# The root-mean-square speed of a gas is found to be 391.2 m/s at 270 K. The gas is ?

Mar 6, 2015

Your gas is any gas that has a molar mass of approximately 44 g/mol. The usual suspects are carbon dioxide, or $C {O}_{2}$, and nitrous oxide, or $N {O}_{2}$. Even propane, ${C}_{3} {H}_{8}$ could fit here, but it's molar mass is closer to $\text{44.1 g/mol}$.

The mathematical expression for the root-mean-square is

${v}_{\text{rms}} = \sqrt{\frac{3 R T}{M} _ m}$, where

$R$ - the universal gas constant;
$T$ - the temperature of the gas in Kelvin;
${M}_{m}$ - the molar mass of the gas;

Two important things to keep in mind for this equation - $R$ is used in Joules per mol K and the molar mass of the gas is expressed in kg per mole, instead of in g per mole.

So, in order to identify the gas, you must determine its molar mass. Use the above equation to solve for ${M}_{m}$

${v}_{\text{rms")^2 = (3RT)/M_m => M_m = (3RT)/v_("rms}}^{2}$

Plug in your values and solve for ${M}_{m}$

M_m = (3 * 8.31446"J"/("mol" * "K") * "270 K")/(391.2^(2) "m"^2 * "s"^(-2)

${M}_{m} = 0.044 {\text{J"/"mol" * "s"^(2)/"m}}^{2}$ $\to$ rounded to two sig figs.

Use the fact that ${\text{Joule" = ("kg" * "m"^2)/"s}}^{2}$ to get

M_m = 0.044("kg" * "m"^(2))/("mol" * "s"^(2)) * "m"^(2)/s^(2) = "0.044 kg/mol"

Transform this value into a more familiar one

$0.044 \text{kg"/"mol" * "1000 g"/"1 kg" = "44 g/mol}$

Therefore, your unknown gas has a molar mass of approximately $\text{44 g/mol}$.