1. Given data

• Number of moles: $n=1$
• Initial temperature: $T_1=T$
• Adiabatic expansion to double volume: $V_2=2V_1$
• Adiabatic index: $\gamma=\dfrac{3}{2}$

We need the work done by the gas in this adiabatic process.

2. Use adiabatic relation between $T$ and $V$

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

So, $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$

Substitute $\gamma=\dfrac{3}{2}$: $\gamma-1=\frac{1}{2}$

Hence, $T_1V_1^{1/2}=T_2V_2^{1/2}$

Since $V_2=2V_1$, $T\,V_1^{1/2}=T_2(2V_1)^{1/2}$ $T\,V_1^{1/2}=T_2\sqrt{2}\,V_1^{1/2}$

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

3. Apply first law for adiabatic process

For adiabatic expansion: $Q=0$

From first law: $\Delta U = Q-W$ $\Delta U = -W$

So, $W=-\Delta U=nC_V(T_1-T_2)$

Now, $\gamma=\frac{C_P}{C_V}, \qquad C_P-C_V=R$

Using $\gamma=\frac{3}{2}$ we get $\frac{C_P}{C_V}=\frac{3}{2}, \qquad C_P=C_V+R$

So, $\frac{C_V+R}{C_V}=\frac{3}{2}$ $1+\frac{R}{C_V}=\frac{3}{2}$ $\frac{R}{C_V}=\frac{1}{2}$ $C_V=2R$

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

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

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

4. Match with options

The work done by the gas is $\boxed{RT(2-\sqrt{2})}$

This corresponds to Option B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.