N molecules each of mass m of gas A and 2N mo | vedclass

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(D) The mean square velocity of a gas molecule is given by $v^2 = \frac{3kT}{m}$.For gas $A$,the mean square velocity is $v_A^2 = \frac{3kT}{m}$.Since

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$\sqrt{\frac{2}{3}}$

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(D) The mean square velocity of a gas molecule is given by $v^2 = \frac{3kT}{m}$.For gas $A$,the mean square velocity is $v_A^2 = \frac{3kT}{m}$.Since $v^2 = v_x^2 + v_y^2 + v_z^2$ and due to symmetry $v_x^2 = v_y^2 = v_z^2$,we have $v_x^2 = \frac{v^2}{3}$.Thus,$V_1^2 = v_{Ax}^2 = \frac{v_A^2}{3} = \frac{3kT}{3m} = \frac{kT}{m}$.Therefore,$V_1 = \sqrt{\frac{kT}{m}}$. $(1)$For gas $B$,the mean square velocity is $V_2^2 = \frac{3kT}{2m}$.Therefore,$V_2 = \sqrt{\frac{3kT}{2m}}$. $(2)$Dividing equation $(1)$ by $(2)$,we get:$\frac{V_1}{V_2} = \frac{\sqrt{kT/m}}{\sqrt{3kT/2m}} = \sqrt{\frac{kT}{m} \cdot \frac{2m}{3kT}} = \sqrt{\frac{2}{3}}$.

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