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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 $\frac{T_2}{T_1}$ is
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The correct answer is:
$\left[\frac{\mathrm{L}_1}{\mathrm{~L}_2}\right]^{2 / 3}$
In an adiabatic process $\mathrm{T}_1 \mathrm{~V}_1^{\gamma-1}=\mathrm{T}_2 \mathrm{~V}_2^{\gamma-1}$
For an ideal mono atomic gas the no of degree of freedom is 3
$Y=1+\frac{2}{f}=1+\frac{2}{3}$
$\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{\mathrm{Y}-1}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{2 / 3}=\left(\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right)^{2 / 3}$
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
$Y=1+\frac{2}{f}=1+\frac{2}{3}$
$\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{\mathrm{Y}-1}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{2 / 3}=\left(\frac{\mathrm{L}_2}{\mathrm{~L}_1}\right)^{2 / 3}$
Since volume is proportional to length due to area being constant.
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