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A monoatomic ideal gas, initially at temperature ' $\mathrm{T}_1$ ' is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature ' $\mathrm{T}_2$ ' by releasing the piston suddenly $L_1$ and $L_2$ are the lengths of the gas columns before and after the expansion respectively. The $\frac{\mathrm{T}_2}{\mathrm{~T}_1}$ is
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The correct answer is:
$\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{2 / 3}$
In an adiabatic process $\mathrm{T}_1 \mathrm{~V}_1^{-1}=\mathrm{T}_2 \mathrm{~V}_2^{-1}$
For an ideal mono atomic gas the no of degree of freedom is 3
$$
\begin{aligned}
& \mathrm{Y}=1+\frac{2}{\mathrm{f}}=1+\frac{2}{3} \\
& \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{\mathrm{Y}-1}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{2 / 3}=\left(\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right)^{2 / 3}
\end{aligned}
$$
Since volume is proportional to length due to area being constant.
For an ideal mono atomic gas the no of degree of freedom is 3
$$
\begin{aligned}
& \mathrm{Y}=1+\frac{2}{\mathrm{f}}=1+\frac{2}{3} \\
& \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{\mathrm{Y}-1}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{2 / 3}=\left(\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right)^{2 / 3}
\end{aligned}
$$
Since volume is proportional to length due to area being constant.
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