Question

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:

Answer 1

$V_{{rms}_{\left(Xgas\right)\left(400K\right)}}=V_{{mp}_{\left(Ygas\right)\left(60K\right)}}$
M.W. ($X$ gas) $=$ 40; M.W. ($Y$ gas) $=$ $x$
$\sqrt{\frac{3R{T}_1}{M_1}}=\sqrt{\frac{2R{T}_2}{M_2}}$

$\sqrt{\frac{400\times 3}{40}}=\sqrt{\frac{2\times 60}{x}}$
$30=\frac{120}{x}$
$x=4$