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, T_He = –20°C= 253 K
Atomic mass of argon, M_Ar = 39.9 u
Atomic mass of helium, M_He = 4.0 u
Let, (v_rms)_Ar be the rms speed of argon.
Let (v_rms)_He be the rms speed of helium.
The rms speed of argon is given by:
(v_rms)_Ar = (3RT_Ar / M_Ar)^1/2  … (i)
Where,
R is the universal gas constant
T_Ar is temperature of argon gas
The rms speed of helium is given by:
(v_rms)_He = (3RT_He / M_He)^1/2     .....(ii)
It is given that:
(v_rms)_Ar = (v_rms)_He
(3RT_Ar / M_Ar)^1/2 = (3RT_He/M_He)^1/2
T_Ar / M_Ar = T_He / M_He
T_Ar = ( T_He / M_He) X M_Ar
= (253 / 4) X 39.9
= 2523.675 = 2.52 × 10³ K
Therefore, the temperature of the argon atom is 2.52 × 10³ K.