A flask contains argon and oxygen in the ratio of 3: 2 in mass and the mixture is kept at 27° C. The ratio of their average kinetic energy per molecule respectively will be :

Question

A flask contains argon and oxygen in the ratio of 3: 2 in mass and the mixture is kept at 27° C. The ratio of their average kinetic energy per molecule respectively will be : (A) 3: 2 (B) 9: 4 (C) 2: 3 (D) 1: 1

Tardigrade Admin · Accepted Answer

Average K.E./molecule =( f /2) kT So, ( K Ar / K O 2)=((3/2) KT /(5)2 KT )=(3/5) If we consider only translational kinetic energy of a molecule, then ( K Ar / K O 2)=((3/2) kT /(3)2 kT )=(1/1)

Q. A flask contains argon and oxygen in the ratio of $3 : 2$ in mass and the mixture is kept at $2 7^{\circ} C$ . The ratio of their average kinetic energy per molecule respectively will be :

$3 : 2$

$9 : 4$

$2 : 3$

$1 : 1$

Average K.E./molecule $= \frac{f}{2} k T$
So, $\frac{K _{A r}}{K _{O_{2}}} = \frac{\frac{3}{2} K T}{\frac{5}{2} K T} = \frac{3}{5}$
If we consider only translational kinetic energy of a molecule, then
$\frac{K _{A r}}{K _{O_{2}}} = \frac{\frac{3}{2} k T}{\frac{3}{2} k T} = \frac{1}{1}$

Q. A flask contains argon and oxygen in the ratio of 3:2 in mass and the mixture is kept at 27∘C. The ratio of their average kinetic energy per molecule respectively will be :

3:2

9:4

2:3

1:1