1. Use the adiabatic relation

For an adiabatic process of an ideal gas, $PV^\gamma=\text{constant}$ Also, $TV^{\gamma-1}=\text{constant}$

Given:

• number of moles, $n=1$
• initial temperature $=T$
• final volume is double the initial volume, so $V_2=2V_1$
• $\gamma=\frac{3}{2}$

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

Substitute $\gamma-1=\frac{1}{2}$: $TV_1^{1/2}=T_2(2V_1)^{1/2}$ $TV_1^{1/2}=T_2\sqrt{2}\,V_1^{1/2}$

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

2. Find work done in adiabatic expansion

For an adiabatic process, $Q=0$ By first law of thermodynamics, $\Delta U=Q-W=-W$ So, $W=-\Delta U=nC_V(T_1-T_2)$

Now use $\gamma=\frac{C_P}{C_V}, \qquad C_P-C_V=R$ Hence, $C_V=\frac{R}{\gamma-1}$ Since $\gamma-1=\frac{1}{2}$, $C_V=\frac{R}{1/2}=2R$

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

3. Match with options

The work done is $\boxed{RT(2-\sqrt{2})}$ This corresponds to Option B.

4. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So the derived answer agrees with the stored answer.