JEE Physics Kinetic Theory MCQs Set 15

Multiple Choice Questions (MCQs) for JEE Physics: Kinetic Theory of gases

Access targeted multiple-choice questions for Kinetic Theory of gases designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Practice Kinetic Theory of gases MCQs for JEE Physics

View or download the dedicated Kinetic Theory of gases MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Physics evaluations.

Multiple Choice Questions

Question 1: The temperature of a gas at pressure \( P \) and volume \( V \) is \( 27^\circ C \). Keeping its volume constant, if its temperature is raised to \( 927^\circ C \), then its pressure will be
(a) \( 2P \)
(b) \( 3P \)
(c) \( 4P \)
(d) \( 6P \)
Show Answer & Explanation

Answer: (c) \( 4P \)

Explanation:
1. Constant volume: \( \dfrac{P_2}{P_1} = \dfrac{T_2}{T_1} = \dfrac{1200}{300} = 4 \).

Teacher's Note:
1. Always convert Celsius to kelvin before using gas laws.

Question 2: 4 moles of an ideal gas are at \( 0^\circ C \). At constant pressure it is heated to double its volume. The final temperature will be
(a) \( 0^\circ C \)
(b) \( 273^\circ C \)
(c) \( 546^\circ C \)
(d) \( 136.5^\circ C \)
Show Answer & Explanation

Answer: (b) \( 273^\circ C \)

Explanation:
1. Charles' law: \( V\propto T \), so \( T_2 = 2\times273 = 546 \) K.
2. \( 546-273 = 273^\circ C \).

Teacher's Note:
1. Doubling the volume doubles the kelvin temperature, not the Celsius one.

Question 3: Every real gas behaves as an ideal gas
(a) at high temperature and low pressure
(b) at low temperature and high pressure
(c) at normal temperature and pressure
(d) none of the above
Show Answer & Explanation

Answer: (a) at high temperature and low pressure

Explanation:
1. At high temperature and low pressure the molecules are far apart and move fast, so intermolecular forces and molecular volume become negligible.

Teacher's Note:
1. Deviation from ideal behaviour is largest at low temperature and high pressure.

Question 4: Boyle's law holds for an ideal gas during
(a) isobaric changes
(b) isothermal changes
(c) isochoric changes
(d) isotonic changes
Show Answer & Explanation

Answer: (b) isothermal changes

Explanation:
1. Boyle's law states \( PV = \) constant at constant temperature, i.e. an isothermal change.

Teacher's Note:
1. Boyle: constant T. Charles: constant P. Gay-Lussac: constant V.

Question 5: The SI unit of the universal gas constant is
(a) cal/\( ^\circ \)C
(b) J/mol
(c) J mol\( ^{-1} \) K\( ^{-1} \)
(d) J/kg
Show Answer & Explanation

Answer: (c) J mol\( ^{-1} \) K\( ^{-1} \)

Explanation:
1. \( R = \dfrac{PV}{nT} \) has units \( \dfrac{\text{J}}{\text{mol K}} \).

Teacher's Note:
1. R = 8.31 J per mole per kelvin.

Question 6: Molecules of a gas behave like
(a) inelastic rigid spheres
(b) perfectly elastic non-rigid spheres
(c) perfectly elastic rigid spheres
(d) inelastic non-rigid spheres
Show Answer & Explanation

Answer: (c) perfectly elastic rigid spheres

Explanation:
1. Kinetic theory models gas molecules as perfectly elastic rigid spheres, so collisions conserve kinetic energy.

Teacher's Note:
1. No energy is lost in collisions in the ideal-gas model.

Question 7: An ideal gas follows the process \( P\,V^{1/3} = \) constant. The initial temperature and volume of the gas are \( T \) and \( V \). If the gas expands to \( 27V \), its temperature will become
(a) \( T \)
(b) \( 9T \)
(c) \( 27T \)
(d) \( T/9 \)
Show Answer & Explanation

Answer: (b) \( 9T \)

Explanation:
1. \( P = kV^{-1/3} \) and \( PV = RT \Rightarrow T\propto V^{2/3} \).
2. \( \dfrac{T_2}{T_1} = (27)^{2/3} = 9 \).

Teacher's Note:
1. Combine the process equation with PV = nRT to find T versus V.

