Chapter Exam

Unit 9: Kinetic Theory of Gases — Online MCQ Test

Q: The r.m.s speed of oxygen molecules at 47°C is equal to that of the hydrogen molecules kept at ____ °C. (Mass of oxygen molecule/mass of hydrogen molecule=32/2)

Answer: C. -253 — v_rms = sqrt(3RT/M). Setting v_rms_O2 = v_rms_H2 implies T_O2/M_O2 = T_H2/M_H2. Given T_O2 = 320 K (47+273), M_O2 = 32, M_H2 = 2. Then T_H2 = (320 * 2) / 32 = 20 K. Converting to Celsius: 20 - 273 = -253°C.

Q: An insulated cylinder of volume 60 cm^3 is filled with a gas at 27°C and 2 atmospheric pressure. The gas is compressed to 20 cm^3 while the temperature rises to 77°C. The final pressure is α atmospheric pressure. Find α.

Answer: C. 7 — Using (P1V1)/T1 = (P2V2)/T2. P1 = 2 atm, V1 = 60 cm^3, T1 = 300 K. V2 = 20 cm^3, T2 = 350 K. P2 = (P1 * V1 * T2) / (T1 * V2) = (2 * 60 * 350) / (300 * 20) = 42000 / 6000 = 7 atm.

Q: The mean free path of a molecule of diameter 5 × 10^-10 m at the temperature 41°C and pressure 1.38 × 10^5 Pa is given as m. (Given k_B = 1.38 × 10^-23 J/K)

Answer: C. 2 × 10^-8 — The mean free path formula is λ = k_B * T / (√2 * π * σ² * P). Substituting σ = 5 × 10^-10 m, T = 314 K, P = 1.38 × 10^5 Pa, λ ≈ 2 × 10^-8 m.

Q: A cylindrical tube AB of length , closed at both ends contains an ideal gas of 1 mol having molecular weight M. The tube is rotated in a horizontal plane with constant angular velocity  about an axis perpendicular to AB and passing through the edge at end A. If PA and PB are the pressures at A and B respectively, then (Consider the temperature is same at all points in the tube)

Answer: A. PB = PA * exp(Mω^2 * ^2 / 2RT) — Integrating the pressure variation due to centrifugal force: dP = rho * omega^2 * x * dx. Using rho = PM / RT, dP/P = (M * omega^2 / RT) * x * dx. Integrating from 0 to L gives ln(PB/PA) = M * omega^2 * L^2 / 2RT.

Q: Match List I with List II : List I (A) 3 Translational degrees of freedom (B) 3 Translational, 2 rotational degrees of freedoms (C) 3 Translational, 2 rotational and 1 vibrational degrees of freedom (D) 3 Translational, 3 rotational and more than one vibrational degrees of freedom List II (I) Monoatomic gases (II) Polyatomic gases (III) Rigid diatomic gases (IV) Nonrigid diatomic gases Choose the correct answer from the options given below :

Answer: B. (A)-(I), (B)-(III), (C)-(IV), (D)-(II) — Monoatomic gases have only $3$ translational degrees of freedom, so $(A)\to(I)$. Rigid diatomic gases have $3$ translational and $2$ rotational degrees of freedom, so $(B)\to(III)$. Nonrigid diatomic gases additionally have vibrational degrees of freedom, so $(C)\to(IV)$. Polyatomic gases generally have $3$ translational, $3$ rotational, and more than one vibrational degree of freedom, so $(D)\to(II)$. Therefore the correct matching is option (b).

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Practice Unit 9: Kinetic Theory of Gases with a free chapter-wise online MCQ test for JEE JEE MAINS 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 ga.... 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 JEE JEE MAINS PHYSICS — Unit 9: Kinetic Theory of Gases, with answers and explanations. These are sample questions; the exam has its own separate question set.

