Calculate the rms speed of an oxygen gas molecule, O_{2}, at 21.0 °C.

Express your answer numerically in meters per second.

In a given sample of gas, the particles move at varying speeds. The root mean square speed (rms speed) of particles in a gas sample, u, is given by the formula

$\mathbf{u}\mathbf{=}\sqrt{\frac{\mathbf{3}\mathbf{RT}}{\mathbf{M}}}$

where *T* is the Kelvin temperature, *M* is the molar mass in kg/mol, and *R* = 8.314 J/(mol ⋅ K) is the gas constant. Effusion is the escape of gas molecules through a tiny hole into a vacuum.

The rate of effusion of a gas is directly related to the rms speed of the gas molecules, so it's inversely proportional to the square root of its mass. The rms speed is related to kinetic energy, rather than average speed, and is the speed of a molecule possessing a kinetic energy identical to the average kinetic energy of the sample.

Given its relationship to the mass of the molecule, you can conclude that the lighter the molecules of the gas, the more rapidly it effuses. Mathematically, this can be expressed as

$\mathbf{effusion}\mathbf{}\mathbf{rate}\mathbf{\propto}\frac{\mathbf{1}}{\sqrt{\mathbf{m}}}$

The relative rate of effusion can be expressed in terms of molecular masses *m*_{A} and *m*_{B} as

$\frac{\mathbf{rate}\mathbf{}\mathbf{of}\mathbf{}\mathbf{gas}\mathbf{}\mathbf{A}\mathbf{}\mathbf{effusion}}{\mathbf{rate}\mathbf{}\mathbf{of}\mathbf{}\mathbf{gas}\mathbf{}\mathbf{B}\mathbf{}\mathbf{effusion}}\mathbf{=}\sqrt{\frac{{\mathbf{m}}_{\mathbf{B}}}{{\mathbf{m}}_{\mathbf{A}}}}$

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Root Mean Square Speed concept. You can view video lessons to learn Root Mean Square Speed. Or if you need more Root Mean Square Speed practice, you can also practice Root Mean Square Speed practice problems.