For an adiabatic process, the work done by the gas can be found using the formula:

$$ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} $$

Given that the volume is doubled ($$V_2 = 2V_1$$) and the adiabatic constant $$\gamma = \frac{3}{2}$$, we can manipulate the ideal gas law and the adiabatic process relationship to find an expression for work done in terms of the initial conditions and $$\gamma$$.

Recall, for an adiabatic process, $$PV^{\gamma} = \text{constant}$$, so we can write:

$$ P_1V_1^{\gamma} = P_2V_2^{\gamma} $$

Using the fact that $$V_2 = 2V_1$$, we can express $$P_2$$ in terms of $$P_1$$ and $$V_1$$ as follows:

$$ P_1V_1^{\gamma} = P_2(2V_1)^{\gamma} $$

$$ P_2 = P_1 \left( \frac{V_1}{2V_1} \right)^{\gamma} $$

$$ P_2 = P_1 \left( \frac{1}{2} \right)^{\gamma} $$

The work done then becomes:

$$ W = \frac{P_1 V_1 - P_1 \left( \frac{1}{2} \right)^{\gamma} \cdot 2V_1}{\gamma - 1} $$

$$ W = P_1V_1 \frac{1 - \left( \frac{1}{2} \right)^{\gamma} \cdot 2}{\gamma - 1} $$

Plugging in $$\gamma = \frac{3}{2}$$, we get:

$$ W = P_1V_1 \frac{1 - \left( \frac{1}{2} \right)^{\frac{3}{2}} \cdot 2}{\frac{3}{2} - 1} $$

$$ W = P_1V_1 \frac{1 - \sqrt{\frac{1}{2}} \cdot 2}{\frac{1}{2}} $$

$$ W = 2P_1V_1 (1 - \sqrt{\frac{1}{2}}) $$

Since $$P_1V_1 = nRT$$ for 1 mole ($$n = 1$$) of gas at temperature $$T$$, we can further simplify:

$$ W = 2RT(1 - \sqrt{\frac{1}{2}}) $$

$$ W = 2RT(1 - \frac{1}{\sqrt{2}}) $$

$$ W = RT(2 - \sqrt{2}) $$

Therefore, the correct option is:

Option B $$\mathrm{RT}[2-\sqrt{2}]$$.