Question 8: One mole of a monatomic ideal gas is mixed with one mole of a diatomic ideal gas. The molar specific heat of the mixture at constant volume is
(a) \( 8R \)
(b) \( \dfrac32R \)
(c) \( 2R \)
(d) \( 2.5R \)
Show Answer & Explanation

Answer: (c) \( 2R \)

Explanation:
1. \( C_V \) (mono) \( = \dfrac32R \), \( C_V \) (di) \( = \dfrac52R \).
2. Mixture: \( \dfrac{1\times\frac32R+1\times\frac52R}{2} = 2R \).

Teacher's Note:
1. Mole-weighted average of the molar specific heats.

Question 9: When the temperature of a gas is raised from \( 27^\circ C \) to \( 90^\circ C \), the percentage increase in the r.m.s. velocity of the molecules will be
(a) 10%
(b) 15%
(c) 20%
(d) 17.5%
Show Answer & Explanation

Answer: (a) 10%

Explanation:
1. \( v_{rms}\propto\sqrt T \): \( \dfrac{v_2}{v_1} = \sqrt{\dfrac{363}{300}} = 1.1 \).
2. Increase \( = 10\% \).

Teacher's Note:
1. Use kelvin temperatures: 300 K and 363 K.

Question 10: A gas is enclosed in a closed pot. On keeping this pot in a train moving with high speed, the temperature of the gas
(a) will increase
(b) will decrease
(c) will remain the same
(d) will change according to the nature of the gas
Show Answer & Explanation

Answer: (c) will remain the same

Explanation:
1. Temperature depends only on the random (disordered) motion of the molecules, not on the ordered motion of the container.

Teacher's Note:
1. Bulk motion does not change temperature.

Question 11: Two spherical vessels of equal volume are connected by a narrow tube. The apparatus contains an ideal gas at one atmosphere and 300 K. Now one vessel is immersed in a bath of constant temperature 600 K and the other in a bath of constant temperature 300 K. The common pressure will be
(a) 1 atm
(b) \( \dfrac45 \) atm
(c) \( \dfrac43 \) atm
(d) \( \dfrac34 \) atm
Show Answer & Explanation

Answer: (c) \( \dfrac43 \) atm

Explanation:
1. Total moles are conserved: \( \dfrac{2P_0V}{R\,300} = \dfrac{PV}{R\,600}+\dfrac{PV}{R\,300} \).
2. \( \dfrac{2}{300} = P\left(\dfrac{1}{600}+\dfrac1{300}\right) = \dfrac{3P}{600} \Rightarrow P = \dfrac43 \) atm.

Teacher's Note:
1. The pressure is the same in both vessels; the number of moles changes.

Question 12: The r.m.s. velocity of a gas at a certain temperature is \( \sqrt2 \) times that of the oxygen molecules at that temperature. The gas can be
(a) \( H_2 \)
(b) He
(c) \( CH_4 \)
(d) \( SO_2 \)
Show Answer & Explanation

Answer: (c) \( CH_4 \)

Explanation:
1. \( v_{rms}\propto1/\sqrt M \): \( \sqrt{\dfrac{32}{M}} = \sqrt2 \Rightarrow M = 16 \).
2. The gas is methane \( CH_4 \).

Teacher's Note:
1. Lighter gases have larger r.m.s. speed at the same temperature.

Question 13: At what temperature will the mean kinetic energy of \( O_2 \) be the same as that of \( H_2 \) molecules at \( -73^\circ C \)?
(a) \( 127^\circ C \)
(b) \( 527^\circ C \)
(c) \( -73^\circ C \)
(d) \( -173^\circ C \)
Show Answer & Explanation

Answer: (c) \( -73^\circ C \)

Explanation:
1. Mean kinetic energy \( =\tfrac32kT \) depends on temperature only.
2. So \( O_2 \) must be at the same temperature, \( -73^\circ C \).

Teacher's Note:
1. Mean translational KE is independent of the type of gas.

Question 14: The volume of a gas at pressure \( 21\times10^4\ \text{N/m}^2 \) and temperature \( 27^\circ C \) is 83 litres. If \( R = 8.3 \) J/mol/K, the quantity of gas in gram-moles will be
(a) 15
(b) 42
(c) 7
(d) 14
Show Answer & Explanation

Answer: (c) 7

Explanation:
1. \( n = \dfrac{PV}{RT} = \dfrac{21\times10^4\times83\times10^{-3}}{8.3\times300} = 7 \).

Teacher's Note:
1. Convert litres to cubic metres first.

