ThermodynamicsNEET MCQs with solutions
Thermodynamics in Physics covers the zeroth, first and second laws, isothermal/adiabatic/isochoric/isobaric processes, Carnot engine, heat engine efficiency and refrigerator COP. NEET tests PV diagram work calculations, first law application and Carnot efficiency — both conceptual and numerical.
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- 11 Physics
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- 24 questions
- In the RankUp app
- 617 questions
- ELITE questions
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- NCERT topics
- 12
Practise 24 questions
Tap an option to check it. Questions from every NCERT topic in this chapter, from easy to hard.
Q1Thermodynamic Processes
A gas is compressed to half its original volume while its temperature remains unchanged. Which thermodynamic process is this?
Not quite — the answer is A.
Constant temperature during compression defines an isothermal process. For an ideal gas, PV = nRT; since T is constant, pressure doubles as volume halves. Temperature change would occur in adiabatic compression, ruling out option D.
Q2Thermodynamic Processes
500 J of heat is supplied to a gas kept in a rigid container. What are the work done by the gas and the change in its internal energy?
Not quite — the answer is B.
Rigid container means constant volume (isochoric): ΔV = 0, so W = 0. First Law: ΔU = Q − W = 500 − 0 = 500 J. Option A sets W = Q — confusing isochoric with isothermal expansion where Q = W. Option D halves both quantities without physical basis.
Q3Specific Heat Capacity of Gases
For an ideal gas, the specific heat at constant pressure is always
Not quite — the answer is A.
At constant pressure, supplied heat raises internal energy and does expansion work W=PΔV. At constant volume no work is done. Therefore Cp > Cv and Cp − Cv = R > 0.
Q4Specific Heat Capacity of Gases
According to Mayer's relation for one mole of an ideal gas
Not quite — the answer is B.
Mayer's relation is Cp − Cv = R, derived from the first law at constant pressure: nCpΔT = nCvΔT + nRΔT. Option A confuses ratio with difference; Option C reverses the subtraction giving a negative value.
Q5First Law of Thermodynamics
The mathematical statement of the First Law of Thermodynamics as per the NCERT sign convention is
Not quite — the answer is A.
First Law: ΔU = Q − W, expressing conservation of energy. Option B uses the alternate convention where W is work done on the system. Option C reverses both signs. Option D is an incorrect rearrangement with a sign error.
Q6First Law of Thermodynamics
In an isochoric process, 500 J of heat is supplied to an ideal gas. The increase in internal energy is
Not quite — the answer is B.
Isochoric: ΔV = 0, so W = 0. First Law: ΔU = Q − W = 500 − 0 = 500 J. Option A confuses isochoric with isothermal (ΔU = 0). Option D wrongly applies a negative sign to heat supplied into the system.
Q7Heat Engines, Refrigerators and Heat Pumps
A heat engine converts
Not quite — the answer is A.
A heat engine absorbs Q₁ from a hot source, converts part into work W, and rejects Q₂ to a cold sink where Q₁ = W + Q₂. Complete conversion (Q₂ = 0) is forbidden by the Kelvin-Planck statement of the Second Law.
Q8Heat Engines, Refrigerators and Heat Pumps
The efficiency of a heat engine is given by
Not quite — the answer is C.
Efficiency η = W/Q₁ = (Q₁ − Q₂)/Q₁, where Q₁ is heat absorbed and W is net work output. Option A (Q₂/Q₁) gives the fraction of heat rejected — the complement (1 − η). Option B is the reciprocal of efficiency. Option C inverts the denominator.
Q9Mixed Revision
A gas is taken through a cyclic process. The net work done by the gas during one complete cycle is equal to
Not quite — the answer is A.
In a cyclic process, ΔU = 0 (state function). Net work = net heat = area enclosed by the P–V curve. Option D is wrong — enclosed area is non-zero for a real cycle; zero work only if the cycle is degenerate.
Q10Mixed Revision
In an isothermal expansion of an ideal gas, which quantity remains constant?
Not quite — the answer is B.
For an ideal gas, internal energy depends only on temperature. In isothermal expansion, ΔT = 0 so ΔU = 0. Pressure and density decrease; volume increases. Internal energy alone stays constant.
Q11Carnot Engine and Maximum Efficiency
A Carnot engine operates between two reservoirs at temperatures T₁ and T₂ (T₁ > T₂). Its efficiency is
Not quite — the answer is A.
η = 1 − T₂/T₁ where T₁ is source and T₂ is sink, both in Kelvin. Option B inverts the ratio, giving a value greater than 1 for T₁ > T₂, which is physically impossible.
Q12Carnot Engine and Maximum Efficiency
In the Carnot efficiency formula η = 1 − T₂/T₁, temperatures must be expressed in
Not quite — the answer is B.
The formula is derived from thermodynamic principles anchored at absolute zero; only the Kelvin scale satisfies this. Using Celsius gives a wrong numerical result since Celsius has an arbitrary zero.
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Get RankUp on Google PlayQ13Grand Test
A gas absorbs 600 J of heat and performs 250 J of work. The increase in its internal energy is
Not quite — the answer is A.
ΔU = Q − W = 600 − 250 = 350 J by the First Law. Option D (850 J) arises from incorrectly adding W instead of subtracting. Option C confuses ΔU with W alone.
Q14Grand Test
Which thermodynamic process for an ideal gas satisfies both ΔU = 0 and Q = W?
