A flask contains argon and chlorine gas in the ratio 2:1 by . . .

Question : A flask contains argon and chlorine gas in the ratio 2:1 by mass. The temperature of the mixture is 27°C. Obtain the ratio of :

i) mean translational kinetic energy of molecules. 

ii) mean kinetic energy of molecules

iii) root mean square velocity (V_rms) of the molecules of two gases. 

Atomic mass of Argon = 39.9 u

Atomic mass of Chlorine = 70.9 u

Doubt by Saif 

Solution : 

Argon (Monoatomic Gas)
Chlorine (Diatomic Gas)

M_Argon : M_Chlorine = 2:1

T₁=T₂=27°C
=(27+273) = 300 K

i) We know,
Mean Translational Kinetic Energy of the molecule (Average Kinetic Energy Per Molecule of the gas)  is given by (3/2)K_BT

where K_B is Boltzmann Constant and T is absolute temperature of the gas. It does not depend on the nature of the gas whether it is monoatomic or diatomic. 

Here the temperature of both the gases is same hence the Mean Translational Kinetic Energy of the molecule of each gas would be same. So the required ratio would be 1:1.

ii) Mean Kinetic Energy of the molecules
[Here rotational Kinetic Energy is also included]

As per the Law of Equipartition of Energy, the energy associated with each degree of freedom per molecule is 1/2K_BT. 

For Argon (Monoatomic Gas)
Degree of Freedom = 3
E₁=3×(1/2)KBT
E₁=3/2K_BT

For Chlorine (Diatomic Gas)

Degree of Freedom = 5
E₂=5×(1/2)K_BT
E₂=5/2K_BT

So the required ration of E₁:E₂ is 3:5

iii) We know,