Question 15: The pressure and temperature of an ideal gas in a closed vessel are 720 kPa and \( 40^\circ C \) respectively. If \( \dfrac14 \)th of the gas is released from the vessel and the temperature of the remaining gas is raised to \( 353^\circ C \), the final pressure of the gas is
(a) 1440 kPa
(b) 1080 kPa
(c) 720 kPa
(d) 540 kPa
Show Answer & Explanation

Answer: (b) 1080 kPa

Explanation:
1. At constant volume \( P\propto nT \).
2. \( P_2 = 720\times\dfrac34\times\dfrac{626}{313} = 1080 \) kPa.

Teacher's Note:
1. Remaining moles are 3/4 of the original.

Question 16: An air bubble doubles its radius on rising from the bottom of a water reservoir to the surface. If the atmospheric pressure is equal to 10 m of water, the height of water in the reservoir is
(a) 10 m
(b) 20 m
(c) 70 m
(d) 80 m
Show Answer & Explanation

Answer: (c) 70 m

Explanation:
1. Radius doubles, so volume becomes 8 times and pressure falls to 1/8 (isothermal).
2. \( P_{bottom} = 8\times10 = 80 \) m of water (absolute).
3. Depth \( = 80-10 = 70 \) m.

Teacher's Note:
1. Subtract atmospheric pressure to get the water depth.

Question 17: If the r.m.s. velocity of a gas at a given temperature (kelvin scale) is 300 m/s, the r.m.s. velocity of a gas having twice the molecular weight and half the temperature on the kelvin scale will be
(a) 300 m/s
(b) 600 m/s
(c) 75 m/s
(d) 150 m/s
Show Answer & Explanation

Answer: (d) 150 m/s

Explanation:
1. \( v\propto\sqrt{T/M} \): factor \( = \sqrt{\dfrac{1/2}{2}} = \dfrac12 \).
2. \( v = 150 \) m/s.

Teacher's Note:
1. r.m.s. speed is proportional to \( \sqrt{T/M} \).

Question 18: The energy (translational kinetic) of a gas per litre is 300 joules. Its pressure will be
(a) \( 3\times10^5\ \text{N/m}^2 \)
(b) \( 6\times10^5\ \text{N/m}^2 \)
(c) \( 10^5\ \text{N/m}^2 \)
(d) \( 2\times10^5\ \text{N/m}^2 \)
Show Answer & Explanation

Answer: (d) \( 2\times10^5\ \text{N/m}^2 \)

Explanation:
1. \( P = \dfrac23\dfrac EV = \dfrac23\times\dfrac{300}{10^{-3}} = 2\times10^5\ \text{N/m}^2 \).

Teacher's Note:
1. For a gas, \( P = \tfrac23 \times \) (translational KE per unit volume).

Question 19: A cylinder of fixed capacity 44.8 litres contains a monatomic gas at standard temperature and pressure. The amount of heat required to raise the temperature of the gas in the cylinder by \( 10^\circ C \) will be (\( R \) = universal gas constant)
(a) \( R \)
(b) \( 10R \)
(c) \( 20R \)
(d) \( 30R \)
Show Answer & Explanation

Answer: (d) \( 30R \)

Explanation:
1. \( n = \dfrac{44.8}{22.4} = 2 \) mol.
2. \( Q = nC_V\Delta T = 2\times\tfrac32R\times10 = 30R \).

Teacher's Note:
1. At constant volume use \( C_V = \tfrac32R \) for a monatomic gas.

Question 20: A pressure cooker contains air at 1 atm and \( 30^\circ C \). If the safety valve of the cooker blows when the inside pressure is 3 atm, then the maximum temperature of the air inside the cooker can be
(a) \( 90^\circ C \)
(b) \( 636^\circ C \)
(c) \( 909^\circ C \)
(d) \( 363^\circ C \)
Show Answer & Explanation

Answer: (b) \( 636^\circ C \)

Explanation:
1. Constant volume: \( T_2 = 3\times303 = 909 \) K.
2. \( 909-273 = 636^\circ C \).

Teacher's Note:
1. Use kelvin, then convert back.

Question 21: One mole of an ideal monatomic gas requires 210 J of heat to raise its temperature by 10 K when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by 10 K, the heat required is
(a) 238 J
(b) 126 J
(c) 210 J
(d) 350 J
Show Answer & Explanation

Answer: (b) 126 J

Explanation:
1. \( \dfrac{Q_V}{Q_P} = \dfrac{C_V}{C_P} = \dfrac1\gamma = \dfrac35 \) for a monatomic gas.
2. \( Q_V = 0.6\times210 = 126 \) J.