Not quite — the answer is B.
For an ideal gas, ΔT = 0 in an isothermal process, so ΔU = 0. First Law gives Q = W. In adiabatic Q = 0; in isochoric W = 0; in isobaric both Q and ΔU are non-zero.
Q15NCERT Summary & Key Concepts
Which thermodynamic process has zero heat exchange between the system and surroundings?
Not quite — the answer is B.
In an adiabatic process, Q = 0 by definition. The First Law then gives ΔU = −W — all work done by or on the gas comes entirely at the expense of internal energy. Option A is the trap: isothermal has Q = W ≠ 0 for an ideal gas.
Q16NCERT Summary & Key Concepts
Which of the following correctly explains why the internal energy of an ideal gas depends only on temperature?
Not quite — the answer is C.
In an ideal gas, intermolecular interactions are zero by assumption, eliminating all potential energy contributions. Internal energy is therefore entirely translational (and rotational/vibrational) kinetic energy, which depends only on absolute temperature.
Q17Introduction to Thermodynamics
Which branch of physics deals with the relationship among heat, work and internal energy of macroscopic systems?
Not quite — the answer is A.
Thermodynamics is the branch that studies energy transfer via heat and work in macroscopic systems. Mechanics deals with motion and forces, not energy in thermal form.
Q18Introduction to Thermodynamics
Which of the following is the best example of a closed thermodynamic system?
Not quite — the answer is B.
A closed system exchanges energy but not mass with surroundings. The sealed cylinder traps a fixed mass while allowing heat or work exchange across its walls.
Q19Internal Energy
Internal energy of a thermodynamic system is the sum of which of the following?
Not quite — the answer is A.
Internal energy includes all microscopic kinetic energies (translational, rotational, vibrational) and intermolecular potential energies. It excludes macroscopic kinetic and potential energies of the system moving as a whole.
Q20Internal Energy
Which of the following does NOT contribute to the internal energy of a gas?
Not quite — the answer is D.
Internal energy consists only of microscopic energies. The macroscopic kinetic energy of the gas sample moving as a whole (bulk motion) is not part of internal energy — it is a mechanical energy of the system, not a thermodynamic property.
Q21Heat and Work
Heat is defined as
Not quite — the answer is A.
Heat is energy in transit flowing between bodies due to a temperature difference alone. Internal energy is the energy stored within a body, not heat. Average kinetic energy relates to temperature, not to heat transfer.
Q22Heat and Work
According to NCERT thermodynamic sign convention, which of the following statements about work is correct?
Not quite — the answer is B.
NCERT convention: work done by the system is positive (energy leaves system); work done on the system is negative. This sign convention is applied consistently in the First Law: ΔU = Q − W.
Q23Thermal Equilibrium & State Variables
Two bodies are said to be in thermal equilibrium when they have the same
Not quite — the answer is B.
Thermal equilibrium means no net heat flows between bodies because their temperatures are equal. Two bodies can have equal temperatures yet different internal energies depending on mass and molecular nature.
Q24Thermal Equilibrium & State Variables
Which law of thermodynamics forms the basis of temperature measurement?
Not quite — the answer is C.
The Zeroth Law states that if A is in thermal equilibrium with C and B is in thermal equilibrium with C, then A and B are in equilibrium with each other. This transitivity makes temperature a measurable and universally comparable property.
ELITE question · AIR under 50 level
This chapter has 208 ELITE questions for students aiming at the very top. They are only in the app.
Unlock ELITE questions in the appKey Thermodynamic Processes
Quick revision: most questions in this chapter test these facts.
| Process | Key Fact |
|---|---|
| Isothermal | T = constant; ΔU = 0; W = nRT ln(V₂/V₁); PV = constant |
| Adiabatic | Q = 0; PVᵞ = constant; TV^(γ−1) = constant; W = (P₁V₁−P₂V₂)/(γ−1) |
| Isochoric | V = constant; W = 0; Q = nCᵥΔT = ΔU |
| Isobaric | P = constant; W = PΔV = nRΔT; Q = nCₚΔT |
| Carnot efficiency | η = 1 − T₂/T₁; maximum possible efficiency between two temperatures |
| First law | Q = ΔU + W; heat = change in internal energy + work done by gas |
What the app covers in this chapter
617 questions in total, each with a detailed explanation.
| Thermodynamic Processes | 60 |
| Specific Heat Capacity of Gases | 59 |
| First Law of Thermodynamics | 58 |
| Heat Engines, Refrigerators and Heat Pumps | 58 |
| Mixed Revision | 58 |
| Carnot Engine and Maximum Efficiency | 56 |
| Grand Test | 55 |
| NCERT Summary & Key Concepts | 54 |
| Introduction to Thermodynamics | 40 |
| Internal Energy | 40 |
| Heat and Work | 40 |
| Thermal Equilibrium & State Variables | 39 |
Questions students ask
Is Thermodynamics (Physics) important for NEET?
Yes — first law application, PV diagram work calculations and Carnot engine efficiency are tested every year. Both conceptual and numerical questions appear.
Which topics should I revise first?
Master all four processes with their conditions and work formulas, first law application, PV diagram area = work, Carnot efficiency formula, and the difference between Cₚ and Cᵥ.
How many questions from this chapter are on RankUp?
The RankUp app has 617 questions on Thermodynamics, including 208 ELITE questions. Every question has a detailed explanation.
