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Unit 9: Kinetic Theory of Gases — Online MCQ Test

PHYSICS · NEET (All) · NEET

Practice Unit 9: Kinetic Theory of Gases with a free chapter-wise online MCQ test for NEET NEET (All) PHYSICS. This chapter covers: Equation of state of a perfect gas work done on compressing a gas Kinetic theory of gases - assumptions the concept of pressure. Kinetic interpretation of temperature: RMS speed of.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Unit 9: Kinetic Theory of Gases — Important Questions & Answers (FAQ)

Frequently asked questions from NEET NEET (All) PHYSICS — Unit 9: Kinetic Theory of Gases, with answers and explanations. These are sample questions; the exam has its own separate question set.

A cylinder contains hydrogen gas at a pressure of 249 kPa and temperature 27°C. Its density is: (R=8.3 J mol⁻¹ K⁻¹)
  • A. 0.2 kg/m³ ✓
  • B. 0.1 kg/m³
  • C. 0.02 kg/m³
  • D. 0.5 kg/m³
Answer: A. 0.2 kg/m³
Using PV = nRT = (m/M)RT, then P = (ρ/M)RT where ρ is density and M is molar mass of H₂ (2x10⁻³ kg/mol). ρ = PM/RT = (249x10³ * 2x10⁻³) / (8.3 * 300) = 498 / 2490 = 0.2 kg/m³.
The mean free path for a gas, with molecular diameter d and number density n can be expressed as:
  • A. 1 / (√2 nπd²) ✓
  • B. 1 / (√2 n²πd²)
  • C. 1 / (√2 n²π²d²)
  • D. 1 / (√2 nπd)
Answer: A. 1 / (√2 nπd²)
The standard kinetic theory expression for mean free path λ is 1 / (√2 nσ), where σ = πd² is the collision cross-section.
The average thermal energy for a mono-atomic gas is:
  • A. 3/2 kBT ✓
  • B. 5/2 kBT
  • C. kBT
  • D. 7/2 kBT
Answer: A. 3/2 kBT
According to the equipartition of energy, each degree of freedom contributes 1/2 kBT. A monoatomic gas has 3 degrees of freedom, so energy = 3 * 1/2 kBT = 3/2 kBT.
A container has two chambers of volumes V1 = 2 litres and V2 = 3 litres separated by a partition made of a thermal insulator. The chambers contains n1 = 5 and n2 = 4 moles of ideal gas at pressures p1 = 1 atm and p2 = 2 atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of:
  • A. 1.3 atm
  • B. 1.6 atm ✓
  • C. 1.4 atm
  • D. 1.8 atm
Answer: B. 1.6 atm
Using Dalton's Law or P_final = (P1V1 + P2V2) / (V1 + V2). PV = nRT, so P1V1 = n1RT and P2V2 = n2RT. P_final = (n1RT + n2RT) / (V1 + V2) = (5 + 4)RT / (2+3)L = 9RT/5. Given n1=5, V1=2, P1=1 -> PV/n = 1*2/5 = 0.4. P_final = (n1+n2)RT / (V1+V2) = 9 * 0.4 / 5 = 1.6 atm.
The correction factor ‘a’ to the ideal gas equation corresponds to:
  • A. Density of the gas molecules
  • B. Forces of attraction between the gas molecules ✓
  • C. Electric field present between the gas molecules
  • D. Volume of the gas molecules
Answer: B. Forces of attraction between the gas molecules
In the van der Waals equation (P + an^2/V^2)(V-nb) = nRT, the term 'a' represents the magnitude of intermolecular forces of attraction.

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