeroth Law · Isothermal Process

8 free Thermodynamics practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · medium · Work and P-V diagrams

   A refrigerator's working substance traces an anticlockwise loop on its P-V diagram over one cycle, while a petrol engine's working gas traces a clockwise loop. What does this tell us about the net work in each case?

   • A.In the engine, the gas does positive net work on the surroundings (delivering useful work); in the refrigerator, net work is done ON the gas by an external agent (e.g. compressor)
   • B.Both deliver positive net work to the surroundings, just via different mechanisms
   • C.The direction of the loop has no bearing on whether work is delivered or consumed; only the enclosed area matters
   • D.Both consume net work from the surroundings, since both require an energy input to run
   Show answer and explanation

   Answer: A. In the engine, the gas does positive net work on the surroundings (delivering useful work); in the refrigerator, net work is done ON the gas by an external agent (e.g. compressor)

   A clockwise P-V loop (engine) does net positive work on the surroundings; an anticlockwise loop (refrigerator/heat pump) has net work done on the working substance by an external agent.

   The direction in which a cyclic process is traced on a P-V diagram determines the sign of the net work. In a clockwise cycle, expansion legs occur at higher pressure than compression legs, so the gas does more work expanding than is done on it while compressing — net positive work is delivered to the surroundings, exactly the goal of a heat engine like a petrol engine. In an anticlockwise cycle, the reverse holds: compression happens at higher pressure than expansion, so net work must be supplied BY an external agent (the compressor) TO the working substance — this is how a refrigerator or heat pump operates, using external work to move heat from a colder region to a hotter one. Options B and D misstate which device delivers versus consumes net work; option C ignores that the loop's direction (not just its area) fixes the sign of the net work.

   Common mistake: Assuming loop area alone (without direction) tells you whether work is delivered or consumed.

2. Question 2 · easy · Second law and reversibility

   A hot cup of tea left in a room always cools down to room temperature. The reverse — the tea spontaneously heating up by drawing heat from the cooler room air — is never observed, even though it would not violate energy conservation. The principle that decides this natural direction of heat flow is:

   • A.the first law of thermodynamics
   • B.the second law of thermodynamics
   • C.the zeroth law of thermodynamics
   • D.the principle of calorimetry
   Show answer and explanation

   Answer: B. the second law of thermodynamics

   Energy conservation (first law) would be satisfied either way. It is the second law that forbids heat from flowing spontaneously from a colder body to a hotter one, fixing the direction of natural processes.

   The first law is a bookkeeping statement: energy is conserved whether heat flows hot-to-cold or cold-to-hot, so it cannot explain why only one direction occurs. The second law supplies the missing directionality — in Clausius's form, heat cannot of itself pass from a colder to a hotter body. That is exactly what rules out the tea spontaneously reheating. The zeroth law merely establishes that thermal equilibrium defines a common temperature, and calorimetry is an application of energy conservation, so neither addresses direction.

   Common mistake: Crediting the first law, since students associate all heat problems with energy conservation.

3. Question 3 · medium · Second law and reversibility

   A domestic refrigerator continuously transfers heat from its cold interior to the warmer kitchen. This does NOT violate the Clausius statement of the second law because:

   • A.the amount of heat transferred per cycle is very small
   • B.the Clausius statement applies only to solids, not to refrigerant gases
   • C.the transfer is not self-acting: external electrical work drives the heat from cold to hot
   • D.the kitchen air eventually returns the heat to the interior, completing the cycle
   Show answer and explanation

   Answer: C. the transfer is not self-acting: external electrical work drives the heat from cold to hot

   The Clausius statement forbids heat flowing from cold to hot WITHOUT any other effect. A refrigerator consumes external electrical work, so the transfer is driven, not spontaneous — no violation.

