NEET (UG) · Physics, Class 11
Thermodynamics: NEET PYQ, 2018–2026
Thermodynamics (Physics, Class 11) has been asked 9 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 1.3 questions a paper, 3% of all Physics questions. It is the 19th most-asked of 20 Physics chapters. NEET 2026 had 2 from it. Last checked 28 October 2026.
Questions per paper
Year by year.
| Year | Questions |
|---|---|
| 2018 | 3 |
| 2020 | 1 |
| 2021 | 0 |
| 2023 | 1 |
| 2024 | 1 |
| 2025 | 1 |
| 2026 | 2 |
From 2021 to 2024 each paper had 200 questions (a Section B with a choice); the others had 180.
Inside the chapter
Topics asked most.
- Thermodynamic processes4
- First law3
- Second law2
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10 Thermodynamics questions to practise.
Our own questions, written in the NEET pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Physics · Thermodynamics
The zeroth law of thermodynamics leads to the concept of:
Answer: (C)
The zeroth law says that two systems each in thermal equilibrium with a third are in thermal equilibrium with each other. The property they then share is temperature.
Q2 · Physics · Thermodynamics
A platinum resistance thermometer follows R_t = R_0(1 + αt) with t in °C. Its resistance is 100.0 Ω at the ice point and 138.5 Ω at the steam point. When placed in a bath it reads 123.1 Ω. What is the temperature of the bath?
Answer: (A)
Because R varies linearly with t, (R_t − R_0)/(R_100 − R_0) = t/100. So t = 100 × (123.1 − 100.0)/(138.5 − 100.0) = 100 × 23.1/38.5 = 60 °C. The two fixed points calibrate the thermometer, and the zeroth law lets its reading stand for the bath's temperature.
Q3 · Physics · Thermodynamics
One mole of an ideal monatomic gas (Cv = 3R/2) expands quasi-statically from state A (pressure 3P₀, volume V₀) to state B (pressure P₀, volume 3V₀) along a straight line on the P–V diagram. Over this path, the gas absorbs heat in one part and rejects heat in another. What is the total heat absorbed by the gas, counting only the part of the path where it takes in heat?
Answer: (A)
On the line, P = P₀(4 − V/V₀). Write x = V/V₀. Then nRT = PV = P₀V₀(4x − x²), so nR dT = P₀(4 − 2x)dV. The heat is dQ = (3/2)nR dT + P dV = P₀[(3/2)(4 − 2x) + (4 − x)]dV = P₀(10 − 4x)dV. This is positive for x < 2.5 and negative for x > 2.5. So heat is absorbed from V₀ to 2.5V₀, and the maximum temperature at x = 2 does not mark where absorption ends. Then Q_abs = P₀V₀ ∫ from 1 to 2.5 of (10 − 4x)dx = P₀V₀[10(1.5) − 2(6.25 − 1)] = 4.5 P₀V₀. As a check, ΔU = 1.5(3.75 − 3)P₀V₀ = 1.125 P₀V₀ and W = 3.375 P₀V₀, which sum to 4.5 P₀V₀. The heat rejected over the rest of the path is 0.5 P₀V₀, so the net is 4.0 P₀V₀.
Q4 · Physics · Thermodynamics
A gas is taken through a cyclic process and returns to its initial state. During the cycle the gas absorbs a net 300 J of heat from its surroundings. What is the net work done by the gas over one complete cycle?
Answer: (B)
Principle: the internal energy is a state function, so over any closed cycle the change in internal energy is zero: ΔU = 0. First law: ΔU = Q − W, where W is the work done by the gas. For a cycle, 0 = Q_net − W_net, so W_net = Q_net. Given Q_net = +300 J (heat absorbed), the net work done by the gas is 300 J.
Q5 · Physics · Thermodynamics
One mole of an ideal gas is taken around a cycle on a P–V diagram: A (P₀, V₀) → B (3P₀, V₀) at constant volume, then B → C (3P₀, 2V₀) at constant pressure, then C → D (P₀, 2V₀) at constant volume, then D → A (P₀, V₀) at constant pressure. What is the net heat absorbed by the gas in one complete cycle?
