At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the r.m.s. speed of a helium gas atom at `-20^(@) C` ? (Atomic mass of Ar = 39.9 u, of He = 4.0 u).

Explanation

Temperature of the helium atom, `T_He=-20^(@) C=253 K`
Atomic mass of argon,` M_(Ar) = 39.9 u`
Atomic mass of helium, `M_(He) = 4.0 u`
Let, `(v_(rms))`.Ar be the rms speed of argon.
Let `(v_(rms))`He be the rms speed of helium.
The rms speed of argon is given by: `(v_(rms))_(Ar)=sqrt((3RT_Ar)/(M_(Ar))`
Where,
R is the universal gas constant `T_(Ar)` is temperature of argon gas
The rms speed of helium is given by:
`v_(rms)_He=sqrt((3RT_(He))/(M_(He))`
It is given that: `=sqrt((3RT_(Ar))/(M_(Ar)))=sqrt((3RT_(He))/(M_(He)))`
` (T_(Ar))/(M_Ar)=(T_(He))/(M_He)`
`T_(Ar)=(T_(He))/(M_(He))xxM_(Ar)`
`=253/4xx39.9`
` = 2523.675 = 2.52 xx 10^3 K `
Therefore, the temperature of the argon atom is `2.52 xx 10^3 K`.