91Ó°ÊÓ

A formula connects the temperature to the root average square velocity.

$v_{rms}=\sqrt{\frac{3k_{B}T}{m}}$

where $m=\frac{M}{N_A}$The mass of a single molecule is The Boltzmann constant may be observed as a result., and the temperature $t$remains constant. Let As a result, we'll set those constants aside and see what happens.

$3k_{B}T=\frac{v_{\mathrm{Ï„}ms}^2}{m}$

The constant to the left will be the same for both gases. As a result, a link between the masses of neon and argon atoms may be established. their $rms$speeds as

$\frac{v_{rms,Ag}^2}{m_{Ag}}=\frac{v_{rms,Ne}^2}{m_{Ne}}$

The square root and express the $rms$speed of the argon atoms as

$v_{rms,Ag}=\sqrt{\frac{m_{Ne}}{m_{Ag}}}{v}_{rms,Ne}$

Remembering how $m$is given, we can rearrange our expression to

$v_{\mathrm{Ï„}ms,Ag}=\sqrt{\frac{M_{Ne}}{M_{Ag}}{v}_{rms,Ne}}$

Numerically, remembering that the molar mass of argon is $40g/mol$and that of neon is $20g/mol$, we have

$v_{rms,Ag}=400\sqrt{\frac{20}{40}}$

$=283m/s$

The $rms$ velocity of argon ions is$283m/s$.