 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2017/paper-2/section-c/c12/solution

For a classical dilute ideal gas in thermal equilibrium, the probability of a single-particle velocity is proportional to the Boltzmann factor for kinetic energy,

e^{- m v^{2} / (2 k_{B} T)}

. Velocities with speed between

v

and

v + d v

4 π v^{2} d v

. The normalized Maxwell speed distribution is therefore

f (v) = 4 π (\frac{m}{2 π k _{B} T})^{3/2} v^{2} e^{- m v^{2} / (2 k_{B} T)}, v \geq 0.

The factor

v^{2}

counts states, rather than representing an extra Boltzmann energy factor.

α = m / (2 k_{B} T)

and use

I_{j} = \int_{0}^{\infty} v^{j} e^{- α v^{2}} d v

. Normalization cancels in ratios, so the given recurrence gives

\overline{v} = \frac{I _{3}}{I _{2}} = \frac{2}{π α} = \frac{8 k _{B} T}{πm}, \overline{v^{2}} = \frac{I _{4}}{I _{2}} = \frac{3}{2 α} = \frac{3 k _{B} T}{m} .

Consequently

\overline{v^{2}} = \frac{3 π}{8} \overline{v}^{2} .

The root mean square speed is larger than the mean speed, consistent with the positive variance of the speed distribution.

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