A monoatomic ideal gas, initially at temperature $T_{1}$ , is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature $T_{2}$ by releasing the piston suddenly. If $L_{1}$ and $L_{2}$ are lengths of gas column before and after expansion respectively, then $\frac{T_{1}}{T_{2}}$ is given by

{(\frac{L_{1}}{L_{2}})}^{2 / 3}

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\frac{L_{1}}{L_{2}}

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\frac{L_{2}}{L_{1}}

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{(\frac{L_{2}}{L_{1}})}^{2 / 3}

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Solution

The correct option is A ${(\frac{L_{2}}{L_{1}})}^{2 / 3}$
In an adiabatic process $T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$
For an ideal mono atomic gas the no of degrees of freedom is 3.
$γ = 1 + \frac{2}{f} = 1 + \frac{2}{3}$
$\frac{T_{1}}{T_{2}} = (\frac{V_{2}}{V_{1}})^{γ - 1} = (\frac{V_{2}}{V_{1}})^{\frac{2}{3}} = (\frac{L_{2}}{L_{1}})^{\frac{2}{3}}$ since volume is proportional to length due to area being constant.

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A monoatomic ideal gas, initially at temperature $T_{1}$ is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature. $T_{2}$ by releasing the piston suddenly. If $L_{1}$ and $L_{2}$ are the lengths of the gas column before and after expansion respectively, then $\frac{T_{1}}{T_{2}}$ is given by