A Level Physics 15.3 Kinetic Theory of Gases MCQs – 30 Practice Questions with Answers (Cambridge 9702)
This page covers Section 15.3, Kinetic Theory of Gases, from the Cambridge International AS & A Level Physics (9702) syllabus – an A Level (A2) topic. Work through 30 original multiple-choice questions on the kinetic theory assumptions, the equation pV = (1/3)Nm<c2>, root-mean-square speed and the link between temperature and mean molecular kinetic energy, each with a fully worked answer.
Formula recap: pV = (1/3) N m <c2> | crms = √<c2> | average translational kinetic energy of one molecule = (1/2) m <c2> = (3/2) kT = (3/2) (R / NA) T | k = 1.38 × 10−23 J K−1, R = 8.31 J mol−1 K−1. For the full set of equations, see the A Level Physics Formula Sheet.
Kinetic Theory Assumptions
Q1. Which statement is one of the assumptions of the kinetic theory of an ideal gas?
- (A) The molecules attract each other with strong forces at all distances.
- (B) Collisions between molecules are inelastic, so kinetic energy is lost.
- (C) The total volume of the molecules is negligible compared with the volume of the container.
- (D) At a given temperature, every molecule moves with exactly the same speed.
Show Answer
Answer: C – The molecules are treated as tiny points whose own volume can be ignored next to the volume of the container. There are no intermolecular forces except during collisions, collisions are perfectly elastic, and the molecules have a spread (distribution) of speeds.
Q2. In the kinetic theory, the collisions between gas molecules, and between molecules and the walls, are assumed to be perfectly elastic. This means that in each collision:
- (A) no kinetic energy is lost.
- (B) the molecules rebound with a greater speed than before.
- (C) the molecules stick together for a short time.
- (D) kinetic energy is permanently changed into potential energy.
Show Answer
Answer: A – A perfectly elastic collision conserves total kinetic energy. This is why, at constant temperature, the molecules do not gradually slow down and the pressure stays constant.
Q3. The kinetic theory assumes that molecules exert no forces on each other except during collisions. This assumption is consistent with the observation that a gas:
- (A) liquefies easily when it is cooled slightly.
- (B) exerts a pressure that falls to zero at high temperature.
- (C) has molecules that are always touching each other.
- (D) expands to fill the whole of any container it is placed in.
Show Answer
Answer: D – With no attractive forces pulling them together between collisions, the molecules move freely and randomly until they hit the walls, so the gas spreads throughout the available volume.
Q4. In Brownian motion, tiny smoke particles suspended in air are seen to jiggle about randomly. This is evidence for the kinetic theory because the jiggling is caused by:
- (A) the smoke particles attracting and repelling each other.
- (B) uneven bombardment of the smoke particles by invisible air molecules in continuous random motion.
- (C) convection currents that can only flow upwards.
- (D) the smoke particles being alive.
Show Answer
Answer: B – The much smaller air molecules hit each smoke particle from all sides. Because the hits are random and not exactly balanced at any instant, the particle is knocked about in a jerky, random path.
Q5. Individual molecular collisions with the walls of a container are brief and separate, yet the measured pressure of a gas is steady. This is because:
- (A) the number of molecules is so large that collisions with the walls occur extremely frequently, giving an effectively constant average force.
- (B) each molecule collides with a wall exactly once every second.
- (C) the molecules pull continuously on the walls.
- (D) the walls of the container vibrate to smooth out the force.
Show Answer
Answer: A – A macroscopic sample contains an enormous number of molecules, so billions of wall collisions occur every second. Their combined effect is a steady average force per unit area, which is the pressure.
Q6. The kinetic theory assumes that the duration of a collision is negligible compared with the time between collisions. A consequence of this assumption is that the molecules:
- (A) never actually collide with each other.
- (B) are in contact with each other for most of the time.
- (C) spend nearly all of their time moving freely, not interacting with each other.
- (D) come to rest during each collision.
Show Answer
Answer: C – Collisions are treated as instantaneous events. Almost all of the time each molecule is in free flight, so the theory can ignore the details of what happens during the very short collisions.
Q7. Which of the following is NOT an assumption of the kinetic theory of an ideal gas?
- (A) The molecules exert no forces on each other except during collisions.
- (B) All the molecules have exactly the same speed at a given temperature.
- (C) Collisions between molecules are perfectly elastic.
