Question

Calculate the mean-square speed, of krypton atoms in a sample of gas at a temperature of 22ºC?

Mass of 1 mole of krypton = 0.084kg

Answer 1

`< v^2 > ~~8.8xx10^4"m"//"s"`

Explanation:

Both kinetic energy and the ideal gas law provide expressions for pressure. Those expressions must be numerically equal, so it must be that

`1/2m< v^2 > =3/2kT`

We can solve for the mean-square speed:

`< v^2 > =(3kT)/m`

where:

• `k` is Boltzmann's constant `(k_b=1.381xx10^(-23)m^2kg//s^2K)`

• `T` is the temperature of the gas (in Kelvin)

• `m` is the mass of one molecule of the gas (in kg)

Then `T=22^oC=(273+22)K=295K`

Given that one mole of the gas has `m=0.084"kg"` we can find the mass of one molecule using Avagadro's number.

`(0.084"kg")/"mole"*(1"mole")/(6.022*10^23"molecules")`

`=>m=1.395xx10^-25"kg"`

And so we have:

`< v^2 > = (3(1.381xx10^(-23)m^2kg//s^2K)(295"K"))/(1.395xx10^-25"kg")`

`=>< v^2 > ~~8.8xx10^4"m"//"s"`

Usually, you are asked for the root mean square speed, which is the square root of the above. That would give `~~296"m"//"s"`.

You may also see the equation written in terms of the universal gas constant `R` and the mass of one mole of the gas `M`. The calculation is similar.

Here is a question I have answered previously on Socratic pertaining to this expression and similar expressions using the maxwell speed distribution.