Chapter 17, Problem 76P

The Maxwell-Boltzmann distribution, plotted in Fig. 17.4, is given by

$N(v)\Delta v=4\pi N{\left(\frac{m}{2\pi KT}\right)}^{3/2}{v}^{2}e-m{v}^2/2kT\Delta v$

where N(v) Δv is the number of molecules in a small speed range Δv around speed v, N is the total number of molecules in the gas, m is the molecular mass, k is Boltzmann’s constant, and T is the temperature. Use this equation to show that the most probable speed for a gas molecule—the speed at the peak of the curves in Fig. 17.4—is $\sqrt{2kT/m}$. Note that the thermal speed (Equation 17.4), which is the average molecular speed, is a factor of $\sqrt{3/2}$or about 20% greater than the most probable speed—a fact that reflects the long, high-energy “tail” of the Maxwell-Boltzmann distribution.