Unit 9: Kinetic Theory of Gases — Online MCQ Test
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Practice Unit 9: Kinetic Theory of Gases with a free chapter-wise online MCQ test.
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....
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Unit 9: Kinetic Theory of Gases — Important Questions & Answers
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.
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 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.
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.
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.
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).
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.
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).
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.
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.
In van der Waals equation, 'a' represents the magnitude of intermolecular forces of attraction.