- (D) The volume of the molecules is negligible compared with the volume of the gas.
Show Answer
Answer: B – The molecules have a range of speeds that continually changes as they collide. The theory uses averages, such as the mean-square speed <c2>, rather than one single speed for all molecules.
Q8. To derive the pressure of a gas from the motion of its molecules, the kinetic theory assumes that the molecules obey:
- (A) the laws of thermodynamics only, not mechanics.
- (B) quantum rules that forbid definite paths.
- (C) the theory of relativity.
- (D) Newton’s laws of motion.
Show Answer
Answer: D – The derivation treats each molecule as a tiny particle whose momentum changes at a wall collision are calculated with Newton’s laws. Adding up these momentum changes gives the force, and hence the pressure.
Q9. A real gas at very high pressure deviates noticeably from the predictions of the ideal kinetic theory, mainly because at high pressure:
- (A) the volume of the molecules themselves is no longer negligible compared with the volume of the container.
- (B) collisions between molecules suddenly become inelastic.
- (C) the molecules stop moving altogether.
- (D) Newton’s laws no longer apply to the molecules.
Show Answer
Answer: A – At high pressure the molecules are packed close together, so their own volume is a significant fraction of the total volume, and intermolecular forces also become important. Both effects break the ideal-gas assumptions.
Q10. Between one collision and the next, a gas molecule moves in a straight line at constant speed. This is because, between collisions:
- (A) the pull of the container walls balances the push of other molecules.
- (B) it is continuously pushed from behind by neighbouring molecules.
- (C) no resultant force acts on it, in accordance with Newton’s first law.
- (D) there is no gravity inside a gas.
Show Answer
Answer: C – With no intermolecular forces except during collisions, the resultant force on a molecule between collisions is zero, so by Newton’s first law it travels in a straight line at constant speed until the next collision.
Pressure & Molecular Speeds
Q11. A gas exerts a pressure on the walls of its container because the molecules colliding with the walls:
- (A) attract the walls towards them.
- (B) change momentum at the walls, so the walls experience a force equal to the rate of change of momentum of the molecules.
- (C) rest on the walls, and pressure is simply the weight of the gas.
- (D) stick to the walls and build up in layers.
Show Answer
Answer: B – Each molecule rebounds from a wall with its momentum reversed, so it undergoes a change of momentum. By Newton’s second and third laws, the wall experiences a force; the total force per unit area from the huge number of collisions is the pressure.
Q12. Four molecules have speeds of 300 m s−1, 400 m s−1, 500 m s−1 and 600 m s−1. The root-mean-square speed of the molecules is:
- (A) 450 m s−1
- (B) 460 m s−1
- (C) 475 m s−1
- (D) 464 m s−1
Show Answer
Answer: D – <c2> = (3002 + 4002 + 5002 + 6002) / 4 = (90 000 + 160 000 + 250 000 + 360 000) / 4 = 860 000 / 4 = 215 000 m2 s−2. Then crms = √215 000 ≈ 464 m s−1. Note that 450 m s−1 is just the mean speed; the r.m.s. speed is always found by squaring first, then averaging, then taking the square root.
Q13. A gas has density 0.90 kg m−3 and the mean-square speed of its molecules is <c2> = 2.5 × 105 m2 s−2. Using p = (1/3) ρ <c2>, the pressure of the gas is:
- (A) 7.5 × 104 Pa
- (B) 2.25 × 105 Pa
- (C) 2.5 × 104 Pa
- (D) 1.5 × 105 Pa
Show Answer
Answer: A – p = (1/3) × 0.90 × 2.5 × 105 = (1/3) × 2.25 × 105 = 7.5 × 104 Pa. Option B is the value obtained if the factor of 1/3 is forgotten.
Q14. For the molecules of a gas, <c2> = 4.9 × 105 m2 s−2. The root-mean-square speed of the molecules is:
- (A) 350 m s−1
- (B) 700 m s−1
- (C) 2.45 × 105 m s−1
- (D) 4.9 × 105 m s−1
Show Answer
Answer: B – crms = √<c2> = √(4.9 × 105) = 700 m s−1.
Q15. A fixed volume of gas is kept at constant temperature and more molecules are added so that the number density (number of molecules per unit volume) doubles, while <c2> stays the same. The pressure of the gas:
- (A) stays the same.
- (B) increases by a factor of 4.
