A sample of gas at temperature $T$ is adiabatically expanded to double its volume. The work done by the gas in the process is $\left(\mathrm{given}, \gamma=\frac{3}{2}\right)$ :
Solution
$\gamma=\frac{3}{2}$
<br/><br/>$$
\begin{aligned}
& W =\frac{n R \Delta T}{1-\gamma}=\frac{n R T_{f}-n R T_{i}}{1-\gamma} \\\\
& =\frac{(P V)_{f}-\left(P V_{i}\right)}{1-\gamma} \quad \ldots \text { (1) } \\\\
& P V^{\gamma}=\text { constant } \\\\
& P_{i} V_{i}^{\gamma}=P_{f}\left(2 V_{i}\right)^{\gamma} \Rightarrow P_{f}=\frac{P_{i}}{2^{\gamma}}=\frac{P_{i}}{2 \sqrt{2}} ......(2)
\end{aligned}
$$
<br/><br/>From (1) and (2)
<br/><br/>$$
\begin{aligned}
& W=\frac{\frac{P_{i}}{2 \sqrt{2}} 2 V_{i}-P_{i} V_{i}}{1-\gamma}=\frac{P_{i} V_{i}}{-1 / 2}\left(\frac{1}{\sqrt{2}}-1\right) \\\\
& =-n R T(\sqrt{2}-2) \\\\
& =n R T(2-\sqrt{2})
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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