Question

If the rms speed of nitrogen molecules is $490m{s}^{-1}$ at 273K, find the rms speed of hydrogen molecules at the same temperature -

• A. 1580 m/s
• B. 1830 m/s
• C. 1700 m/s
• D. 2150 m/s

Answer 1

The correct option is B

1830 m/s

The molecular weight of nitrogen $\left(N_2\right)is28gmo{l}^{-1}$ and that of hydrogen $\left(H_2\right)is2gmo{l}^{-1}$. Let $m_1$ and $m_2$ be the masses, $v_1$ and $v_2$ be the rms speeds of a nitrogen molecule and a hydrogen molecule respectively. Then $m_1=14m_2$. Since the kinetic energies of the hydrogen and nitrogen molecules will be the same at the same temperature -

$\frac{1}{2}{m}_1{v}_1^2=\frac{1}{2}{m}_2{v}_2^2$

or, $v_2=v_1\sqrt{\frac{m_1}{m_2}}=490m{s}^{-1}\times \sqrt{14}=1830m{s}^{-1}$.