JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 13)
A sample of gas at temperature $$T$$ is adiabatically expanded to double its volume. The work done by the gas in the process is $$\left(\mathrm{given}, \gamma=\frac{3}{2}\right)$$ :
$$W=T R[\sqrt{2}-2]$$
$$W=\frac{T}{R}[\sqrt{2}-2]$$
$$W=\frac{R}{T}[2-\sqrt{2}]$$
$$W=R T[2-\sqrt{2}]$$
Explanation
$\gamma=\frac{3}{2}$
$$ \begin{aligned} & W =\frac{n R \Delta T}{1-\gamma}=\frac{n R T_{f}-n R T_{i}}{1-\gamma} \\\\ & =\frac{(P V)_{f}-\left(P V_{i}\right)}{1-\gamma} \quad \ldots \text { (1) } \\\\ & P V^{\gamma}=\text { constant } \\\\ & P_{i} V_{i}^{\gamma}=P_{f}\left(2 V_{i}\right)^{\gamma} \Rightarrow P_{f}=\frac{P_{i}}{2^{\gamma}}=\frac{P_{i}}{2 \sqrt{2}} ......(2) \end{aligned} $$
From (1) and (2)
$$ \begin{aligned} & W=\frac{\frac{P_{i}}{2 \sqrt{2}} 2 V_{i}-P_{i} V_{i}}{1-\gamma}=\frac{P_{i} V_{i}}{-1 / 2}\left(\frac{1}{\sqrt{2}}-1\right) \\\\ & =-n R T(\sqrt{2}-2) \\\\ & =n R T(2-\sqrt{2}) \end{aligned} $$
$$ \begin{aligned} & W =\frac{n R \Delta T}{1-\gamma}=\frac{n R T_{f}-n R T_{i}}{1-\gamma} \\\\ & =\frac{(P V)_{f}-\left(P V_{i}\right)}{1-\gamma} \quad \ldots \text { (1) } \\\\ & P V^{\gamma}=\text { constant } \\\\ & P_{i} V_{i}^{\gamma}=P_{f}\left(2 V_{i}\right)^{\gamma} \Rightarrow P_{f}=\frac{P_{i}}{2^{\gamma}}=\frac{P_{i}}{2 \sqrt{2}} ......(2) \end{aligned} $$
From (1) and (2)
$$ \begin{aligned} & W=\frac{\frac{P_{i}}{2 \sqrt{2}} 2 V_{i}-P_{i} V_{i}}{1-\gamma}=\frac{P_{i} V_{i}}{-1 / 2}\left(\frac{1}{\sqrt{2}}-1\right) \\\\ & =-n R T(\sqrt{2}-2) \\\\ & =n R T(2-\sqrt{2}) \end{aligned} $$
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