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Step-by-Step Solution
Step 1: Identify the Adiabatic Process
An adiabatic process is one in which no heat is exchanged with the surroundings. For an ideal gas, the adiabatic condition relates pressure and volume through
$P \, V^\gamma = \text{constant},$
where $P$ is the pressure, $V$ is the volume, and $\gamma$ is the ratio of specific heats ($\gamma = \frac{C_p}{C_v}$). Given here, $\gamma = \frac{3}{2}.$
Step 2: Establish Initial and Final States
Let the initial pressure and volume be $P_i$ and $V_i$, and the final pressure and volume be $P_f$ and $V_f$. It is given that the volume doubles, so
$V_f = 2\,V_i$.
Also, $\gamma = \frac{3}{2}.$
Step 3: Apply the Adiabatic Condition
Using $P_i \, V_i^\gamma = P_f \, V_f^\gamma$:
\[
P_i \, V_i^{\gamma} = P_f \, (2\,V_i)^\gamma.
\]
Hence,
\[
P_f = P_i \,\frac{V_i^\gamma}{(2\,V_i)^\gamma} = \frac{P_i}{2^\gamma}.
\]
Substitute $\gamma = \frac{3}{2}$:
\[
P_f = \frac{P_i}{2^{\frac{3}{2}}} = \frac{P_i}{2\sqrt{2}}.
\]
Step 4: Write the Formula for Adiabatic Work
The work done by an ideal gas in an adiabatic expansion from $(P_i, V_i)$ to $(P_f, V_f)$ is:
\[
W = \frac{P_f\,V_f - P_i\,V_i}{1 - \gamma}.
\]
Step 5: Substitute and Simplify
Using $P_f = \frac{P_i}{2\sqrt{2}}$ and $V_f = 2V_i$:
\[
W = \frac{\left(\frac{P_i}{2\sqrt{2}}\right)\,(2V_i) - P_i\,V_i}{1 - \frac{3}{2}}
= \frac{\frac{P_i\,V_i}{\sqrt{2}} - P_i\,V_i}{1 - \frac{3}{2}}.
\]
Combine terms inside the numerator:
\[
= \frac{P_i\,V_i\left(\frac{1}{\sqrt{2}} - 1\right)}{1 - \frac{3}{2}}
= \frac{P_i\,V_i\left(\frac{1}{\sqrt{2}} - 1\right)}{-\frac{1}{2}}
= -2\,P_i\,V_i\left(\frac{1}{\sqrt{2}} - 1\right).
\]
Rewriting:
\[
W = P_i\,V_i\,(2 - \sqrt{2}).
\]
Step 6: Express $P_i\,V_i$ in Terms of $R\,T$
For an ideal gas with 1 mole at the initial temperature $T$:
\[
P_i\,V_i = R\,T.
\]
Thus,
\[
W = (R\,T)\,(2 - \sqrt{2}).
\]
Therefore, the work done by the gas is:
\[
\boxed{W = R\,T\,[\,2 - \sqrt{2}\,]}.
\]
Final Answer
$\displaystyle W = R\,T\,[2 - \sqrt{2}].$
