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
$K.E_{A} = \frac{1}{2} m \left[\left(\nu_{r.ms}\right)^{2}_{x} + \left(\nu_{r.ms}\right)^{2}_{y} + \left(\nu_{r.ms}\right)^{2}_{z}\right]$
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} \times2 m \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} $