# Air is approximately 21% O_2 and 78% N_2 by mass. What is the root-mean-square speed of each gas at 273 K?

Dec 28, 2017

The $r m s$ speed for oxygen is 461 m/s and for nitrogen, 493 m/s

#### Explanation:

The relation between molecular kinetic energy and temperature is given by the equation

$\frac{1}{2} m {v}^{2} = \frac{3}{2} k T$

where $k$ is the Boltzmann constant,

Solving this for $v$ gives us the root mean square speed

${v}_{r m s} = \sqrt{\frac{3 k T}{m}} = \sqrt{\frac{3 R T}{M}}$

where the final relation is given in terms of the gas constant $R$ and the molar mass of a gas. (It basically amounts to multiplying top and bottom of the middle relation by the Avogadro constant $\left(6.02 \times {10}^{23}\right)$

Inserting values gives

${v}_{r m s} = \sqrt{\frac{3 \left(8.314\right) \left(273\right)}{M}} = \sqrt{\frac{6809}{M}}$

Using $M = 0.032$ kg we get, for oxygen, ${v}_{r m s} = 461 \frac{m}{s}$

and using 0.028 kg for nitrogen ${v}_{r m s} = 493 \frac{m}{s}$

Note that the molar mass had to be changed to kg per mole in order to be consistent with the other units in the equation.