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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 the expansion respectively, then the value of $\frac{T_{1}}{T_{2}}$ will be

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$\frac{l_{1}}{l_{2}}$

${(\frac{l_{1}}{l_{2}})}^{\frac{2}{3}}$

${(\frac{l_{2}}{l_{1}})}^{\frac{2}{3}}$

$\frac{l_{2}}{l_{1}}$

answer is B.

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Detailed Solution

As We know that, In adiabatic condition

$\begin{array}{l} {TV}^{γ - 1} = constant where, A is constant \\ T (Al)^{γ - 1} = constant \\ \Rightarrow {Tl}^{γ - 1} = constant γ = adiabatic constant \end{array}$

For monoatomic gas,

$γ = 5 / 3$

$W_{AB} = ΔQ - ΔU = {nC}_{p} dT - {nC}_{V}$

$\Rightarrow T_{1} l_{1}^{\frac{5}{3} - 1} = T_{2} l_{2}^{\frac{5}{3} - 1} \Rightarrow \frac{T_{1}}{T_{2}} = {(\frac{l_{2}}{l_{1}})}^{\frac{2}{3}}$

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