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, (given $$\gamma = \frac{3}{2}$$) :

A sample of gas (1 mole) at temperature $$T$$ is adiabatically expanded to double its volume. We need to find the work done, given $$\gamma = \frac{3}{2}$$.

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

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

Substituting $$T_1=T$$, $$V_2=2V_1$$, and $$\gamma-1=\tfrac{1}{2}$$ gives

$$T\,V_1^{1/2}=T_2\,(2V_1)^{1/2}\quad\Longrightarrow\quad T_2=T\,\frac{V_1^{1/2}}{(2V_1)^{1/2}}=\frac{T}{\sqrt{2}}\,. $$

The work done by the gas in an adiabatic expansion is

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

and with $$n=1$$, $$T_1=T$$, $$T_2=T/\sqrt{2}$$, and $$\gamma-1=\tfrac12$$ this becomes

$$W=\frac{R\bigl(T-\tfrac{T}{\sqrt{2}}\bigr)}{\tfrac12} =2RT\Bigl(1-\frac{1}{\sqrt{2}}\Bigr) =RT\,(2-\sqrt{2})\,. $$

Thus, the correct answer is Option D: $$W=RT(2-\sqrt{2})\,. $$