# 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

Dec 21, 2017

$< {v}^{2} > \approx 8.8 \times {10}^{4} \text{m"//"s}$

#### Explanation:

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

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

We can solve for the mean-square speed:

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

where:

• $k$ is Boltzmann's constant $\left({k}_{b} = 1.381 \times {10}^{- 23} {m}^{2} k g / {s}^{2} K\right)$

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

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

Then $T = {22}^{o} C = \left(273 + 22\right) K = 295 K$

Given that one mole of the gas has $m = 0.084 \text{kg}$ we can find the mass of one molecule using Avagadro's number.

$\left(0.084 \text{kg")/"mole"*(1"mole")/(6.022*10^23"molecules}\right)$

$\implies m = 1.395 \times {10}^{-} 25 \text{kg}$

And so we have:

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

$\implies < {v}^{2} > \approx 8.8 \times {10}^{4} \text{m"//"s}$

Usually, you are asked for the root mean square speed, which is the square root of the above. That would give $\approx 296 \text{m"//"s}$.

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.