The r.m.s speed of oxygen molecules at 47°C is equal to that of the hydrogen molecules kept at ____ °C. (Mass of oxygen molecule/mass of hydrogen molecule=32/2)

A. -235
B. -100
C. -253 ✓
D. -20

Answer: C. -253
v_rms = sqrt(3RT/M). Setting v_rms_O2 = v_rms_H2 implies T_O2/M_O2 = T_H2/M_H2. Given T_O2 = 320 K (47+273), M_O2 = 32, M_H2 = 2. Then T_H2 = (320 * 2) / 32 = 20 K. Converting to Celsius: 20 - 273 = -253°C.

An insulated cylinder of volume 60 cm^3 is filled with a gas at 27°C and 2 atmospheric pressure. The gas is compressed to 20 cm^3 while the temperature rises to 77°C. The final pressure is α atmospheric pressure. Find α.

A. 5
B. 6
C. 7 ✓
D. 8

Answer: C. 7
Using (P1V1)/T1 = (P2V2)/T2. P1 = 2 atm, V1 = 60 cm^3, T1 = 300 K. V2 = 20 cm^3, T2 = 350 K. P2 = (P1 * V1 * T2) / (T1 * V2) = (2 * 60 * 350) / (300 * 20) = 42000 / 6000 = 7 atm.

The mean free path of a molecule of diameter 5 × 10^-10 m at the temperature 41°C and pressure 1.38 × 10^5 Pa is given as m. (Given k_B = 1.38 × 10^-23 J/K)

A. 2 × 10^-10
B. 10^-8
C. 2 × 10^-8 ✓
D. 10^-8

Answer: C. 2 × 10^-8
The mean free path formula is λ = k_B * T / (√2 * π * σ² * P). Substituting σ = 5 × 10^-10 m, T = 314 K, P = 1.38 × 10^5 Pa, λ ≈ 2 × 10^-8 m.

A cylindrical tube AB of length , closed at both ends contains an ideal gas of 1 mol having molecular weight M. The tube is rotated in a horizontal plane with constant angular velocity  about an axis perpendicular to AB and passing through the edge at end A. If PA and PB are the pressures at A and B respectively, then (Consider the temperature is same at all points in the tube)

A. PB = PA * exp(Mω^2 * ^2 / 2RT) ✓
B. PB = PA
C. PB = PA * exp(Mω^2 * ^2 / 3RT)
D. PB = PA * exp(Mω^2 * ^2 / RT)

Answer: A. PB = PA * exp(Mω^2 * ^2 / 2RT)
Integrating the pressure variation due to centrifugal force: dP = rho * omega^2 * x * dx. Using rho = PM / RT, dP/P = (M * omega^2 / RT) * x * dx. Integrating from 0 to L gives ln(PB/PA) = M * omega^2 * L^2 / 2RT.

Match List I with List II : List I (A) 3 Translational degrees of freedom (B) 3 Translational, 2 rotational degrees of freedoms (C) 3 Translational, 2 rotational and 1 vibrational degrees of freedom (D) 3 Translational, 3 rotational and more than one vibrational degrees of freedom List II (I) Monoatomic gases (II) Polyatomic gases (III) Rigid diatomic gases (IV) Nonrigid diatomic gases Choose the correct answer from the options given below :

A. (A)-(I), (B)-(IV), (C)-(III), (D)-(II)
B. (A)-(I), (B)-(III), (C)-(IV), (D)-(II) ✓
C. (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
D. (A)-(IV), (B)-(II), (C)-(I), (D)-(III)

Answer: B. (A)-(I), (B)-(III), (C)-(IV), (D)-(II)
Monoatomic gases have only $3$ translational degrees of freedom, so $(A)\to(I)$. Rigid diatomic gases have $3$ translational and $2$ rotational degrees of freedom, so $(B)\to(III)$. Nonrigid diatomic gases additionally have vibrational degrees of freedom, so $(C)\to(IV)$. Polyatomic gases generally have $3$ translational, $3$ rotational, and more than one vibrational degree of freedom, so $(D)\to(II)$. Therefore the correct matching is option (b).

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