Chapter-12 Kinetic Theory — Online MCQ Test
PHYSICS · Grade 11 · CBSE(NCERT)
Practice Chapter-12 Kinetic Theory with a free chapter-wise online MCQ test.
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.
Chapter-12 Kinetic Theory — Important Questions & Answers
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.
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.
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 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.
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.
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).
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.
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).
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.
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.
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.