Q. At 400 K, the root mean square (rms) speed of a gas $X$ (molecular weight = 40) is equal to the most probable speed of gas $Y\ at\ 60\ K.$ The molecular weight of the gas $Y$ is
A) 2
B) 4
C) 6
D) 8
Explanation
(b)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V_{rms} = {\ V}_{mps}$
$\ \ \ \ \ \ \ \ \ \ \ \ \sqrt{\frac{3RT}{M\ (X)}} = \sqrt{\frac{2RT'}{M\ (Y)}}$
$\Rightarrow \sqrt{\frac{3R \times 400}{40}} = \sqrt{\frac{2R \times 60}{M\ (Y)}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ M(Y) = 4\$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.