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A monoatomic ideal gas initially at temperature ' $\mathrm{T}_1$ ' is enclosed in a cylinder fitted with massless, frictionless piston. By releasing the piston suddenly the gas is allowed to expand to adiabatically to a temperature ' $\mathrm{T}_2$ '. If ' $\mathrm{L}_1$ ' and ' $\mathrm{L}_2$ ' are the lengths of the gas columns before and after expansion respectively, then $\frac{T_2}{T_1}$ is
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Verified Answer
The correct answer is:
$\left(\frac{L_1}{L_2}\right)^{2 / 3}$
For an adiabatic process
$$
\begin{aligned}
& \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\
\therefore \quad & \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}
\end{aligned}
$$
For a monoatomic gas, $r=\frac{5}{3}$
$$
\begin{aligned}
& \Rightarrow \gamma-1=\frac{5}{3}-1=\frac{2}{3} \\
& \mathrm{~V}_1=\mathrm{AL}_1 \text { and } \mathrm{V}_2=\mathrm{AL}_2 \\
\therefore \quad & \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left[\frac{\mathrm{AL}_1}{\mathrm{AL}_2}\right]^{2 / 3}=\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{2 / 3} \\
& \gamma-1=\frac{2}{3}
\end{aligned}
$$
For an adiabatic process,
$$
\begin{aligned}
& \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\frac{2}{3}} \\
\mathrm{~V} & \propto \mathrm{L} \\
\therefore \quad \frac{\mathrm{T}_2}{\mathrm{~T}_1} & =\left(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right)^{\frac{2}{3}}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1} \\
\therefore \quad & \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}
\end{aligned}
$$
For a monoatomic gas, $r=\frac{5}{3}$
$$
\begin{aligned}
& \Rightarrow \gamma-1=\frac{5}{3}-1=\frac{2}{3} \\
& \mathrm{~V}_1=\mathrm{AL}_1 \text { and } \mathrm{V}_2=\mathrm{AL}_2 \\
\therefore \quad & \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left[\frac{\mathrm{AL}_1}{\mathrm{AL}_2}\right]^{2 / 3}=\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{2 / 3} \\
& \gamma-1=\frac{2}{3}
\end{aligned}
$$
For an adiabatic process,
$$
\begin{aligned}
& \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\gamma-1}=\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)^{\frac{2}{3}} \\
\mathrm{~V} & \propto \mathrm{L} \\
\therefore \quad \frac{\mathrm{T}_2}{\mathrm{~T}_1} & =\left(\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right)^{\frac{2}{3}}
\end{aligned}
$$
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