A sample of gas at temperature $$T$$ is adiabatically expanded to double its volume. The work done by the gas in the process is, (given $$\gamma = \frac{3}{2}$$):

We have a gas at temperature $$T$$ that is adiabatically expanded to double its volume, with $$\gamma = \frac{3}{2}$$.

For an adiabatic process, $$TV^{\gamma - 1} = \text{constant}$$. With $$\gamma - 1 = \frac{1}{2}$$:

$$T_1 V_1^{1/2} = T_2 V_2^{1/2}$$

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

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

Now the work done by the gas in an adiabatic process (for 1 mole) is

$$W = \frac{R(T_1 - T_2)}{\gamma - 1}$$

Substituting:

$$W = \frac{R\left(T - \frac{T}{\sqrt{2}}\right)}{\frac{1}{2}} = 2RT\left(1 - \frac{1}{\sqrt{2}}\right) = 2RT \cdot \frac{\sqrt{2} - 1}{\sqrt{2}}$$

Simplifying (by multiplying numerator and denominator by $$\sqrt{2}$$):

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

Hence, the correct answer is $$W = RT(2 - \sqrt{2})$$.