Kinetic Theory of Gases — NEET Physics MCQs with answers

Kinetic Theory of Gases is a Class 11 Physics chapter in the NEET (UG) syllabus. NEET720 has 501 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Avogadro's law and Avogadro's number · Kinetic energy of gas molecules and temperature · Ideal gas equation and gas laws · Kinetic interpretation of pressure and rms speed · Degrees of freedom and equipartition · Mean free path · Kinetic interpretation of temperature · Pressure of an ideal gas from kinetic theory · Specific heats from kinetic theory · Law of equipartition of energy · Kinetic theory postulates · Ideal gas equation and kinetic theory · RMS, average and most probable speed · Pressure and mean free path from kinetic theory · Maxwell-Boltzmann speed distribution · Specific heats of gases from kinetic theory · Degrees of freedom and specific heats · Diffusion · Specific heats of gases · Pressure of an ideal gas · Brownian motion · Kinetic energy and temperature · Specific heat capacity · Speed of gas molecules · Pressure from kinetic theory · Ideal gas laws · Assumptions of kinetic theory · Kinetic energy of gas molecules · Pressure of a gas · Real gases · Internal energy · Gas constants · Maxwell speed distribution · Dalton's law · Kinetic Theory · KE and Temperature · Avogadro's Law · Real Gas · Maxwell Distribution · Graham's Law

8 free Kinetic Theory of Gases practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · easy · Degrees of freedom and equipartition

   The number of degrees of freedom of a monoatomic gas molecule (e.g. helium), treated as a point mass, is

   • A.2
   • B.4
   • C.3
   • D.6
   Show answer and explanation

   Answer: C. 3

   A monoatomic molecule, modelled as a point mass, can only translate along the three independent x, y, z directions, giving 3 degrees of freedom.

   Degrees of freedom count the independent ways a molecule can store energy. A monoatomic gas molecule (like He, Ne, Ar) is modelled as a point particle with no internal structure to rotate or vibrate. It can only move independently along three mutually perpendicular directions (x, y, z), so it has exactly 3 translational degrees of freedom and no rotational or vibrational ones.

   Common mistake: Undercounting the number of independent translational directions.

2. Question 2 · easy · Kinetic interpretation of pressure and rms speed

   According to the kinetic theory of gases, the pressure exerted by a gas on the walls of its container arises mainly because

   • A.molecules briefly stick to the wall, building up an attractive force before leaving
   • B.molecules near the wall repel each other, pushing outward molecules into it
   • C.molecules continuously collide elastically with the wall, transferring momentum to it
   • D.the weight of the gas column above presses down on the wall, as in a static fluid
   Show answer and explanation

   Answer: C. molecules continuously collide elastically with the wall, transferring momentum to it

   Kinetic theory models pressure as the rate of momentum transfer to the walls from a very large number of elastic molecular collisions per second.

   In kinetic theory, gas molecules move randomly and collide elastically with the container walls. Each collision changes a molecule's momentum component normal to the wall, and by Newton's third law, an equal and opposite momentum is delivered to the wall. Summed over the enormous number of collisions per second over the wall area, this constant bombardment produces a steady macroscopic force per unit area, which we call pressure. There is no sticking, no repulsion between molecules causing pressure, and no hydrostatic weight effect at the molecular level (that effect is negligible for typical gas samples).

   Common mistake: Thinking of gas pressure as arising from a static, weight-like effect rather than dynamic momentum transfer.

3. Question 3 · medium · RMS speed

   At the same temperature, the ratio of the rms speed of hydrogen (M=2 g/mol) to that of oxygen (M=32 g/mol) is:

   • A.4:1
   • B.16:1
   • C.1:4
   • D.2:1
   Show answer and explanation

   Answer: A. 4:1

   v_rms ∝ 1/√M at fixed T; ratio = √(M_O2/M_H2) = √(32/2) = √16 = 4.

   Since v_rms = √(3RT/M), at the same temperature v_rms ∝ 1/√M. So v_rms(H2)/v_rms(O2) = √(M_O2/M_H2) = √(32/2) = √16 = 4. Hence the ratio is 4:1.

   Common mistake: Using the mass ratio directly (16:1) instead of its square root

4. Question 4 · medium · Kinetic energy of gas molecules

   The average kinetic energy of a gas molecule at 500 K is (k = 1.38×10⁻²³ J/K):

   • A.1.04×10⁻²⁰ J
   • B.6.9×10⁻²¹ J
   • C.2.07×10⁻²⁰ J
   • D.3.45×10⁻²¹ J
   Show answer and explanation

   Answer: A. 1.04×10⁻²⁰ J

   KE_avg = (3/2)kT = 1.5×1.38×10⁻²³×500 ≈ 1.04×10⁻²⁰ J.

