A mono-atomic ideal gas, initially at temperature T_1 is enclosed...

For a reversible adiabatic process, the relationship between temperature and volume is given by $TV^{\gamma - 1} = \text{constant}$.

Applying this to the initial and final states of the gas:

$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$

$\dfrac{T_1}{T_2} = \left(\dfrac{V_2}{V_1}\right)^{\gamma - 1}$

Let $A$ be the cross-sectional area of the cylinder. The volume of the gas column of length $L$ is $V = AL$.

Substituting $V_1 = AL_1 and$V_2 = AL_2$:<br><br>$\dfrac{T_1}{T_2} = \left(\dfrac{AL_2}{AL_1}\right)^{\gamma - 1} = \left(\dfrac{L_2}{L_1}\right)^{\gamma - 1}$<br><br>Given that the gas is mono-atomic,$\gamma = \dfrac{5}{3}$.<br><br>$\gamma - 1 = \dfrac{5}{3} - 1 = \dfrac{2}{3}$<br><br>Substituting this value into the equation:<br><br>$\dfrac{T_1}{T_2} = \left(\dfrac{L_2}{L_1}\right)^{\dfrac{2}{3}}$<br><br>Answer: \left(\dfrac{L_2}{L_1}\right)^{\frac{2}{3}}$