JEE (Advanced) · Chemistry, Class 11
States of Matter: JEE Advanced PYQ, 2007–2026
States of Matter (Chemistry, Class 11) has been asked 2 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.1 questions a paper, 0% of all Chemistry questions. It is the 27th most-asked of 31 Chemistry chapters. The two 2026 papers had 0 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 0in 2 papers |
| 2008 | 0in 2 papers |
| 2009 | 0in 2 papers |
| 2010 | 0in 2 papers |
| 2011 | 0in 2 papers |
| 2012 | 0in 2 papers |
| 2013 | 0in 2 papers |
| 2014 | 0in 2 papers |
| 2015 | 0in 2 papers |
| 2016 | 0in 2 papers |
| 2017 | 0in 2 papers |
| 2018 | 0in 2 papers |
| 2019 | 0in 2 papers |
| 2020 | 0in 2 papers |
| 2021 | 0in 2 papers |
| 2022 | 0in 2 papers |
| 2023 | 1in 2 papers |
| 2024 | 0in 2 papers |
| 2025 | 1in 2 papers |
| 2026 | 0in 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.
- Gas laws2
Try it
10 States of Matter 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 · Chemistry · States of Matter
A sample of an ideal gas occupies 2.0 L at 27 °C under a constant pressure. The gas is heated at constant pressure until its temperature becomes 327 °C. What is the final volume of the gas?
Answer: (A)
Charles's law states that for a fixed amount of gas at constant pressure, V is directly proportional to absolute temperature: V1/T1 = V2/T2. Convert to kelvin first: T1 = 27 + 273 = 300 K and T2 = 327 + 273 = 600 K. Then V2 = 2.0 L x (600 K / 300 K) = 4.0 L. Doubling the absolute temperature doubles the volume.
Q2 · Chemistry · States of Matter
A sealed rigid steel container holds a gas at 1 atm and 27 °C. If the container is heated to 327 °C, what pressure will the gas exert, assuming ideal behaviour?
Answer: (A)
For a fixed amount of gas in a rigid container, P ∝ T in kelvin (Gay-Lussac's law). T1 = 300 K and T2 = 600 K, so P2 = P1 × T2/T1 = 1 atm × 600/300 = 2 atm.
Q3 · Chemistry · States of Matter
Hydrogen is collected over water at 27 °C. The moist gas occupies 246 mL at a total pressure of 760 mm Hg. The aqueous tension of water at 27 °C is 27 mm Hg. What volume would the dry hydrogen occupy at 27 °C and 760 mm Hg?
Answer: (D)
By Dalton's law, P_dry = P_total − aqueous tension = 760 − 27 = 733 mm Hg, and this is the pressure of the hydrogen in the 246 mL. At constant temperature, P₁V₁ = P₂V₂ gives V₂ = 733 × 246/760 ≈ 237 mL. Compressing the gas from 733 to 760 mm Hg reduces the volume.
Q4 · Chemistry · States of Matter
The density of an ideal gas is 1.25 g L\u207b\u00b9 at STP (273 K and 1 atm). Taking R = 0.0821 L atm K\u207b\u00b9 mol\u207b\u00b9, what is the molar mass of the gas?
Answer: (A)
Rearranging the ideal gas equation PV = nRT with n = m/M and d = m/V gives PM = dRT, so M = dRT/P. Substituting d = 1.25 g L\u207b\u00b9, R = 0.0821 L atm K\u207b\u00b9 mol\u207b\u00b9, T = 273 K and P = 1 atm gives M = 1.25 x 0.0821 x 273 = 28.0 g mol\u207b\u00b9. The answer should be checked against the units: g L\u207b\u00b9 x L atm mol K\u207b\u00b9 x K / atm gives g mol\u207b\u00b9.
Q5 · Chemistry · States of Matter
Under identical conditions of temperature and pressure, equal volumes of hydrogen (molar mass 2 g mol⁻¹) and an unknown gas X effuse through the same small orifice. Gas X takes 3 times as long as hydrogen. What is the molar mass of X in g mol⁻¹?
