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

1 Answer
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 CO_2, and nitrous oxide, or NO_2. Even propane, C_3H_8 could fit here, but it's molar mass is closer to "44.1 g/mol".

The mathematical expression for the root-mean-square is

v_("rms") = sqrt((3RT)/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_("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 "J"/"mol" * "s"^(2)/"m"^(2) -> rounded to two sig figs.

Use the fact that "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"kg"/"mol" * "1000 g"/"1 kg" = "44 g/mol"

Therefore, your unknown gas has a molar mass of approximately "44 g/mol".