Teacher's Note:
1. For a monatomic gas \( \gamma = 5/3 \).

Question 22: A cylinder contains 10 kg of gas at a pressure of \( 10^7\ \text{N/m}^2 \). The quantity of gas taken out of the cylinder, if the final pressure is \( 2.5\times10^6\ \text{N/m}^2 \), will be (temperature of the gas is constant)
(a) 15.2 kg
(b) 3.7 kg
(c) zero
(d) 7.5 kg
Show Answer & Explanation

Answer: (d) 7.5 kg

Explanation:
1. At constant T and V the mass is proportional to pressure.
2. Remaining mass \( = 10\times\dfrac{2.5\times10^6}{10^7} = 2.5 \) kg.
3. Taken out \( = 10-2.5 = 7.5 \) kg.

Teacher's Note:
1. Mass of gas is proportional to pressure at fixed volume and temperature.

Question 23: A certain amount of an ideal gas is contained in a closed vessel. The vessel is moving with a constant velocity \( v \). The molecular mass of the gas is \( M \). The rise in temperature of the gas when the vessel is suddenly stopped is (\( \gamma = C_P/C_V \))
(a) \( \dfrac{Mv^2}{2R(\gamma-1)} \)
(b) \( \dfrac{Mv^2(\gamma-1)}{2R} \)
(c) \( \dfrac{Mv^2}{2R(\gamma+1)} \)
(d) \( \dfrac{Mv^2(\gamma-1)}{2R(\gamma+1)} \)
Show Answer & Explanation

Answer: (b) \( \dfrac{Mv^2(\gamma-1)}{2R} \)

Explanation:
1. Kinetic energy per mole \( \tfrac12Mv^2 \) becomes internal energy: \( \tfrac12Mv^2 = C_V\Delta T \).
2. \( C_V = \dfrac{R}{\gamma-1} \), so \( \Delta T = \dfrac{Mv^2(\gamma-1)}{2R} \).

Teacher's Note:
1. Use \( C_V = R/(\gamma-1) \).

Question 24: Air is filled at \( 60^\circ C \) in a vessel with an open mouth. The vessel is heated to a temperature \( T \) so that \( \dfrac14 \)th part of the air escapes. Assuming the volume of the vessel remains constant, the value of \( T \) is
(a) \( 80^\circ C \)
(b) \( 444^\circ C \)
(c) \( 333^\circ C \)
(d) \( 171^\circ C \)
Show Answer & Explanation

Answer: (d) \( 171^\circ C \)

Explanation:
1. Open vessel: pressure and volume constant, so \( n\propto1/T \).
2. \( \dfrac{3}{4} = \dfrac{333}{T} \Rightarrow T = 444 \) K \( = 171^\circ C \).

Teacher's Note:
1. Fraction of air left equals \( T_1/T_2 \).

Free study material for Physics

Download Chapter MCQs: JEE Physics Kinetic Theory of gases

Chapter MCQs with Answers for JEE Physics

Review structured objective questions for JEE Physics Kinetic Theory of gases. Built according to official JEE guidelines, these MCQ sets support daily revision and core concept reinforcement.

Expert Practice Material for JEE Physics

Each question includes structured solution keys mapped directly to standard JEE textbooks, helping students evaluate their reasoning and correct mistakes early in their revision.

Complete Your Chapter Revision

Wrap up your chapter revision by testing your knowledge against standard question formats. Everything on our platform is provided free of charge.

FAQs

Where can I access latest JEE Physics Kinetic Theory MCQs Set 15?

You can get most exhaustive JEE Physics Kinetic Theory MCQs Set 15 for free on StudiesToday.com. These MCQs for JEE Physics are updated for the 2026-27 academic session as per JEE examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Physics JEE material?

Yes, our JEE Physics Kinetic Theory MCQs Set 15 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the JEE paper is now competency-based.

How do practicing Physics MCQs help in scoring full marks in JEE exams?

By solving our JEE Physics Kinetic Theory MCQs Set 15, JEE students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Physics.

Do you provide answers and explanations for JEE Physics Kinetic Theory MCQs Set 15?

Yes, Physics MCQs for JEE have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused JEE exams.

Can I practice these Physics JEE MCQs online?

Yes, you can also access online interactive tests for JEE Physics Kinetic Theory MCQs Set 15 on StudiesToday.com as they provide instant answers and score to help you track your progress in Physics.