At what temperature is the root mean square speed of a an atom is an argon gas cylinder equal to the rms speed of helium gas at `- 20^(@)C` ? (atomic mass of Ar = 39.9 u and of He = 4.0u)

Explanation

We know that `V_(mn)=((-2)/(v))^(1/2)`
`V_(mn)prop sqrt((T)/(m)`
and m `prop` where M is molar mass
`therefore " " (T_(1))/(M_(1))=(T_(1))/(M_(2)), " " T_(1)=273-20=253K`
`or " " T_(2)=(T_(2))=(T_(1))((M_(2))/(M_(1)))=253((39.9)/(4))`
`T_(2)=2523.6K`
`or " " t_(2)=2250.6^(@)C.`