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).

Options

A. `2.52xx10^(3)K`
B. `2.52xx10^(2)K`
C. `4.03xx10^(3)K`
D. `4.03xx10^(2)K`

Answer

A

Explanation

Let 1 and 2 represent for Argon atom and Helium atom.
rms speed of Argon, `v_(rms_(1))=sqrt((3RT_(1))/(M_(1)))`
rms speed of Helium `v_(rms_(2))=sqrt((3RT_(2))/(M_(2)))`
According to question,
`v_(rms_(1))=v_(rms_(2))`
`thereforesqrt((3RT_(1))/(M_(1)))=sqrt((3RT_(2))/(M_(2))),(T_(1))/(M_(1))=(T_(2))/(M_(2))`
`or T_(1)(T_(2))/(M_(2))xxM_(1)=(253)/(4)xx39.9=2.52xx10^(3)K`