Questions: Maxwell-Boltzmann Distribution and Classical Limit

Question 1

The Maxwell-Boltzmann most probable speed is v_mp = sqrt(2kT/m). If the temperature of a gas is doubled from T to 2T, by what factor does the most probable speed change?

• A. It doubles (factor of 2)
• B. It increases by a factor of sqrt(2)
• C. It stays the same
• D. It quadruples (factor of 4)

Answer: (B)

v_mp ∝ sqrt(T), so doubling T gives v_mp → sqrt(2T/T) · v_mp = sqrt(2) · v_mp ≈ 1.41 · v_mp. Speed scales as the square root of temperature, not linearly. A common error is assuming a linear relationship because kinetic energy is proportional to T — but speed is the square root of kinetic energy.

Question 2

At thermal equilibrium, most molecules in an ideal gas have kinetic energy equal to (3/2)kT.

• T. True
• F. False

Answer: False

(3/2)kT is the average kinetic energy per molecule. Individual molecules are distributed across a broad range of energies according to the Maxwell-Boltzmann distribution — some have much more, some much less. Only the mean value equals (3/2)kT. The distribution has a long high-energy tail, which is physically important: those rare fast molecules are responsible for chemical reactions, evaporation, and atmospheric escape.

Question 3

The Boltzmann factor exp(-E/kT) makes high-energy states less probable. Why does this arise naturally from the behavior of a system in contact with a heat reservoir?

Model answer: When the system occupies a high-energy state, the reservoir must supply that energy and therefore loses entropy. Because the combined entropy of system plus reservoir is maximized at equilibrium, states that reduce the reservoir's entropy are suppressed exponentially, giving the Boltzmann factor.

This is the central argument of statistical mechanics. The reservoir has an enormous number of microstates, and its entropy decreases when it gives energy to the system. The probability of the system having energy E is proportional to the number of microstates available to the reservoir after giving up E, which is exp(-E/kT) by the definition of temperature as ∂S/∂E = 1/T. The partition function Z normalizes this over all possible states.