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Kinetic Theory: questions with answers

Class 11 Physics, Chapter 12 (NCERT). 15 questions by concept, 3 with pictures. Open “Show answer” for the answer and why.

How to teach this chapter · Common misconceptions & worked visual for this chapter

Molecular matter and dynamic equilibrium

Question 1

A sealed vessel of nitrogen has steady pressure and temperature and no bulk flow. What does kinetic theory predict about its molecules?

  1. (a)They have stopped moving because the gas is in equilibrium.
  2. (b)They all move with the same constant velocity in one direction.
  3. (c)They continue random motion and collisions while macroscopic averages remain steady.
Show answer

(c) They continue random motion and collisions while macroscopic averages remain steady.

Thermal equilibrium is a statistical steady state. A steady macroscopic state does not require each particle to be stationary.

Question 2

A gas in equilibrium has constant pressure, yet individual molecular velocities change after collisions. Which explanation is correct?

  1. (a)Every individual velocity change must produce a permanent rise of the gas temperature.
  2. (b)Individual changes are compatible with a steady distribution of molecular velocities.
  3. (c)This is impossible because equilibrium requires molecular rest and no collisions.
Show answer

(b) Individual changes are compatible with a steady distribution of molecular velocities.

Collisions redistribute individual energies and directions. Steady statistical averages allow continuing molecular motion.

Molecular volume, number density and spacing

Question 3

At fixed molecule number, a gas sample’s volume is multiplied by 1000. In the simple three-dimensional spacing estimate, by what factor does mean separation increase?

  1. (a)100
  2. (b)10
  3. (c)1000
Show answer

(b) 10

Volume per molecule is proportional to the cube of typical separation. Take the cube root of the volume factor.

Question 4

A fixed number of molecules has its typical separation increased by a factor 2 in all three directions. By what factor does the occupied volume increase?

  1. (a)8
  2. (b)2
  3. (c)4
Show answer

(a) 8

The volume represented by one typical spacing is a three-dimensional quantity. Cube the linear factor to obtain the volume factor.

Avogadro law and partial pressures

Question 5

Two non-reacting ideal gases are in the same vessel at the same temperature. Their partial pressures are in the ratio P1/P2 = 1/2, and their molar masses are 4 and 40 g/mol. What is their mass ratio m1/m2?

  1. (a)1/20
  2. (b)5
  3. (c)1/2
Show answer

(a) 1/20

Partial-pressure ratio equals mole-number ratio at the common volume and temperature. Multiply by the molar-mass ratio to convert moles into masses.

Ideal-gas equation, amount and pressure conventions

Question 6

A fixed amount of ideal gas occupies 2 L at 27°C. At unchanged absolute pressure it is heated to 327°C. Taking 0°C=273 K, what is the final volume?

  1. (a)4 L
  2. (b)218/9 L
  3. (c)1 L
Show answer

(a) 4 L

At fixed pressure and amount, V/T is constant only with T in kelvin. Convert both readings before taking the temperature ratio.

Ideal-gas limits and real-gas graphs

Question 7

The graph shows the same real-gas sample at temperatures T1 and T2 in the illustrated attraction-dominated regime. Which temperature ordering is consistent with the curves?

idealT2T1PV/T (J/K)P0
  1. (a)T1 > T2
  2. (b)T2 > T1
  3. (c)T1 = T2
Show answer

(b) T2 > T1

Higher temperature reduces the attraction-driven departure from the ideal line in this comparison. The shallower dip belongs to T2, which is the hotter state.

Question 8

For the same real gas and molecule count, curve B has a small dip below ideal PV/T, while curve A has a deeper attraction-driven dip over the same pressure range. Which inference is consistent with the chapter’s temperature comparison?

  1. (a)State A is hotter than state B.
  2. (b)State B is hotter than state A.
  3. (c)The different departures require equal temperatures.
Show answer

(b) State B is hotter than state A.

The smaller negative departure indicates weaker relative attraction effects. In the given comparison this is the higher-temperature curve.

