Let's start by writing down the equations for the root mean square (rms) speed and the most probable speed. The root mean square speed $$ v_{rms} $$ of a gas with molecular weight $$ M $$ at a temperature $$ T $$ in Kelvin is given by the formula:

$$ v_{rms} = \sqrt{\frac{3kT}{M}} $$

where $$ k $$ is the Boltzmann constant.

The most probable speed $$ v_{mp} $$ of a gas is given by:

$$ v_{mp} = \sqrt{\frac{2kT}{M}} $$

According to the problem statement, at 400 K, the $$ v_{rms} $$ speed of gas X (with molecular weight $$ M_X = 40 $$ g/mol) is equal to the $$ v_{mp} $$ of gas Y at 60 K. We set the equations equal to each other:

$$ \sqrt{\frac{3k \times 400}{40}} = \sqrt{\frac{2k \times 60}{M_Y}} $$

We simplify this equation. First, we can cancel $$ k $$ from both sides:

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

Simplify further:

$$ \sqrt{\frac{1200}{40}} = \sqrt{\frac{120}{M_Y}} $$

$$ \sqrt{30} = \sqrt{\frac{120}{M_Y}} $$

Squaring both sides gives:

$$ 30 = \frac{120}{M_Y} $$

Rearrange to solve for $$ M_Y $$:

$$ M_Y = \frac{120}{30} = 4 $$

So, the molecular weight of gas Y is 4 g/mol.