   Clausius's form of the second law: heat cannot, of itself, pass from a colder body to a hotter body — 'of itself' meaning with no other change anywhere. In a refrigerator, the compressor does electrical work on the refrigerant every cycle, and the heat rejected to the kitchen exceeds the heat extracted from the interior by exactly that work input. The transfer is thus paid for by an external agency, precisely the loophole the statement allows. Options A and B misread the law as quantitative or substance-specific; option D invents a compensating flow that does not exist.

   Common mistake: Thinking any cold-to-hot heat transfer whatsoever contradicts the second law.

4. Question 4 · medium · Second law and reversibility

   For a thermodynamic process to be reversible, the essential requirements are that it must be:

   • A.quasi-static (infinitely slow, through equilibrium states) and free of all dissipative effects such as friction and viscosity
   • B.carried out rapidly, so that no heat has time to leak to the surroundings
   • C.one in which the system finally returns to its initial state, whatever happens to the surroundings
   • D.conducted at a constant temperature throughout
   Show answer and explanation

   Answer: A. quasi-static (infinitely slow, through equilibrium states) and free of all dissipative effects such as friction and viscosity

   Reversibility needs (i) quasi-static execution, so the system is always essentially in equilibrium and can be retraced, and (ii) absence of dissipative effects, which would otherwise leave permanent changes in the surroundings.

   A reversible process must be traceable backwards through exactly the same intermediate states, leaving no net change anywhere. That demands the process be quasi-static — each intermediate state an equilibrium state, achieved only in the infinitely slow limit — and free of dissipation (friction, viscosity, Joule heating, inelasticity), since dissipated work cannot be recovered from the surroundings. Speed (option B) works against equilibrium; a system-only return (option C) ignores the surroundings, which is precisely why cyclic processes are not automatically reversible; and constant temperature (option D) describes only one family of processes — a slow frictionless adiabatic is perfectly reversible with changing temperature.

   Common mistake: Believing an isothermal process is automatically reversible, or that speed does not matter.

5. Question 5 · medium · Second law and reversibility

   When a car brakes to a stop, its entire kinetic energy is converted into heat in the brake discs. Could some cyclic device, in principle, absorb exactly this heat from the discs and convert all of it back into kinetic energy of the car?

   • A.Yes — energy is conserved, so complete recovery is possible in principle
   • B.Yes — provided the device is perfectly frictionless
   • C.No — the first law forbids creating work from heat
   • D.No — a cyclic device converting all the heat from a single source into work would violate the second law
   Show answer and explanation

   Answer: D. No — a cyclic device converting all the heat from a single source into work would violate the second law

   Braking degrades ordered kinetic energy into disordered heat — an irreversible one-way step. Recovering it completely would require a cyclic device turning all heat from one reservoir into work, which the Kelvin-Planck statement forbids.

   Work converts to heat completely and spontaneously (friction in the brakes), but the reverse conversion is fundamentally limited: by the Kelvin-Planck statement, no cyclic device can absorb heat from a single reservoir (the hot discs) and deliver an equivalent amount of work with no other effect. At best a fraction could be recovered by running an engine between the discs and something colder, with the rest necessarily rejected. This asymmetry — complete in one direction, incomplete in the other — is the everyday signature of the second law and of the irreversibility of frictional dissipation. Options A and B ignore the second-law restriction; option C wrongly blames the first law, which is silent on direction.

   Common mistake: Concluding from energy conservation that full recovery of dissipated heat is possible.

6. Question 6 · hard · Second law and reversibility

   On a hot day, a student tries to cool a sealed, thermally insulated kitchen by leaving the refrigerator door wide open. After the refrigerator runs for a long time, the average temperature of the room:

   • A.falls steadily, since the refrigerator is a cooling device
   • B.rises, because the refrigerator rejects into the room all the heat it extracts PLUS the electrical work consumed
   • C.stays exactly constant, because heat extracted equals heat rejected
   • D.falls until it equals the normal operating temperature of the freezer compartment
   Show answer and explanation

   Answer: B. rises, because the refrigerator rejects into the room all the heat it extracts PLUS the electrical work consumed

   With the door open, the 'cold interior' and the room are the same air mass. Each cycle the fridge rejects (heat extracted + electrical work) into the room while extracting only the heat term — a net addition equal to the work input. The insulated room warms up.

