At `127^(@)C` and `1.00 xx 10^(-2)` atm pressure, the density of a gas is `1.24 xx 10^(-2) kg m^(-3)`.
a. Find `v_(rms)` for the gas molecules.
b. Find the molecular weight of the gas and identify it.

Explanation

`p = (1)/(3) rho v_(rms)^(2) implies v_(rms) = sqrt((3 P)/(rho))`
`:. V_(rms) = sqrt((3 xx 1.00 xx 10^(-2) xx 1.013 xx 10^(5))/(1.24 xx 10^(-2)))`
`= 495 m//s`
Again, `v_(rms) = sqrt((3 RT)/(M))`
or `M = (3RT)/(v_(rms)^(2))`
`= (3 xx 8.3 xx 400)/((495)^(2)) = 40.6 g//mol`
This is the molecular weight of argon and so the gas is argon.