The Distribution of Molecular Speeds — Question 1

Question 1

Normalizing a speed density fixes both its coefficient and its units

For a classical equilibrium gas, write the speed density as $f(v)=Cv^2e^{-av^2}$ for $v\ge0$, where $a=m/(2k_BT)$. You may use $I_2=\int_0^\infty v^2e^{-av^2}dv=\sqrt\pi/(4a^{3/2})$ and $I_4=3\sqrt\pi/(8a^{5/2})$.

1. Find $C$ by normalization and show that $\langle v^2\rangle=3/(2a)$.

2. Determine the units of $a$, $C$ and $f$. Does the condition $f(0)=0$ imply that no molecule can have an arbitrarily small positive speed?

Question 1 - Solution

Normalizing a speed density fixes both its coefficient and its units

(a) Method and calculation

$$1=CI_2\ \Rightarrow\ C=\frac{4a^{3/2}}{\sqrt\pi}.$$$$\langle v^2\rangle=CI_4=\frac{4a^{3/2}}{\sqrt\pi}\frac{3\sqrt\pi}{8a^{5/2}}=\frac3{2a}=\frac{3k_BT}{m}.$$

(b) Interpretation and further calculation

$$[a]=\mathrm{s^2/m^2},\quad[C]=\mathrm{s^3/m^3},\quad[f]=\mathrm{s/m}.$$$$P(0<v<\epsilon)=\int_0^\epsilon f(v)dv\simeq\frac{C\epsilon^3}{3}>0\quad(a\epsilon^2\ll1).$$A density evaluated at a single speed is not a finite-interval probability.

Physical check

Normalization makes the total area unity, independently of the plotted peak height.