Question 1.11: MEAN AND RMS SPEEDS OF MOLECULES Given the Maxwell Boltzmann...
MEAN AND RMS SPEEDS OF MOLECULES Given the Maxwell Boltzmann distribution law for the velocities of molecules in a gas, derive expressions for the mean speed (v_{av}), most probable speed (v^{*}), and rms velocity (v_{rms}) of the molecules and calculate the corresponding values for a gas of noninteracting electrons.
The number of molecules with speeds in the range v to (v + dv) is
d N=n_{v} d v=4 \pi N\left(\frac{m}{2 \pi k T}\right)^{3 / 2} v^{2} \exp \left(-\frac{m v^{2}}{2 k T}\right) d v
We know that n_{v}∕N is the probability per unit speed that a molecule has a speed in the range v to (v + dv). By definition, then, the mean speed is given by
v_{ av }=\frac{\int v d N}{\int d N}=\frac{\int v n_{v} d v}{\int n_{v} d v}=\sqrt{\frac{8 k T}{\pi m}} Mean speed
where the integration is over all speeds (v = 0 to ∞). The mean square velocity is given by
\overline{v^{2}}=\frac{\int v^{2} d N}{\int d N}=\frac{\int v^{2} n_{v} d v}{\int n_{v} d v}=\frac{3 k T}{m}
so the rms velocity is
v_{ rms }=\sqrt{\frac{3 k T}{m}} Root mean square velocity
Differentiating n_{v} with respect to v and setting this to zero, dn_{v}∕dv = 0, gives the position of the peak of n_{v} versus v, and thus the most probable speed v^{*},
V^{*}=\left[\frac{2 k T}{m}\right]^{1 / 2} Most probable speed
Substituting m = 9.1 × 10^{−31} kg for electrons and using T = 300 K, we find v^{*} = 95.3 km s^{−1} , v_{av} = 108 km s^{−1}, {\text and} v_{rms} = 117 km s^{−1}, all of which are close in value. We often use the term thermal velocity to describe the mean speed of particles due to their thermal random motion. Also, the integrations shown above are not trivial and they involve substitution and integration by parts.