A sample of 1 mole gas at temperature T is adiabatically expanded to double its volume. If adiabatic constant for the gas is $$\gamma = \frac{3}{2}$$, then the work done by the gas in the process is:

For an adiabatic process with $$n = 1$$ mole, $$\gamma = \frac{3}{2}$$, initial temperature $$T$$, volume doubles.

For an adiabatic process: $$TV^{\gamma-1} = \text{constant}$$.

$$T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$$

$$T \cdot V^{1/2} = T_2 \cdot (2V)^{1/2}$$

$$T_2 = \frac{T}{\sqrt{2}} = \frac{T\sqrt{2}}{2}$$

Work done in an adiabatic process:

$$W = \frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{1 \cdot R\left(T - \frac{T}{\sqrt{2}}\right)}{\frac{1}{2}} = 2R\left(T - \frac{T}{\sqrt{2}}\right)$$

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

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

The correct answer is Option 3: $$RT[2 - \sqrt{2}]$$.