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A sample of gas at temperature $\mathrm{T}$ is adiabatically expanded to double its volume. The work done by the gas in the process is $\left(\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\gamma=\frac{3}{2}\right)(\mathrm{R}=$ gas constant $)$
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Verified Answer
The correct answer is:
$\mathrm{RT}(2-\sqrt{2})$
Using the formula for adiabatic expansion,
$\begin{aligned}
& \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\
& \mathrm{~T}_1 \mathrm{~V}_1^{(1 / 2)}=\mathrm{T}_2\left(2 \mathrm{~V}_1^{1 / 2}\right) \quad \ldots .\left(\text { Given: } \gamma=\frac{3}{2}\right) \\
\therefore \quad & \mathrm{T}_1=\mathrm{T}_2(\sqrt{2}) \\
\therefore \quad & \mathrm{T}_2=\frac{\mathrm{T}}{\sqrt{2}}
\end{aligned}$
Work done
$\begin{aligned}
\mathrm{W}_{\mathrm{adi}}= & \frac{\mathrm{R}\left(\mathrm{T}-\mathrm{T}_2\right)}{\gamma-1}=\frac{\mathrm{R}\left(\mathrm{T}-\frac{\mathrm{T}}{\sqrt{2}}\right)}{\frac{1}{2}} \\
& =\frac{\mathrm{R}(\sqrt{2} \mathrm{~T}-\mathrm{T})}{\sqrt{2}} \times 2=\mathrm{RT}(\sqrt{2}-1) \sqrt{2} \\
\therefore \quad \mathrm{W}_{\text {adi }} & =\mathrm{RT}(2-\sqrt{2})
\end{aligned}$
$\begin{aligned}
& \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\
& \mathrm{~T}_1 \mathrm{~V}_1^{(1 / 2)}=\mathrm{T}_2\left(2 \mathrm{~V}_1^{1 / 2}\right) \quad \ldots .\left(\text { Given: } \gamma=\frac{3}{2}\right) \\
\therefore \quad & \mathrm{T}_1=\mathrm{T}_2(\sqrt{2}) \\
\therefore \quad & \mathrm{T}_2=\frac{\mathrm{T}}{\sqrt{2}}
\end{aligned}$
Work done
$\begin{aligned}
\mathrm{W}_{\mathrm{adi}}= & \frac{\mathrm{R}\left(\mathrm{T}-\mathrm{T}_2\right)}{\gamma-1}=\frac{\mathrm{R}\left(\mathrm{T}-\frac{\mathrm{T}}{\sqrt{2}}\right)}{\frac{1}{2}} \\
& =\frac{\mathrm{R}(\sqrt{2} \mathrm{~T}-\mathrm{T})}{\sqrt{2}} \times 2=\mathrm{RT}(\sqrt{2}-1) \sqrt{2} \\
\therefore \quad \mathrm{W}_{\text {adi }} & =\mathrm{RT}(2-\sqrt{2})
\end{aligned}$
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