Question

At $400\text{K}$, the root mean square (rms) speed of a gas $X$ (molecular mass = 40) is equal to the most probable speed of gas $Y$ at $60\text{K}$. The molecular mass (in amu) of the gas $Y$ is:

Answer 1

According to question,
For X, $u_{\text{rms}}=\sqrt{\frac{3R{T}_X}{M_X}}$
For Y, $u_{\text{mp}}=\sqrt{\frac{2R{T}_Y}{M_Y}}$

$\sqrt{\frac{3R{T}_X}{M_X}}$ = $\sqrt{\frac{2R{T}_Y}{M_Y}}$

$\sqrt{\frac{3R\times 400}{40}}=\sqrt{\frac{2R\times 60}{M_Y}}$
$M_Y=\frac{2\times 60\times 40}{3\times 400}=4amu$