   A refrigerator does not destroy heat; it moves it, at the cost of external work. Per cycle it extracts heat from the space in front of its coils and rejects that heat PLUS the compressor's electrical work at the condenser coils behind it. With the door open, both exchanges occur with the same sealed room, so the room's net energy gain per cycle equals the electrical work consumed. In an insulated room the temperature therefore rises. This is the qualitative energy bookkeeping of any refrigerator (rejected heat = extracted heat + work), consistent with the Clausius statement: moving heat 'uphill' always costs work, and that work ends up as heat somewhere. Cooling the room would require rejecting the heat OUTSIDE the room, which is exactly what an air conditioner does.

   Common mistake: Treating a refrigerator as a device that removes heat from existence rather than relocating it.

7. Question 7 · easy · Second law and reversibility

   Which of the following processes can, in principle, be treated as reversible?

   • A.A balloon suddenly bursting and its gas rushing out into the atmosphere.
   • B.Ice cubes melting when dropped into warm water at room temperature.
   • C.A gas compressed extremely slowly (quasi-statically) and without friction, always remaining infinitesimally close to equilibrium.
   • D.A block sliding to rest on a rough floor due to friction.
   Show answer and explanation

   Answer: C. A gas compressed extremely slowly (quasi-statically) and without friction, always remaining infinitesimally close to equilibrium.

   A process is reversible only if it proceeds through a continuous succession of equilibrium states with no dissipative effects (like friction) and can be exactly retraced by an infinitesimal change—quasi-static, frictionless compression is the standard idealised example.

   Reversibility requires two conditions: the process must occur quasi-statically (infinitely slowly, so the system is always essentially in equilibrium) and it must be free of dissipative effects such as friction, viscosity, or unrestrained expansion. Quasi-static, frictionless compression satisfies both, so it can be reversed by an infinitesimal change in the controlling parameter. A bursting balloon (free/uncontrolled expansion), ice melting due to a finite temperature difference (irreversible heat flow), and friction-driven sliding to rest (dissipative) are all classic irreversible processes, even though each eventually settles into a new equilibrium state.

   Common mistake: Assuming any process that eventually reaches equilibrium is reversible, without checking for dissipation or non-quasi-static steps.

8. Question 8 · easy · Second law and reversibility

   A hot cup of tea, left on a table in a room at a lower, constant temperature, gradually cools down until it reaches room temperature. Which statement about this everyday process is correct?

   • A.It is reversible; simply waiting long enough would let the tea spontaneously become hot again.
   • B.It is reversible only if the tea is stirred while cooling.
   • C.Reversibility does not apply to cooling processes at all.
   • D.It is irreversible; the tea will never spontaneously reabsorb heat from the room and return to its original hot state.
   Show answer and explanation

   Answer: D. It is irreversible; the tea will never spontaneously reabsorb heat from the room and return to its original hot state.

   Heat flows spontaneously only from hot to cold across a finite temperature difference; this direction can never spontaneously reverse, which is exactly what makes the process irreversible—a real-world illustration of the second law.

   The tea cools because heat flows from the hotter tea to the cooler room air across a finite temperature difference. By the second law of thermodynamics (Clausius statement), heat never flows spontaneously from a colder to a hotter body without external work being done. Once the tea and room reach the same temperature, there is no spontaneous mechanism for the process to reverse itself—the tea will not spontaneously become hot again while the room cools further. This one-directionality is the hallmark of an irreversible process. Stirring does not change this; it may speed up cooling but does not make the process retraceable.

   Common mistake: Believing any process that settles into a stable final state must be reversible.

Questions about Thermodynamics for NEET

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Questions are original NEET720 compositions reviewed for correctness, syllabus fit and option quality. Counts update as the bank grows (795 active practice questions in this chapter today).