At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at – 20 °C? (atomic mass of Ar = 39.9 u, of He = 4.0 u).
Solution 1
Temperature of the helium atom, THe = –20°C= 253 K
Atomic mass of argon, MAr = 39.9 u
Atomic mass of helium, MHe = 4.0 u
Let, (vrms)Ar be the rms speed of argon.
Let (vrms)He be the rms speed of helium.
The rms speed of argon is given by:
`(v_"rms")_"Ar" = sqrt((3RT_"Ar")/M_"Ar")` .... (1)
Where,
R is the universal gas constant
TAr is temperature of argon gas
The rms speed of helium is given by:
`(v_"rms")_"He" = sqrt((3RT_"He")/M_"He")` ... (ii)
It is given that
(vrms)Ar = (vrms)He
`sqrt((3RT_"Ar")/ M_"Ar")` = `sqrt((3RT_"He")/M_"He")`
`T_"Ar"/M_"Ar" = T_"He"/M_"He"`
`T_"Ar" = T_"He"/M_"He" xx M_"Ar"`
`= 253/4 xx 39.9`
= 2523.675 = 2.52 × 103 K
Therefore, the temperature of the argon atom is 2.52 × 103 K.
Solution 2
Let C and C’ be the rms velocity of argon and a helium gas atoms at temperature T K and T K respectively
Here, M = 39.9; M’ = 4.0; T =?; T = -20 + 273 = 253 K
Now, `C = sqrt((3RT)/M
)= sqrt((3RT)/39.9)` and `C' = sqrt((3RT')/M') = sqrt(3R xx 253)/4`
Since C =C'
Therefore `sqrt((3RT)/39.9) = sqrt(3Rxx253)/4` or `T = (39.9 xx 253)/4 = 2523.7 K`
