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

Velocity of gaseous molecules is increased by increasing temeprature.
Velocity of gaseous molecules is decreased by increasing molecular mass of gas.
$v_{r m s}$ of $X=\sqrt{\frac{3 R T_x}{M_x}} ; v_{m p}$ of $Y=\sqrt{\frac{2 R T_y}{M_y}}$
Given $v_{r m s}=v_{m p}$
$\sqrt{\frac{3 R T_x}{M_x}}=\sqrt{\frac{2 R T_y}{M_y}}$
$M_y=\frac{2 R T_y M_x}{3 R T_x}=\frac{2 \times 60 \times 40}{3 \times 400}=4\,$