Chapter 20, Problem 29PQ

Consider the Maxwell-Boltzmann distribution function plotted in Problem 28. For those parameters, determine the rms velocity and the most probable speed, as well as the values of f(v) for each of these values. Compare these values with the graph in Problem 28.

28. Plot the Maxwell-Boltzmann distribution function for a gas composed of nitrogen molecules (N₂) at a temperature of 295 K. Identify the points on the curve that have a value of half the maximum value. Estimate these speeds, which represent the range of speeds most of the molecules are likely to have. The mass of a nitrogen molecule is 4.68 × 10⁻²⁶ kg.

Equation 20.18 can be used to find the rms velocity given the temperature, Boltzmann’s constant, and the mass of the atom or molecule. The mass of a nitrogen molecule is 4.68 × 10⁻²⁶ kg.

$v_{rms}=\sqrt{\frac{3k_{\text{B}}T}{m}}=\sqrt{\frac{3(1.38\times {10}^{-23}\text{J/K})}{4.68\times {10}^{-26}\text{kg}}}=\boxed{511\text{m/s}}$

Using the results of Problem 28 and the rms velocity, we can calculate the value of f(v).

f(v_rms) = (3.11 × 10⁻⁸)(511)² e $-(5.75\times {10}^{-6} {\left(511\right)}^2)$ = 0.00181

The most probable speed, for which this function has its maximum value, is given by Equation 20.20.

$v_{\text{mp}}=\sqrt{\frac{2k_{\text{B}}T}{m}}=\sqrt{\frac{2(1.38\times {10}^{-23}\text{J/K})(295\text{ K})}{4.68\times {10}^{-26}\text{kg}}}=\boxed{417\text{m/s}}$

f(v_mp) = (3.11×10⁻⁸)(417)2 e⁻ $(5.75\times {10}^{-6}{(417)}^2)$ = 0.00199

We plot these points on the speed distribution. The most probable speed is indeed at the peak of the distribution function. Since the function is not symmetric, the rms velocity is somewhat higher than the most probable speed.

Figure P20.29ANS