Question

N molecules, each of mass m, of gas A and 2 N molecules, each of mass 2 m, of gas B are contained in the same vessel which is maintained at a temperature T. The mean square of the velocity of molecules of B type is denoted by ${\nu }^2$ and the mean square of the X component of the velocity of A type is denoted by ${\omega }^2,{\omega }^2/{\nu }^2$ is

• A. 2
• B. 1
• C. 1/3
• D. 2/3

Answer 1

Answer: (D)

Total K.E. of A type molecules $=\frac{3}{2}m{\omega }^2$
Total K.E. of A type of molecule is
$K.E_A=\frac{1}{2}m\left[{\left({\nu }_{r.ms}\right)}_x^2+{\left({\nu }_{r.ms}\right)}_y^2+{\left({\nu }_{r.ms}\right)}_z^2\right]$
But ${\left({\nu }_{r.ms}\right)}_x=\omega$
So ${\left({\nu }_{r.ms}\right)}_y={\left({\nu }_{r.ms}\right)}_z=\omega$
Total K.E. of B type molecules
$=\frac{1}{2}\times 2m{\nu }^2=m{\nu }^2$
Now, $\frac{3}{2}\times m{\omega }^2=m{\nu }^2\left({\omega }^2/{\nu }^2\right)=\frac{2}{3}$