Answer: (A)
The internal energy is a state function, so over a complete cycle ΔU = 0 and the first law gives net Q = net W. The isochoric steps (A→B and C→D) involve no work. On B→C the gas expands at 3P₀ through a volume change V₀, so W = +3P₀V₀. On D→A the gas is compressed at P₀ through a volume change V₀, so W = −P₀V₀. Net W = 3P₀V₀ − P₀V₀ = 2P₀V₀, which equals the area enclosed by the cycle. Hence net heat absorbed = 2P₀V₀.
Q6 · Physics · Thermodynamics
Two rigid vessels hold different ideal gases and are joined by a fixed, thin, heat-conducting wall that does not let gas pass. Vessel A holds 0.20 mol in a volume of 2.0 L at a pressure of 4.0 × 10⁵ Pa. Vessel B holds 0.050 mol in a volume of 0.40 L. The two gases are in thermal equilibrium with each other. What is the pressure of the gas in vessel B?
Answer: (A)
The zeroth law says that systems in thermal equilibrium share a common temperature. For an ideal gas, PV = nRT, so T = PV/(nR). Equal temperatures therefore require P_A V_A/n_A = P_B V_B/n_B. Then P_B = P_A × (V_A/V_B) × (n_B/n_A) = 4.0 × 10⁵ × (2.0/0.40) × (0.050/0.20) = 4.0 × 10⁵ × 5 × 0.25 = 5.0 × 10⁵ Pa. Check: T = (4.0 × 10⁵)(2.0 × 10⁻³)/(0.20 × 8.314) ≈ 481 K for both gases. Pressure is not the quantity that is equalised.
Q7 · Physics · Thermodynamics
An ideal gas is expanded adiabatically and reversibly. During the expansion, the gas does 250 J of work on its surroundings. What is the change in the internal energy of the gas?
Answer: (C)
The first law is ΔU = Q − W. For an adiabatic process, no heat is exchanged, so Q = 0 and ΔU = −W. Here the gas does W = +250 J of work on the surroundings, so ΔU = −250 J. The internal energy decreases by 250 J, which is why an adiabatically expanding gas cools.
Q8 · Physics · Thermodynamics
A fixed mass of an ideal gas is kept at constant pressure. It is heated from 27 °C to 127 °C. What is the ratio of its final volume to its initial volume?
Answer: (A)
At constant pressure, V ∝ T with T in kelvin (Charles' law). T1 = 27 + 273 = 300 K and T2 = 127 + 273 = 400 K. So V2/V1 = 400/300 = 4/3.
Q9 · Physics · Thermodynamics
Constant-volume gas thermometers filled with different gases, such as hydrogen, nitrogen and helium, give different temperatures for the same bath at ordinary pressures. As the amount of gas, and hence the pressure at the triple point, is reduced, their readings converge to a single value. What is the fundamental reason for this convergence?
Answer: (C)
Each gas thermometer defines T ∝ P at constant volume. At ordinary pressures real gases deviate from ideal behaviour in different ways, because of intermolecular attraction and finite molecular size, so the scales differ slightly. As the density is lowered, all gases approach ideal-gas behaviour, where PV = nRT holds regardless of the gas. The pressure then becomes a gas-independent measure of T, so the readings converge. The zeroth law, molar mass and specific heat do not cause this convergence.
Q10 · Physics · Thermodynamics
Two gases at different temperatures are kept in two compartments separated by a fixed diathermic wall. As time passes:
Answer: (D)
A diathermic wall allows energy to pass as heat, while an adiabatic wall does not. Heat flows from the hotter gas to the colder one until the temperatures are equal, when the two are in thermal equilibrium.
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All 9 real NEET questions from Thermodynamics.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
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About Thermodynamics in NEET.
How many questions come from Thermodynamics in NEET?
Thermodynamics (Physics, Class 11) has been asked 9 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 1.3 questions a paper, 3% of all Physics questions. It is the 19th most-asked of 20 Physics chapters. NEET 2026 had 2 from it.
Which topics of Thermodynamics are asked most in NEET?
Thermodynamic processes (4), First law (3), Second law (2), counted over the papers from 2018–2026.
Is Thermodynamics important for NEET 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 19 of 20 in Physics; use that to decide how much time it gets, not whether to study it.
Where can I solve Thermodynamics NEET previous year questions?
On Nikaalle, free: sign up and practise every past NEET question from Thermodynamics with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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