Molecular Weight of Gas from RMS and Most Probable Speeds

To find the molecular weight of gas $\mathbf{Y}$, we use the formulas for the root mean square (rms) speed and the most probable speed derived from the kinetic theory of gases:

1. Root Mean Square Speed ($v_{\text{rms}}$):
   $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$
   For gas $\mathbf{X}$ at temperature $T_{\mathbf{X}} = 400\text{ K}$ and molecular weight $M_{\mathbf{X}} = 40\text{ g/mol}$:
   $v_{\text{rms}, \mathbf{X}} = \sqrt{\frac{3R \times 400}{40}}$

2. Most Probable Speed ($v_{\text{mp}}$):
   $v_{\text{mp}} = \sqrt{\frac{2RT}{M}}$
   For gas $\mathbf{Y}$ at temperature $T_{\mathbf{Y}} = 60\text{ K}$ and molecular weight $M_{\mathbf{Y}}$:
   $v_{\text{mp}, \mathbf{Y}} = \sqrt{\frac{2R \times 60}{M_{\mathbf{Y}}}}$

According to the given condition:
$v_{\text{rms}, \mathbf{X}} = v_{\text{mp}, \mathbf{Y}}$

Equating the two expressions:
$\sqrt{\frac{3R \times 400}{40}} = \sqrt{\frac{2R \times 60}{M_{\mathbf{Y}}}}$

Squaring both sides:
$\frac{3R \times 400}{40} = \frac{2R \times 60}{M_{\mathbf{Y}}}$

Canceling $R$ from both sides and simplifying:
$3 \times 10 = \frac{120}{M_{\mathbf{Y}}}$
$30 = \frac{120}{M_{\mathbf{Y}}}$
$M_{\mathbf{Y}} = \frac{120}{30} = 4$

Thus, the molecular weight of gas $\mathbf{Y}$ is $4$.