Wall collisions, isotropy and pressure

Question 9

The pictured molecule collides elastically with the stationary smooth wall. What is the magnitude of the impulse delivered to the wall?

m=4×10^-26 kgvx=400 m/svy=300 m/swallP+x+y
  1. (a)3.20 × 10⁻²³ N s
  2. (b)4.00 × 10⁻²³ N s
  3. (c)1.60 × 10⁻²³ N s
Show answer

(a) 3.20 × 10⁻²³ N s

Only the velocity component normal to the wall reverses. The wall receives the opposite of the molecule’s momentum change, with magnitude 2m vx; the tangential component contributes no impulse.

Question 10

An elastic collision of a molecule of mass 2×10⁻²⁶ kg with a stationary smooth wall transfers normal impulse 1.20 × 10⁻²³ N s to the wall. What was the incoming normal speed?

  1. (a)150 m/s
  2. (b)300 m/s
  3. (c)600 m/s
Show answer

(b) 300 m/s

Normal momentum reverses sign at the wall. Use J=2m u_normal and solve for u_normal, rather than dividing J by m alone.

Temperature, translational energy and rms speed

Question 11

A rigid diatomic ideal gas has only its three translations and two rotations active. Its total internal energy is 250 J. What is PV?

  1. (a)100 J
  2. (b)250 J
  3. (c)500/3 J
Show answer

(a) 100 J

For five active quadratic terms, total U=5NkB T/2. PV=NkB T=2U/5; the relation 2E/3 would require translational energy E, not total U.

Active degrees of freedom and equipartition

Question 12

For a rigid linear four-atom molecule in the usual regime where rotations about axes perpendicular to the molecular line are active, how many rotational degrees of freedom contribute? Neglect vibrations.

  1. (a)3
  2. (b)2
  3. (c)0
Show answer

(b) 2

Linearity permits two relevant independent perpendicular rotation axes in this model. Adding atoms along the same line does not create the third rotational contribution of a nonlinear molecule.

Gas and solid heat capacities

Question 13

A sealed rigid cylinder contains 2 mol of ideal helium. Its temperature rises by 15 K. With R=8.31 J/(mol K), what heat is required, to three significant figures?

  1. (a)374 J
  2. (b)249 J
  3. (c)623 J
Show answer

(a) 374 J

The rigid vessel does no expansion work, so use Cv. Helium is monatomic: Cv=3R/2 and Q=n Cv ΔT.

Mean free path and collision times

Question 14

For collisions between the two identical hard-sphere molecules shown, what is the collision cross-section? Express the result in Ų.

Identical moleculesABr = 1 Å
  1. (a)4π Ų
  2. (b)2π Ų
  3. (c)π Ų
Show answer

(a) 4π Ų

A collision occurs when the centre separation is at most the sum of the radii, 2r. The collision cross-section is therefore pi(2r)², rather than pi r².

Question 15

For an approximate collision estimate, take mean free path 2.0 × 10⁻⁷ m, characteristic speed 400 m/s and molecular diameter 2.0 × 10⁻¹⁰ m. What is the estimated time of free flight between collisions?

  1. (a)2.5 × 10⁻³ s
  2. (b)5.0 × 10⁻¹³ s
  3. (c)5.0 × 10⁻¹⁰ s
Show answer

(c) 5.0 × 10⁻¹⁰ s

Flight time is path length divided by characteristic speed, l/v. The much shorter d/v estimates collision duration;1/v alone does not have units of time.

Common misconception in Kinetic Theory

Open full misconception guide →

Does equal temperature mean equal molecular speed?

Wrong

Helium and oxygen molecules have the same average speed at the same temperature.

Why it feels right. Temperature measures molecular motion, so I expect it to fix speed itself.

Right

At the same temperature, ideal gases have the same mean translational kinetic energy per molecule.

Worked example

O₂ speed relative to O₂1
He speed relative to O₂2.83
Normalize oxygen's root-mean-square speed to 1. At equal temperature, helium's is √8, about 2.83 times as large.
  1. O₂ molecules have about eight times the mass of He atoms.
  2. At equal temperature, the root-mean-square speed of He is √8 times that of O₂.