JEE MAIN - Physics (2024 - 4th April Evening Shift - No. 5)
A sample of gas at temperature $$T$$ is adiabatically expanded to double its volume. Adiabatic constant for the gas is $$\gamma=3 / 2$$. The work done by the gas in the process is:
$$(\mu=1 \text { mole })$$
$$R T[2 \sqrt{2}-1]$$
$$R T[2-\sqrt{2}]$$
$$R T[1-2 \sqrt{2}]$$
$$R T[\sqrt{2}-2]$$
Explanation
$$\begin{aligned}
& w=\frac{-n R}{\gamma-1}(\Delta T) \\
&=\frac{-R}{1 / 2}\left(\frac{T}{\sqrt{2}}-T\right) \\
&=2 R\left(\frac{\sqrt{2} T-T}{\sqrt{2}}\right) \\
&=R T(2-\sqrt{2}) \\
& \therefore \quad T V_\gamma^{-1}=\text { cons. } \\
& T V_\gamma^{-1}=T_f(2 V)^{\gamma-1} \\
& T_f=\frac{T}{\sqrt{2}}
\end{aligned}$$
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