Q.22 Consider monoatomic ideal gas molecules of equal mass 𝑚 in thermal equilibrium, at
temperature 𝑇. Which one of the following equations is correct? (the angular brackets denote
average, 𝑘𝐵 is Boltzmann constant, 𝑣 is velocity, and vx is the x-component of velocity)

Introduction

For a monatomic ideal gas, all internal energy is purely translational kinetic energy of point‑like atoms moving randomly in three dimensions. The average kinetic energy per molecule at temperature T follows directly from the equipartition theorem, which assigns ½kBT to each quadratic degree of freedom. Using this principle allows precise evaluation of expressions like ⟨½mv²⟩ and ⟨½mvx²⟩ for gas molecules in thermal equilibrium.

Core physics: equipartition and velocities

A monatomic ideal gas has three translational degrees of freedom (motion in x, y, z) and no rotational or vibrational degrees, so its average kinetic energy per molecule is ³/₂kBT.

The total translational kinetic energy of one molecule is K = ½m(vx² + vy² + vz²) = ½mv², where v² = vx² + vy² + vz².

By equipartition, each squared velocity component contributes ½kBT, so

⟨½mvx²⟩ = ⟨½mvy²⟩ = ⟨½mvz²⟩ = ½kBT

and therefore

⟨½mv²⟩ = ⟨½m(vx² + vy² + vz²)⟩ = 3 × ½kBT = ³/₂kBT.

Option-wise analysis

Let the angular brackets ⟨⋅⟩ denote ensemble average over molecules.