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Indian Navy SSR (Sailor) Kinetic Theory of Gases — practice questions

Explore 6 free Kinetic Theory of Gases questions in Indian Navy SSR (Sailor) 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 1–6 of 6.

The mean translational kinetic energy of one molecule of an ideal gas at temperature $T$ is:Read the answer and worked solution →Internal energy of n moles of ideal gas (monatomic):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 →In adiabatic process for ideal gas: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 →Kinetic energy per molecule of monatomic gas at 300 K (k = 1.38 × 10⁻²³):Read the answer and worked solution →

Worked practice questions

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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Internal energy of n moles of ideal gas (monatomic):

  1. U = (3/2) nRT
  2. U = (5/2) nRT
  3. U = nR
  4. U = nRT
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Correct answer: A. U = (3/2) nRT

**Monatomic:** U = (3/2) nRT. **Diatomic at moderate T:** U = (5/2) nRT. **Triatomic linear:** (5/2)nRT (5 DOF) bent: (3)nRT (6 DOF approximately).

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Two gases of molar masses M₁ and M₂ at same T. Ratio of v_rms:

  1. M₁ + M₂
  2. sqrt(M₂/M₁) (lighter gas has higher v_rms)
  3. M₁/M₂
  4. M₁ × M₂
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Correct answer: B. sqrt(M₂/M₁) (lighter gas has higher v_rms)

v_rms ∝ sqrt(T/M). **Same T:** v₁/v₂ = sqrt(M₂/M₁). So lighter gas (lower M) has higher v_rms. **Hence H₂ molecules move faster than O₂ at same T.**

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