Chapter Exam

Chapter-12 Kinetic Theory — Online MCQ Test

Q: What is Avogadro's number?

Answer: A. 6.02 × 10²³ mol⁻¹ — Avogadro's number is a fundamental constant equal to 6.02 × 10²³ mol⁻¹, representing the number of particles in one mole.

Q: The equation of state for an ideal gas is:

Answer: A. PV = nRT — The ideal gas equation is PV = nRT, where P is pressure, V is volume, n is number of moles, R is gas constant, and T is temperature.

Q: Which of the following is NOT an assumption of the kinetic theory of gases?

Answer: B. Molecules exert attractive forces on each other — The kinetic theory assumes no intermolecular forces; the assumption of attractive forces contradicts this fundamental postulate.

Q: The RMS speed of gas molecules is given by:

Answer: A. v_rms = √(3kT/m) — The root mean square speed is v_rms = √(3kT/m), where k is Boltzmann constant, T is temperature, and m is molecular mass.

Q: For a diatomic gas at room temperature, the degrees of freedom are:

Answer: A. 5 — A diatomic gas has 3 translational and 2 rotational degrees of freedom (rotation about two axes perpendicular to molecular axis), totaling 5.

Q: If the temperature of a gas is doubled, the RMS speed of its molecules will:

Answer: A. Increase by a factor of √2 — Since v_rms ∝ √T, doubling temperature increases RMS speed by √2. This follows from v_rms = √(3kT/m).

Q: A container of volume V₁ at pressure P₁ and temperature T₁ is connected to another container of volume V₂ at pressure P₂ and temperature T₂. After equilibration, the final pressure will be:

Answer: B. (P₁V₁/T₁ + P₂V₂/T₂) × T_f/(V₁ + V₂) — Using ideal gas law, total moles n = (P₁V₁/RT₁ + P₂V₂/RT₂). Final pressure P_f = n_total × R × T_f / (V₁ + V₂).

PHYSICS · CLASS 11 INTER I YEAR · Andhra State Board

Practice Chapter-12 Kinetic Theory with a free chapter-wise online MCQ test for Andhra State Board CLASS 11 INTER I YEAR PHYSICS. This chapter covers: Equation of state - Perfect gas - Kinetic model - Gas pressure - Kinetic interpretation of temperature - RMS speed - Degrees of freedom - Law of equipartition of energy - Specific.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter-12 Kinetic Theory — Important Questions & Answers (FAQ)

Frequently asked questions from Andhra State Board CLASS 11 INTER I YEAR PHYSICS — Chapter-12 Kinetic Theory, with answers and explanations. These are sample questions; the exam has its own separate question set.

What is Avogadro's number?

A. 6.02 × 10²³ mol⁻¹ ✓
B. 6.02 × 10²² mol⁻¹
C. 3.14 × 10²³ mol⁻¹
D. 1.38 × 10⁻²³ mol⁻¹

Answer: A. 6.02 × 10²³ mol⁻¹
Avogadro's number is a fundamental constant equal to 6.02 × 10²³ mol⁻¹, representing the number of particles in one mole.

The equation of state for an ideal gas is:

A. PV = nRT ✓
B. PV = RT
C. P = nRT
D. V = nRT/P²

Answer: A. PV = nRT
The ideal gas equation is PV = nRT, where P is pressure, V is volume, n is number of moles, R is gas constant, and T is temperature.

Which of the following is NOT an assumption of the kinetic theory of gases?

A. Gas molecules are point masses with negligible volume
B. Molecules exert attractive forces on each other ✓
C. Collisions between molecules are perfectly elastic
D. There is random motion of gas molecules

Answer: B. Molecules exert attractive forces on each other
The kinetic theory assumes no intermolecular forces; the assumption of attractive forces contradicts this fundamental postulate.

The RMS speed of gas molecules is given by:

A. v_rms = √(3kT/m) ✓
B. v_rms = √(kT/m)
C. v_rms = √(2kT/m)
D. v_rms = √(3RT/M)

Answer: A. v_rms = √(3kT/m)
The root mean square speed is v_rms = √(3kT/m), where k is Boltzmann constant, T is temperature, and m is molecular mass.

For a diatomic gas at room temperature, the degrees of freedom are:

A. 5 ✓
B. 3
C. 6
D. 7

Answer: A. 5
A diatomic gas has 3 translational and 2 rotational degrees of freedom (rotation about two axes perpendicular to molecular axis), totaling 5.

If the temperature of a gas is doubled, the RMS speed of its molecules will:

A. Increase by a factor of √2 ✓
B. Increase by a factor of 2
C. Remain the same
D. Decrease by a factor of √2

Answer: A. Increase by a factor of √2
Since v_rms ∝ √T, doubling temperature increases RMS speed by √2. This follows from v_rms = √(3kT/m).

Which statement about the relationship between pressure and molecular kinetic energy is correct?

A. Pressure is directly proportional to average kinetic energy of molecules ✓
B. Pressure is inversely proportional to average kinetic energy
C. Pressure is independent of molecular kinetic energy
D. Pressure depends only on the number of molecules, not their energy

Answer: A. Pressure is directly proportional to average kinetic energy of molecules
From kinetic theory, P = (1/3)ρ<c²> ∝ <KE>, showing direct proportionality between pressure and average molecular kinetic energy.

A container of volume V₁ at pressure P₁ and temperature T₁ is connected to another container of volume V₂ at pressure P₂ and temperature T₂. After equilibration, the final pressure will be:

A. (P₁V₁ + P₂V₂)/(V₁ + V₂) × (T_f/T_avg)
B. (P₁V₁/T₁ + P₂V₂/T₂) × T_f/(V₁ + V₂) ✓
C. (P₁ + P₂)/2
D. P₁V₁ + P₂V₂

Answer: B. (P₁V₁/T₁ + P₂V₂/T₂) × T_f/(V₁ + V₂)
Using ideal gas law, total moles n = (P₁V₁/RT₁ + P₂V₂/RT₂). Final pressure P_f = n_total × R × T_f / (V₁ + V₂).

A sealed container with fixed volume contains a gas. If the molecular mass of the gas is doubled while keeping the number of molecules and temperature constant, the pressure will:

A. Remain unchanged because PV = nRT and none of these change ✓
B. Increase by a factor of 2
C. Decrease by a factor of √2
D. Become indeterminate without knowing specific volume

Answer: A. Remain unchanged because PV = nRT and none of these change
Pressure from ideal gas law P = nRT/V depends only on n, R, T, and V—not on molecular mass. Molecular mass affects RMS speed but not pressure.

Two identical containers with different gases A and B at the same temperature and pressure have different molecular masses. Which statement is NOT necessarily true?

A. Gas A and B have the same number of molecules
B. The RMS speeds of A and B are different
C. The average kinetic energies per molecule are equal
D. Gas with smaller molecular mass has higher mean free path ✓

Answer: D. Gas with smaller molecular mass has higher mean free path
Mean free path depends on λ ∝ kT/(√2πd²P), which depends on molecular diameter d, not on mass. Lower mass doesn't guarantee higher mean free path.

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