Question 13.9:
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.
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 KAtomic 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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