Chapter Exam

Unit 9: Kinetic Theory of Gases — Online MCQ Test

Q: 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)

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.

Q: The mean free path for a gas, with molecular diameter d and number density n can be expressed as:

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.

Q: The average thermal energy for a mono-atomic gas is:

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.

Q: 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:

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.

Q: Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape?

Answer: D. 1/8 — Using Graham's Law of diffusion, r(H2)/r(O2) = sqrt(M(O2)/M(H2)) = sqrt(32/2) = 4. If half of H2 escapes, the rate ratio implies the moles of O2 diffused will be 1/4 of the H2 moles, which is 1/8 of the initial total.

Q: The molecules of a given mass of a gas have r.m.s velocity of 200 ms^-1 at 27ºC and 1.0 × 10^5 Nm^-2 pressure. When the temperature and pressure of the gas are respectively, 127ºC and 0.05 × 10^5 Nm^-2, the r.m.s. velocity of its molecules in ms^-1 is:

Answer: A. 400/sqrt(3) — RMS velocity v = sqrt(3RT/M). It is independent of pressure. v1 = 200 at T1 = 300 K. v2 = v1 * sqrt(T2/T1) = 200 * sqrt(400/300) = 200 * sqrt(4/3) = 400/sqrt(3).

Q: The molecules of a given mass of a gas have r.m.s velocity of 200 ms⁻¹ at 27ºC and 1.0 × 10⁵ Nm⁻² pressure. When the temperature and pressure of the gas are respectively, 127ºC and 0.05 × 10⁵ Nm⁻², the r.m.s. velocity of its molecules in ms⁻¹ is:

Answer: A. 400/√3 — RMS velocity v ∝ √T. Initially T1 = 300K, finally T2 = 400K. v2/v1 = √(400/300) = 2/√3. v2 = 200 * (2/√3) = 400/√3.

Q: The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is

Answer: A. h / sqrt(3mkT) — Using de-Broglie wavelength formula λ = h/p and kinetic energy E = 3/2 * kT = p^2/(2m). Thus p = sqrt(3mkT), so λ = h / sqrt(3mkT).

Q: The correction factor ‘a’ to the ideal gas equation corresponds to:

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.

Q: The correction factor 'a' to the ideal gas equation corresponds to:

Answer: C. forces of attraction between the gas molecules — In van der Waals equation, 'a' represents the magnitude of intermolecular forces of attraction.

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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.

Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape?

A. 1/4
B. 3/8
C. 1/2
D. 1/8 ✓

Answer: D. 1/8
Using Graham's Law of diffusion, r(H2)/r(O2) = sqrt(M(O2)/M(H2)) = sqrt(32/2) = 4. If half of H2 escapes, the rate ratio implies the moles of O2 diffused will be 1/4 of the H2 moles, which is 1/8 of the initial total.

The molecules of a given mass of a gas have r.m.s velocity of 200 ms^-1 at 27ºC and 1.0 × 10^5 Nm^-2 pressure. When the temperature and pressure of the gas are respectively, 127ºC and 0.05 × 10^5 Nm^-2, the r.m.s. velocity of its molecules in ms^-1 is:

A. 400/sqrt(3) ✓
B. 2100/sqrt(3)
C. 100/sqrt(3)
D. 2100

Answer: A. 400/sqrt(3)
RMS velocity v = sqrt(3RT/M). It is independent of pressure. v1 = 200 at T1 = 300 K. v2 = v1 * sqrt(T2/T1) = 200 * sqrt(400/300) = 200 * sqrt(4/3) = 400/sqrt(3).

The molecules of a given mass of a gas have r.m.s velocity of 200 ms⁻¹ at 27ºC and 1.0 × 10⁵ Nm⁻² pressure. When the temperature and pressure of the gas are respectively, 127ºC and 0.05 × 10⁵ Nm⁻², the r.m.s. velocity of its molecules in ms⁻¹ is:

A. 400/√3 ✓
B. 2100/√3
C. 100/√3
D. 2100

Answer: A. 400/√3
RMS velocity v ∝ √T. Initially T1 = 300K, finally T2 = 400K. v2/v1 = √(400/300) = 2/√3. v2 = 200 * (2/√3) = 400/√3.

The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is

A. h / sqrt(3mkT) ✓
B. 2h / sqrt(3mkT)
C. 2h / sqrt(mkT)
D. h / sqrt(mkT)

Answer: A. h / sqrt(3mkT)
Using de-Broglie wavelength formula λ = h/p and kinetic energy E = 3/2 * kT = p^2/(2m). Thus p = sqrt(3mkT), so λ = h / sqrt(3mkT).

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.

The correction factor 'a' to the ideal gas equation corresponds to:

A. electric field present between the gas molecules
B. density of the gas molecules
C. forces of attraction between the gas molecules ✓
D. volume of the gas molecules

Answer: C. forces of attraction between the gas molecules
In van der Waals equation, 'a' represents the magnitude of intermolecular forces of attraction.

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