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BIHAR CET Kinetic Theory of Gases — practice questions — page 2

Explore 87 free Kinetic Theory of Gases questions in BIHAR CET 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 31–60 of 87.

If the mean free path of N₂ at 1 atm is ~80 nm, what is it at 0.01 atm (same T)?Read the answer and worked solution →The number of moles of oxygen ($O_2$) gas in a vessel of volume 1 L at STP (273.15 K, 1.013 × 10⁵ Pa) is approximately: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 →Boltzmann's distribution gives fraction of molecules in energy state E:Read the answer and worked solution →During adiabatic compression of ideal gas:Read the answer and worked solution →For an ideal gas in adiabatic process: TV^(γ-1) =Read the answer and worked solution →

Worked practice questions

If the mean free path of N₂ at 1 atm is ~80 nm, what is it at 0.01 atm (same T)?

  1. ~ 8 nm
  2. ~ 80 nm
  3. ~ 800 nm
  4. ~ 8 μm (mean free path ∝ 1/P)
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Correct answer: D. ~ 8 μm (mean free path ∝ 1/P)

λ ∝ T/P. Reducing P by 100× increases λ by 100×: 80 nm × 100 = 8000 nm = 8 μm. (Why vacuum coating works at low pressure — gas molecules don't collide before reaching the substrate.)

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The number of moles of oxygen ($O_2$) gas in a vessel of volume 1 L at STP (273.15 K, 1.013 × 10⁵ Pa) is approximately:

  1. ~ 0.045 mol
  2. ~ 0.045 g
  3. ~ 1.0 mol
  4. ~ 22.4 mol
Show answer and solution

Correct answer: A. ~ 0.045 mol

$n = PV/(RT) = (1.013\times10^5)(10^{-3})/(8.314\times273.15) = 101.3/2270.6 \approx 0.0446$ mol. Equivalently, 22.4 L = 1 mol at STP, so 1 L ≈ 1/22.4 = 0.0446 mol.

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Ideal gas equation:

  1. PV = mR
  2. PV = RT/n
  3. P + V = nRT
  4. PV = nRT
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Correct answer: D. PV = nRT

PV = nRT. n = moles, R = 8.314 J/(mol K). Combines Boyle (P ∝ 1/V at const T), Charles (V ∝ T at const P), Avogadro (V ∝ n).

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