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 vessel which is maintained at a temperature T. The mean square of velocity of the molecules of B type is denoted by $v^2$ and the mean square of the x-component of the velocity of A type is denoted by $w^2$. The ratio of $w^2:v^2$ is?

• A. $3:2$
• B. $1:3$
• C. $2:3$
• D. $1:1$

Answer 1

The correct option is C $2:3$

The mean square velocity of gas molecules is given by $v^2=\frac{3kT}{m}$

For gas A, $v_A^2=\frac{3kT}{m}$ (i)

For a gas molecule, $v^2+v_x^2+v_y^2+v_z^2=3v_x^2$ $(\because v_x^2=v_y^2=v_z^2)$ or $v_x^2=\frac{v^3}{3}$

From eqn, (i), we get

$w^2=v_x^2=\left[\frac{\frac{3kT}{m}}{3}\right]=\frac{kT}{m}$ (ii)

For gas B, $v_B^2=v^2\frac{3kT}{2m}$ ..(iii)

Dividing eqn. (ii) by eqn. (iii), we get

$\frac{w^2}{v^2}=\frac{\frac{kT}{m}}{\frac{3kT}{2m}}=\frac{2}{3}$