JEE (Advanced) · Physics, Class 11
Thermodynamics: JEE Advanced PYQ, 2007–2026
Thermodynamics (Physics, Class 11) has been asked 42 times in the 40 JEE (Advanced) papers from 2007–2026: about 1.1 questions a paper, 6% of all Physics questions. It is the 9th most-asked of 19 Physics chapters. The two 2026 papers had 3 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 0in 2 papers |
| 2008 | 3in 2 papers |
| 2009 | 2in 2 papers |
| 2010 | 2in 2 papers |
| 2011 | 2in 2 papers |
| 2012 | 1in 2 papers |
| 2013 | 1in 2 papers |
| 2014 | 3in 2 papers |
| 2015 | 1in 2 papers |
| 2016 | 1in 2 papers |
| 2017 | 3in 2 papers |
| 2018 | 3in 2 papers |
| 2019 | 4in 2 papers |
| 2020 | 2in 2 papers |
| 2021 | 3in 2 papers |
| 2022 | 2in 2 papers |
| 2023 | 3in 2 papers |
| 2024 | 1in 2 papers |
| 2025 | 2in 2 papers |
| 2026 | 3in 2 papers |
Each year has two papers, and the number of questions has fallen over time, so compare per paper and expect some drift.
Inside the chapter
Topics asked most.
- Thermodynamic processes31
- First law9
- Second law2
Try it
10 Thermodynamics questions to practise.
Our own questions, written in the JEE Advanced pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Physics · Thermodynamics
According to the Carnot theorem, which statement about the efficiency of a Carnot engine operating between two fixed reservoirs is correct?
Answer: (A)
The Carnot efficiency is η = 1 − T_cold/T_hot, with temperatures in kelvin. It depends only on the two reservoir temperatures. The working substance does not enter the formula, which is the content of the Carnot theorem. Raising T_cold lowers η, and raising T_hot raises η. Because T_cold is positive and less than T_hot, η lies between 0 and 1.
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 refrigerator extracts 200 J of heat from its cold interior in one cycle while consuming 50 J of electrical work. What is its coefficient of performance?
Answer: (C)
A refrigerator's performance is measured by the coefficient of performance, β = Q2/W, where Q2 is the heat extracted from the cold body and W is the work input. Here β = 200 J / 50 J = 4. Note that the heat delivered to the hot body, Q1, is larger than Q2 by exactly W, and it is not the numerator of β.
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
A heat pump delivers 1200 J of heat to a warm room in one cycle while consuming 300 J of electrical work. What is its coefficient of performance, defined as heat delivered to the hot reservoir per unit work done on the system?
Answer: (A)
A heat pump is the same cyclic machine as a refrigerator, but its useful output is the heat delivered to the hot reservoir. Its COP is therefore Q_hot/W. Here Q_hot = 1200 J and W = 300 J, so COP = 4. Energy conservation gives the heat taken from the outside as 1200 − 300 = 900 J, which is not the quantity asked for. A refrigerator's COP is Q_cold/W, so the two definitions must be kept separate.
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
A heat engine absorbs 500 J of heat from its hot reservoir and rejects 300 J of heat to the cold reservoir in one complete cycle. What is its efficiency?
Answer: (A)
The efficiency of a heat engine is the ratio of the work output to the heat taken in from the hot reservoir: η = W/Q1. By energy conservation over a cycle, W = Q1 − Q2. Here W = 500 − 300 = 200 J, so η = 200/500 = 0.40, or 40%. Always divide by the heat absorbed from the hot source, and remember that efficiency must lie between 0 and 100%.
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All 42 real JEE Advanced 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 JEE Advanced.
How many questions come from Thermodynamics in JEE Advanced?
Thermodynamics (Physics, Class 11) has been asked 42 times in the 40 JEE (Advanced) papers from 2007–2026: about 1.1 questions a paper, 6% of all Physics questions. It is the 9th most-asked of 19 Physics chapters. The two 2026 papers had 3 from it.
Which topics of Thermodynamics are asked most in JEE Advanced?
Thermodynamic processes (31), First law (9), Second law (2), counted over the papers from 2007–2026.
Is Thermodynamics important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 9 of 19 in Physics; use that to decide how much time it gets, not whether to study it.
Where can I solve Thermodynamics JEE Advanced previous year questions?
On Nikaalle, free: sign up and practise every past JEE Advanced 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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