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"#.