Problem — Speed distribution: hydrogen or oxygen

Problem

A graph shows $f(v)$ for oxygen and hydrogen at the same temperature. One curve has a higher, narrower peak $v_p$, the other a lower peak shifted to the right. Which curve is which, and why?

Solution

At the same temperature the molecules have the same mean kinetic energy, but different speeds because they have different masses. The most probable speed of the Maxwell-Boltzmann distribution is $v_p=\sqrt{\frac{2k_B T}{m}}\quad\Longrightarrow\quad v_p\propto\frac{1}{\sqrt m}$ Hydrogen $H_2$ has molar mass $2$ g/mol, oxygen $O_2$ has $32$ g/mol: the $O_2$ molecule is 16 times heavier. So $\frac{v_{p,\,H_2}}{v_{p,\,O_2}}=\sqrt{\frac{m_{O_2}}{m_{H_2}}}=\sqrt{16}=4$ The hydrogen peak falls at a speed four times greater: its curve is the one shifted to the right.

Why it is wider and lower. Each distribution is normalised: the area under the curve is always 1 (it represents all the molecules). $\int_0^{\infty} f(v)\,\dd v = 1$ If the hydrogen curve spreads over a wider range of speeds (it is wider), to keep the area equal to one it must necessarily be lower. Oxygen, with its peak at low speeds, has a narrow, and therefore tall, distribution.

Conclusion. The narrow, tall curve with the peak on the left is oxygen; the wide, low curve with the peak shifted to the right is hydrogen. $\ev{\text{wide/low/peak on the right} = H_2;\quad \text{narrow/tall/peak on the left} = O_2}$

Links

Topics: Kinetic theory of gases Concepts: Maxwell-Boltzmann distribution · Root-mean-square speed Skills: Reading graphs