Q.Write down the postulates of kinetic theory of gases.

Showing the 12 most recent of 63 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.In an ideal gas, molecules possess ______.

   (a) only kinetic energy

   (b) both kinetic energy and potential energy

   (c) only potential energy

   (d) neither kinetic energy nor potential energy

   ›Reveal solutionSolution

   An ideal gas is defined by point-mass molecules with no intermolecular forces, so they possess only kinetic energy.

   The kinetic theory of gases models an ideal gas on the assumptions that (i) molecules are point masses with negligible volume, and (ii) there are no intermolecular forces of attraction or repulsion between them except during instantaneous elastic collisions.

   …

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2. CBSE 2026Set ANNUAL1 mark
   Q.A monatomic gas molecule has ______ degrees of freedom.

   ›Reveal solutionSolution

   A monatomic gas molecule (a single point-like atom) can only move — translate — along the x, y and z directions, giving it 3 degrees of freedom.

   Degrees of freedom = the number of independent ways a molecule can possess energy (independent coordinates needed to specify its motion). A monatomic molecule (e.g. He, Ne, Ar) is treated as a point mass with negligible size, so it can only undergo translational motion along 3 mutually perpendicular axes. It has no rotational degrees of freedom (a point particle has negligible moment of inertia about any axis through its centre) and, at ordi …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.What is the average kinetic energy of a gas molecule at absolute temperature T?

   (a) (3/2) kT

   (b) kT

   (c) (1/2) kT

   (d) (5/2) kT [option D partly obscured by the page's rotated background watermark digits '301328'; best-effort reading, not fully certain]

   ›Reveal solutionSolution

   The kinetic theory of gases gives the average translational kinetic energy per molecule as (3/2) kT, where k is Boltzmann's constant -- one of the central results of this chapter (also the basis of the law of equipartition of energy: (1/2)kT per translational degree of freedom, and there are 3 such degrees of freedom).

   From the kinetic theory of gases, each molecule (treated as a point particle with 3 translational degrees of freedom -- motion along x, y, z) has, on average, (1/2) kT of kinetic energy associated with EACH degree of freedom (the law of equipartition of energy). With 3 translational degrees of freedom:

   Average KE per molecule = 3 x (1/2) kT = (3/2) kT

   …

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4. CBSE 2026Set ANNUAL1 mark
   Q.Fill in the blank: The upper layers of the atmosphere are ________ hot.

   ›Reveal solutionSolution

   Even though the upper atmosphere feels cold, the gas molecules there actually move at very high average speeds, corresponding to a very high kinetic (rms) temperature.

   Temperature, in the kinetic theory of gases, is a measure of the average translational kinetic energy per molecule: (1/2)m⟨v²⟩ = (3/2)k_BT. In the extremely thin upper layers of the atmosphere, the gas is so rarefied (very low pressure/density) that molecules travel long distances between collisions and can attain very high speeds, corresponding to temperatures of the order of a few thousand kelvin. This is why the upper atmosphere is described as 'hot' in the kinetic-theory sense. However, this does not mean an object placed there would feel hot or would rapidly absorb heat: because the density of molecules is so low, the total heat content (and rate of energy transfer by …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.The average kinetic energy of a molecule of an ideal gas depends on the

   (a) nature of the gas

   (b) pressure

   (c) volume

   (d) temperature

   ›Reveal solutionSolution

   Average molecular KE = (3/2)kT depends only on temperature. Answer (D).

   Kinetic theory gives the average translational kinetic energy of an ideal-gas molecule as:

   (average KE) = (3/2) k_B T,

   …

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6. CBSE 2026Set ANNUAL1 markMCQ
   Q.The mean square speed of the molecules of a gas at absolute temperature T is proportional to

   (a) √T

   (b) T

   (c) T^2

   (d) 1/T

   ›Reveal solutionSolution

   Mean square speed <v^2> is proportional to T. Answer (B).

   From kinetic theory, the average translational kinetic energy of a molecule is (1/2) m <v^2> = (3/2) k_B T.