Answer: (A)
Graham's law: rate ∝ 1/√M. For equal volumes, time is inversely proportional to rate, so t_X/t_H2 = √(M_X/M_H2). Then 3 = √(M_X/2), so M_X/2 = 9 and M_X = 18 g mol⁻¹. The gas is slower, so it is heavier than hydrogen.
Q6 · Chemistry · States of Matter
Two identical bulbs are joined by a narrow tube of negligible volume and contain an ideal gas at 1.0 atm and 300 K. One bulb is then heated to 600 K while the other stays at 300 K. Neglecting expansion of the bulbs, the final pressure in the system is:
Answer: (D)
The total number of moles is fixed. Initially n = (1.0)(2V)/(R × 300). Finally the pressure P is equal in both bulbs, so n = PV/(R × 300) + PV/(R × 600). Equating gives P(1/300 + 1/600) = 2/300, so P(3/600) = 4/600 and P = 4/3 ≈ 1.33 atm. Gas moves to the hotter bulb until pressures equalise, so the amount in each bulb differs.
Q7 · Chemistry · States of Matter
A gas mixture in a 5.0 L rigid vessel at 300 K contains nitrogen and oxygen. The mole fraction of nitrogen is 0.8 and the total pressure is 2.5 atm. What is the partial pressure of nitrogen?
Answer: (A)
Dalton's law of partial pressures states that each gas in a mixture exerts the pressure it would exert alone, and the partial pressure of a component equals its mole fraction multiplied by the total pressure. Here the mole fraction of N2 is 0.8, so P(N2) = 0.8 × 2.5 atm = 2.0 atm. The temperature and volume are not needed for this step.
Q8 · Chemistry · States of Matter
A cylinder of volume 10 L holds an ideal gas at 5 atm. It is connected through a valve to a rigid, completely evacuated vessel of volume 15 L. The valve is opened and the temperature is kept constant. What is the final pressure?
Answer: (A)
At constant temperature and amount of gas, PV is constant (Boyle's law). The gas expands into vacuum and fills both vessels, so V2 = 10 L + 15 L = 25 L. Then P2 = (5 atm × 10 L) / 25 L = 2 atm.
Q9 · Chemistry · States of Matter
What is the ratio of the root mean square speed of H₂ at 300 K to that of O₂ at 1200 K? (Molar masses: H₂ = 2, O₂ = 32 g mol⁻¹)
Answer: (C)
u_rms = √(3RT/M), so u(H₂)/u(O₂) = √[(T₁/M₁)/(T₂/M₂)]. Then T₁/M₁ = 300/2 = 150 and T₂/M₂ = 1200/32 = 37.5. The ratio is √(150/37.5) = √4 = 2. Both M and T lie under the square root. The hotter O₂ moves faster than it would at 300 K, so the ratio falls from 4 to 2.
Q10 · Chemistry · States of Matter
At constant temperature and pressure, 0.50 mol of an ideal gas occupies 5.0 L. What volume will 1.5 mol of the same gas occupy under the same conditions?
Answer: (A)
Avogadro's law states that at constant temperature and pressure, volume is directly proportional to the number of moles: V/n = constant. So V2 = V1 x n2/n1 = 5.0 L x (1.5/0.50) = 15 L. Tripling the moles triples the volume.
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All 2 real JEE Advanced questions from States of Matter.
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 States of Matter in JEE Advanced.
How many questions come from States of Matter in JEE Advanced?
States of Matter (Chemistry, Class 11) has been asked 2 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.1 questions a paper, 0% of all Chemistry questions. It is the 27th most-asked of 31 Chemistry chapters. The two 2026 papers had 0 from it.
Which topics of States of Matter are asked most in JEE Advanced?
Gas laws (2), counted over the papers from 2007–2026.
Is States of Matter important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 27 of 31 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve States of Matter JEE Advanced previous year questions?
On Nikaalle, free: sign up and practise every past JEE Advanced question from States of Matter 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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