- (C) increases by a factor of 2.
- (D) increases by a factor of √2.
Show Answer
Answer: C – From pV = (1/3) N m <c2>, at constant V the pressure is proportional to N. Doubling the number density doubles the number of wall collisions per second, so the pressure doubles.
Q16. The mean-square speed <c2> of the molecules of a gas is increased by a factor of 4, with the number density unchanged. The pressure of the gas increases by a factor of:
- (A) 2
- (B) 1 (no change)
- (C) 16
- (D) 4
Show Answer
Answer: D – p = (1/3) (N/V) m <c2>, so the pressure is directly proportional to <c2>. Multiplying <c2> by 4 multiplies the pressure by 4 (each collision transfers more momentum and collisions occur more often).
Q17. A gas is compressed to half of its original volume at constant temperature. Which statement correctly describes what happens?
- (A) <c2> is unchanged, the number density doubles, and the pressure doubles.
- (B) <c2> doubles, and this alone causes the pressure to double.
- (C) The pressure is unchanged because the temperature is unchanged.
- (D) The pressure increases by a factor of 4.
Show Answer
Answer: A – At constant temperature the mean kinetic energy, and so <c2>, is unchanged. Halving V doubles N/V in p = (1/3)(N/V)m<c2>, so the pressure doubles – in agreement with Boyle’s law.
Q18. The number of molecules in a container is doubled and, at the same time, the volume of the container is doubled, the temperature being unchanged. The pressure of the gas:
- (A) doubles.
- (B) increases by a factor of 4.
- (C) stays the same.
- (D) halves.
Show Answer
Answer: C – The number density N/V is unchanged, and at constant temperature <c2> is unchanged. Since p = (1/3)(N/V)m<c2>, the pressure is unchanged.
Q19. A molecule of mass m approaches a wall at right angles with speed v and rebounds elastically with the same speed. The magnitude of the change in momentum of the molecule is:
- (A) zero.
- (B) 2mv.
- (C) mv.
- (D) (1/2)mv2.
Show Answer
Answer: B – Taking the direction towards the wall as positive, the momentum changes from +mv to −mv, a change of −2mv, whose magnitude is 2mv. Option D is a kinetic energy, not a momentum. It is this momentum change, repeated by vast numbers of molecules, that produces the pressure.
Q20. Which of the following changes would NOT increase the pressure of a fixed mass of ideal gas?
- (A) Adding more molecules at constant volume and temperature.
- (B) Raising the temperature at constant volume.
- (C) Reducing the volume at constant temperature.
- (D) Increasing the number of molecules and the volume in the same proportion, at constant temperature.
Show Answer
Answer: D – In (D) the number density N/V is unchanged and, at constant temperature, <c2> is unchanged, so p = (1/3)(N/V)m<c2> is unchanged. Each of (A), (B) and (C) increases either N/V or <c2>, and so increases the pressure.
Temperature & Mean Kinetic Energy
Q21. The average translational kinetic energy of one molecule of an ideal gas at a temperature of 300 K is: (k = 1.38 × 10−23 J K−1)
- (A) 2.1 × 10−21 J
- (B) 4.1 × 10−21 J
- (C) 6.2 × 10−21 J
- (D) 1.2 × 10−20 J
Show Answer
Answer: C – Average kinetic energy = (3/2) kT = 1.5 × 1.38 × 10−23 × 300 = 6.21 × 10−21 J ≈ 6.2 × 10−21 J. Option B is kT without the factor of 3/2.
Q22. The temperature of an ideal gas is raised from 300 K to 600 K. The average translational kinetic energy of one molecule becomes:
- (A) 1.24 × 10−20 J
- (B) 6.2 × 10−21 J
- (C) 2.5 × 10−20 J
- (D) 3.1 × 10−21 J
Show Answer
Answer: A – Average kinetic energy = (3/2) kT = 1.5 × 1.38 × 10−23 × 600 = 1.242 × 10−20 J ≈ 1.24 × 10−20 J. Doubling the kelvin temperature doubles the mean kinetic energy, from 6.2 × 10−21 J at 300 K.
Q23. Samples of helium and oxygen are held at the same temperature. Which quantity is the same for a helium molecule and an oxygen molecule?
- (A) Their root-mean-square speeds.
- (B) Their momenta.
- (C) The mass of each molecule.