   Solving, <v^2> = 3 k_B T/m, which is directly proportional to the absolute temperature T.

   …

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7. CBSE 2026Set ANN1 markMCQ
   Q.The average kinetic energy of a gas molecule is directly proportional to

   (a) Volume

   (b) Pressure

   (c) Temperature

   (d) Number of moles

   ›Reveal solutionSolution

   The average kinetic energy of a gas molecule is (3/2) kB T, so it is directly proportional to the absolute temperature.

   From the kinetic theory of gases, the mean translational kinetic energy of a single molecule is

   (1/2) m (v-bar squared) = (3/2) kB T, …

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8. CBSE 2025Set ANNUAL1 markMCQ
   Q.According to kinetic theory of gas Kelvin temperature of gas depends upon (A) average translatory kinetic energy of each molecule of gas (B) rotational kinetic energy of each molecule of gas (C) internal potential energy of gas (D) both (A) and (B)

   ›Reveal solutionSolution

   The absolute (Kelvin) temperature of a gas is a measure of the average translational kinetic energy of its molecules.

   From the kinetic theory of gases, the average translational kinetic energy per molecule is directly proportional to the absolute temperature:

   $\dfrac{1}{2}m\overline{v^2} = \dfrac{3}{2}k_BT$

   …

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9. CBSE 2025Set ANNUAL1 markMCQ
   Q.If the temperature of a gas is increased to four times at constant volume then (A) its pressure will become four times (B) root mean square speed of the molecules will become two times (C) mean speed of the molecules will become two times (D) all of these

   ›Reveal solutionSolution

   Quadrupling the temperature at constant volume makes pressure 4× and both rms speed and mean speed 2× — so all three statements are correct.

   At constant volume, from the ideal gas law $PV = nRT$: $P \propto T$, so if $T \to 4T$, then $P \to 4P$ — statement (a) is true.

   Rms speed: $v_{rms} = \sqrt{\dfrac{3RT}{M}} \propto \sqrt{T}$. If $T \to 4T$: $v_{rms} \to \sqrt{4}\,v_{rms} = 2v_{rms}$ — statement (b) is true.

   …

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10. CBSE 2025Set ANNUAL1 markMCQ
    Q.The degree of freedom of a rigid molecule is f and atomicity is n. Which of the following is correct? (A) n = 2, f = 5 (B) n = 3, f = 6 (C) n = 1, f = 3 (D) All of these

    ›Reveal solutionSolution

    All three atomicity–degrees-of-freedom pairings given are correct for rigid molecules, so the answer is "all of these."

    For a rigid molecule (ignoring vibration):

    • Monatomic ($n=1$, e.g. He, Ar): only 3 translational degrees of freedom, $f = 3$ — matches option (c).
    • Diatomic ($n=2$, e.g. O₂, N₂): 3 translational + 2 rotational degrees of freedom (rotation about the bond axis is negligible), $f = 5$ — matches option (a). …

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11. CBSE 2025Set ANNUAL1 markMCQ
    Q.At temperature T the rms speed of molecules of any gas is proportional to (A) T (B) √T (C) T^2 (D) T^3

    ›Reveal solutionSolution

    RMS speed of gas molecules is proportional to the square root of the absolute temperature.

    From kinetic theory: $\dfrac{1}{2}M v_{rms}^2 = \dfrac{3}{2}RT$ (per mole), so:

    $v_{rms} = \sqrt{\dfrac{3RT}{M}} \propto \sqrt{T}$

    …

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12. CBSE 2025Set ANNUAL1 markMCQ
    Q.At absolute zero the speed of molecules of a gas will be (A) very large (B) zero (C) equal to speed at 0°C (D) equal to speed at 100°C

    ›Reveal solutionSolution

    At absolute zero, the (ideal-gas) kinetic-theory speed of gas molecules becomes zero.

    From kinetic theory, $v_{rms} = \sqrt{\dfrac{3RT}{M}}$. As the absolute temperature $T \to 0$ K:

    $v_{rms} \to \sqrt{\dfrac{3R\times0}{M}} = 0$

    …

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