Question

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?

Answer 1

The `rms` 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

`1/2 mv^2 = 3/2 kT`

where `k` is the Boltzmann constant,

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

`v_(rms) = sqrt ((3kT)/m) = sqrt((3RT)/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 `(6.02xx10^(23))`

Inserting values gives

`v_(rms) = sqrt((3(8.314)(273))/M) = sqrt(6809/M)`

Using `M=0.032` kg we get, for oxygen, `v_(rms)=461 m/s`

and using 0.028 kg for nitrogen `v_(rms)=493 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.