A monatomic ideal gas, initially at temperature `T_(1)` is enclosed in a cylinder fitted with a frictionless pistion. The gas is allowed to expand adiabatically to a temperature `T_(2)`. By releasing the piston suddenly. IF `L_(1)` and `L_(2)` are th lengths of the gas column before and after expansion respectively, then `T_(1)//T_(2)` is given by

Options

A. `(L_(1)//L_(2))^(2//3)`
B. `(L_(1)//L_(2))`
C. `(L_(2)//L_(1))`
D. `(L_(2)//L_(1))`

Explanation

To find the ratio \( \frac{T_1}{T_2} \) for a monatomic ideal gas that expands adiabatically, we can follow these steps:

### Step 1: Understand the Adiabatic Process
In an adiabatic process, the relationship between temperature and volume can be expressed as:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
where \( \gamma \) (gamma) is the heat capacity ratio, which for a monatomic ideal gas is \( \frac{5}{3} \).

### Step 2: Relate Volume to Length
Since the volume \( V \) of the gas is related to the length of the gas column \( L \) (assuming the cross-sectional area \( A \) of the cylinder remains constant), we can express the volumes in terms of lengths:
\[ V_1 = A L_1 \quad \text{and} \quad V_2 = A L_2 \]
Thus, we can say:
\[ V_1 = L_1 \quad \text{and} \quad V_2 = L_2 \quad \text{(since area A is constant)} \]

### Step 3: Substitute Volumes into the Adiabatic Equation
Substituting \( V_1 \) and \( V_2 \) into the adiabatic equation gives us:
\[ T_1 (L_1)^{\gamma - 1} = T_2 (L_2)^{\gamma - 1} \]

### Step 4: Rearranging the Equation
We can rearrange this equation to find the ratio of temperatures:
\[ \frac{T_2}{T_1} = \frac{L_1^{\gamma - 1}}{L_2^{\gamma - 1}} \]

### Step 5: Substitute the Value of Gamma
Substituting \( \gamma = \frac{5}{3} \):
\[ \frac{T_2}{T_1} = \frac{L_1^{\frac{5}{3} - 1}}{L_2^{\frac{5}{3} - 1}} = \frac{L_1^{\frac{2}{3}}}{L_2^{\frac{2}{3}}} \]

### Step 6: Simplifying the Ratio
This can be simplified to:
\[ \frac{T_2}{T_1} = \left(\frac{L_1}{L_2}\right)^{\frac{2}{3}} \]

### Step 7: Finding \( \frac{T_1}{T_2} \)
Taking the reciprocal gives us:
\[ \frac{T_1}{T_2} = \left(\frac{L_2}{L_1}\right)^{\frac{2}{3}} \]

### Conclusion
Thus, the final ratio of temperatures is:
\[ \frac{T_1}{T_2} = \left(\frac{L_2}{L_1}\right)^{\frac{2}{3}} \]

### Final Answer
The correct answer is:
\[ \frac{T_1}{T_2} = \left(\frac{L_2}{L_1}\right)^{\frac{2}{3}} \]