- (D) Their average translational kinetic energies.
Show Answer
Answer: D – The average translational kinetic energy of a molecule, (3/2) kT, depends only on the temperature – not on the gas or on the mass of its molecules. The lighter helium molecules make up for their smaller mass with a higher r.m.s. speed.
Q24. At a certain temperature, the root-mean-square speed of oxygen molecules (molecular mass 32) is 480 m s−1. At the same temperature, the root-mean-square speed of hydrogen molecules (molecular mass 2) is:
- (A) 120 m s−1
- (B) 1920 m s−1
- (C) 960 m s−1
- (D) 480 m s−1
Show Answer
Answer: B – At the same temperature both gases have the same mean kinetic energy, so (1/2) m <c2> is the same and crms is proportional to 1/√m. The speed ratio is √(32 / 2) = √16 = 4, so hydrogen molecules have crms = 4 × 480 = 1920 m s−1.
Q25. The root-mean-square speed of the molecules of a gas at 300 K is to be doubled by heating the gas at constant volume. The required final temperature is:
- (A) 600 K
- (B) 900 K
- (C) 1200 K
- (D) 2400 K
Show Answer
Answer: C – crms is proportional to √T, since (1/2) m <c2> = (3/2) kT. To double crms, <c2> must increase by a factor of 4, so T must increase by a factor of 4: 4 × 300 = 1200 K.
Q26. The internal energy of one mole of an ideal gas at 300 K is: (R = 8.31 J mol−1 K−1)
- (A) 3.7 × 103 J
- (B) 2.5 × 103 J
- (C) 7.5 × 103 J
- (D) 1.2 × 103 J
Show Answer
Answer: A – In an ideal gas there are no intermolecular forces, so there is no potential energy; the internal energy is just the total kinetic energy of the molecules. U = NA × (3/2) kT = (3/2) RT = 1.5 × 8.31 × 300 = 3740 J ≈ 3.7 × 103 J.
Q27. An ideal gas has pressure 1.0 × 105 Pa and volume 0.050 m3. The total translational kinetic energy of all the molecules of the gas is:
- (A) 2.5 × 103 J
- (B) 5.0 × 103 J
- (C) 1.5 × 104 J
- (D) 7.5 × 103 J
Show Answer
Answer: D – Combining pV = (1/3) N m <c2> with total kinetic energy = N × (1/2) m <c2> gives total kinetic energy = (3/2) pV = 1.5 × 1.0 × 105 × 0.050 = 7.5 × 103 J. Option B is just pV.
Q28. At the same temperature, the molecules of a heavy gas have a lower root-mean-square speed than the molecules of a light gas. This is because:
- (A) heavy molecules have less kinetic energy at the same temperature.
- (B) the average kinetic energy (3/2) kT is the same, so a larger molecular mass m means a smaller <c2>, since (1/2) m <c2> is fixed by T.
- (C) heavy molecules collide with the walls less often.
- (D) heavy molecules are attracted to each other more strongly.
Show Answer
Answer: B – Temperature fixes the average kinetic energy per molecule, (1/2) m <c2> = (3/2) kT, for every ideal gas. A larger m must therefore be matched by a smaller <c2>, i.e. a lower r.m.s. speed.
Q29. The average translational kinetic energy of one molecule of an ideal gas at a temperature of 400 K is: (k = 1.38 × 10−23 J K−1)
- (A) 8.3 × 10−21 J
- (B) 5.5 × 10−21 J
- (C) 1.1 × 10−20 J
- (D) 2.8 × 10−21 J
Show Answer
Answer: A – Average kinetic energy = (3/2) kT = 1.5 × 1.38 × 10−23 × 400 = 8.28 × 10−21 J ≈ 8.3 × 10−21 J. Option B is kT, without the factor of 3/2.
Q30. The temperature of an ideal gas is raised from 300 K to 900 K. The root-mean-square speed of its molecules increases by a factor of:
- (A) 3
- (B) 1.73
- (C) 9
- (D) 1 (it does not change)
Show Answer
Answer: B – crms is proportional to √T. The temperature increases by a factor of 900 / 300 = 3, so crms increases by a factor of √3 ≈ 1.73. (The mean kinetic energy and <c2> increase by the full factor of 3.)
Keep practising: revise more A Level topics on the Physics MCQs page, and keep the key equations at hand with the A Level Physics Formula Sheet.
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