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

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

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Solution

The correct option is A $(\frac{L_{1}}{L_{2}})^{\frac{L}{3}}$
For adiabatic process $T v^{γ - 1}$ =
Let cross section area be $A$
initial volume $v_{1} = L_{1} A$
Final volume $v_{2} = L_{2} A$
$T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$
$\frac{T_{1}}{T_{2}} = {(\frac{V_{2}}{V_{1}})}^{γ - 1}$
$\frac{T_{1}}{T_{2}} = {(\frac{L_{2} A}{L_{1} A})}^{γ - 1}$
$\frac{T_{1}}{T_{2}} = {(\frac{L_{2}}{4})}^{γ - 1} = {(\frac{L_{2}}{4})}^{3 / 3 - 1} = {(\frac{L_{2}}{L_{1}})}^{2 / 3}$
option D

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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