   Average translational kinetic energy per molecule = (3/2)kT. Substituting k=1.38×10⁻²³ J/K and T=500 K: KE = 1.5 × 1.38×10⁻²³ × 500 = 1.035×10⁻²⁰ J ≈ 1.04×10⁻²⁰ J.

   Common mistake: Using an incorrect numerical factor instead of 3/2 in the kinetic energy formula

5. Question 5 · easy · Degrees of freedom

   Compared to a monoatomic gas (3 translational degrees of freedom), a diatomic gas at moderate temperatures has additional degrees of freedom due to:

   • A.Rotational motion about two axes perpendicular to the line joining the two atoms
   • B.Vibrational motion along all three spatial axes simultaneously, even at room temperature
   • C.Additional translational motion along a fourth spatial dimension
   • D.Electronic transitions occurring continuously at room temperature
   Show answer and explanation

   Answer: A. Rotational motion about two axes perpendicular to the line joining the two atoms

   A diatomic molecule can rotate about 2 independent axes perpendicular to the bond axis, adding 2 rotational degrees of freedom to the 3 translational ones (total 5, at moderate T).

   A diatomic molecule, modeled as two point masses joined by a rigid bond, has 3 translational degrees of freedom (like a monoatomic gas) plus 2 rotational degrees of freedom, corresponding to rotation about the two axes perpendicular to the line joining the atoms (rotation about the bond axis itself is negligible since the moment of inertia there is very small). This gives a total of 5 degrees of freedom at moderate temperatures, with vibrational modes typically activated only at much higher temperatures.

   Common mistake: Assuming vibrational modes are always active at ordinary temperatures

6. Question 6 · medium · Specific heat ratio

   For a diatomic gas (5 degrees of freedom at moderate temperature), the ratio of specific heats γ = Cp/Cv is:

   • A.1.4
   • B.1.67
   • C.1.29
   • D.2.0
   Show answer and explanation

   Answer: A. 1.4

   γ = 1 + 2/f = 1 + 2/5 = 1.4 for f=5 (diatomic gas).

   Using γ = 1 + 2/f, where f is the number of degrees of freedom: for a diatomic gas, f=5, so γ = 1 + 2/5 = 1 + 0.4 = 1.4.

   Common mistake: Using the monoatomic gas's degrees of freedom (f=3) or an incorrect count for diatomic gases

7. Question 7 · medium · Pressure of a gas

   A gas has density 1.2 kg/m³ and rms molecular speed 500 m/s. According to kinetic theory, the pressure exerted by the gas is:

   • A.1×10⁵ Pa
   • B.3×10⁵ Pa
   • C.3×10⁵/2 Pa
   • D.6×10⁵ Pa
   Show answer and explanation

   Answer: A. 1×10⁵ Pa

   P = (1/3)ρv²_rms = (1/3)(1.2)(500)² = 1×10⁵ Pa.

   From kinetic theory, P = (1/3)ρv²_rms. Substituting ρ=1.2 kg/m³ and v_rms=500 m/s: P = (1/3)(1.2)(250000) = (1/3)(300000) = 1×10⁵ Pa.

   Common mistake: Omitting the factor of 1/3 from the kinetic theory pressure formula

8. Question 8 · medium · Mean free path

   At constant temperature, if the pressure of a gas is doubled (keeping the amount of gas fixed), the mean free path of its molecules:

   • A.Becomes half of its original value
   • B.Becomes double its original value
   • C.Remains unchanged
   • D.Becomes four times its original value
   Show answer and explanation

   Answer: A. Becomes half of its original value

   Mean free path λ ∝ 1/n (number density); at constant T, doubling P doubles n (via PV=NkT), so λ halves.

   Mean free path is given by λ = 1/(√2 π d² n), where n is the number density of molecules. At constant temperature, from PV=NkT, number density n = P/(kT) is directly proportional to pressure P. So doubling the pressure doubles n, and since λ ∝ 1/n, the mean free path becomes half its original value.

   Common mistake: Assuming mean free path is unaffected by changes in pressure

Questions about Kinetic Theory of Gases for NEET

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Questions are original NEET720 compositions reviewed for correctness, syllabus fit and option quality. Counts update as the bank grows (501 active practice questions in this chapter today).