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BITSAT Kinetic Theory of Gases — practice questions

Explore 76 free Kinetic Theory of Gases questions in BITSAT Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Kinetic Theory of Gases questions and exercises with answers

Showing questions 1–30 of 76.

The root-mean-square (rms) speed of gas molecules is given by:Read the answer and worked solution →The mean translational kinetic energy of one molecule of an ideal gas at temperature $T$ is:Read the answer and worked solution →If the absolute temperature of an ideal gas is doubled, the rms speed of its molecules:Read the answer and worked solution →The mean free path $\lambda$ of a gas molecule of diameter $d$ at number density $n$ is approximately:Read the answer and worked solution →For an ideal gas, the relationship between $C_p$ and $C_v$ is (Mayer's relation):Read the answer and worked solution →Ideal gas equation:Read the answer and worked solution →Postulate of kinetic theory: gas molecules:Read the answer and worked solution →Avogadro's number:Read the answer and worked solution →R / N_A =Read the answer and worked solution →Root-mean-square speed of gas molecules (mass m):Read the answer and worked solution →Average kinetic energy per molecule (monatomic gas):Read the answer and worked solution →Degrees of freedom for monatomic gas (e.g., He, Ar):Read the answer and worked solution →Most probable speed v_mp:Read the answer and worked solution →Average speed v_avg of molecules:Read the answer and worked solution →Maxwell speed distribution is:Read the answer and worked solution →Internal energy of n moles of ideal gas (monatomic):Read the answer and worked solution →Ratio of specific heats γ for monatomic gas:Read the answer and worked solution →For ideal gas, C_p - C_v =Read the answer and worked solution →Two gases of molar masses M₁ and M₂ at same T. Ratio of v_rms:Read the answer and worked solution →Find v_rms of O₂ at 300 K (M = 32 g/mol, R = 8.314):Read the answer and worked solution →Graham's law of effusion: rate of effusion ∝Read the answer and worked solution →For Maxwell-Boltzmann distribution, fraction of molecules with KE > E behaves as:Read the answer and worked solution →Boyle's law derivation from KTG: at constant T, doubling V causes:Read the answer and worked solution →For monatomic ideal gas, U/T (heat capacity at const V) per mole:Read the answer and worked solution →In adiabatic process for ideal gas:Read the answer and worked solution →For mixture of two ideal gases, total pressure = (Dalton's law):Read the answer and worked solution →Equipartition: at temperature T, each quadratic DOF carries energy:Read the answer and worked solution →Mean free path doubles when:Read the answer and worked solution →At what temperature does v_rms of H₂ equal v_rms of O₂ at 300 K?Read the answer and worked solution →For diatomic gas at very high T (vibrational modes active), DOF = ?Read the answer and worked solution →

Worked practice questions

The root-mean-square (rms) speed of gas molecules is given by:

  1. $v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$
  2. $v_{\text{rms}} = \sqrt{\dfrac{RT}{M}}$
  3. $v_{\text{rms}} = \sqrt{\dfrac{2RT}{M}}$
  4. $v_{\text{rms}} = \sqrt{\dfrac{8RT}{\pi M}}$
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Correct answer: A. $v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$

From kinetic theory, the average translational kinetic energy of a molecule is $\dfrac{1}{2} m \overline{v^2} = \dfrac{3}{2} kT$ giving $\overline{v^2} = \dfrac{3kT}{m} = \dfrac{3RT}{M}$, where $M$ is molar mass. The **rms** speed is the square root: $v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$. Other useful speeds: most probable speed $v_p = \sqrt{\dfrac{2RT}{M}}$ (option C) average speed $\langle v\rangle = \sqrt{\dfrac{8RT}{\pi M}}$ (option D). Order: $v_p < \langle v\rangle < v_{\text{rms}}$.

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The mean translational kinetic energy of one molecule of an ideal gas at temperature $T$ is:

  1. $\dfrac{1}{2} kT$
  2. $kT$
  3. $\dfrac{3}{2} kT$
  4. $3 kT$
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Correct answer: C. $\dfrac{3}{2} kT$

Equipartition theorem: each translational degree of freedom contributes $\dfrac{1}{2} kT$ of energy. There are 3 translational degrees ($x, y, z$) so total mean translational KE per molecule is $\dfrac{3}{2} kT$. For a mole: $\dfrac{3}{2} RT$. If the molecule has internal degrees (rotation, vibration), more energy is stored — but those don't add to **translational** KE specifically.

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If the absolute temperature of an ideal gas is doubled, the rms speed of its molecules:

  1. Doubles
  2. Increases by a factor of $\sqrt{2}$
  3. Quadruples
  4. Halves
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Correct answer: B. Increases by a factor of $\sqrt{2}$

$v_{\text{rms}} \propto \sqrt{T}$ (in Kelvin). Doubling $T$ scales the rms speed by $\sqrt{2}$. Intuition: kinetic energy is proportional to $T$, and KE $\propto v^2$, so $v \propto \sqrt{T}$. Option A (doubles) would be true if $v \propto T$, but it's $\sqrt{T}$.

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