Kinetic Theory NCERT Solutions Class 11 Physics - Solved Exercise Question 13.9

Question 13.9:
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:
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:
(vrms)Ar = (3RTAr / MAr)1/2  … (i)
Where,
R is the universal gas constant
TAr is temperature of argon gas
The rms speed of helium is given by:
(vrms)He = (3RTHe / MHe)1/2     .....(ii)
It is given that:
(vrms)Ar = (vrms)He
(3RTAr / MAr)1/2 = (3RTHe/MHe)1/2
TAr / MAr = THe / MHe
TAr = ( THe / MHe) X MAr
= (253 / 4) X 39.9
= 2523.675 = 2.52 × 103 K
Therefore, the temperature of the argon atom is 2.52 × 103 K.

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