A monoatomic ideal gas, initially at temperature `T_1,` is enclosed in a cylinder fitted with a friction less 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 length of the gas column before expansion respectively, then `(T_1)/(T_2)` is given by

Options

A. (a) `((L_1)/(L_2))^(2/3)`
B. (b) `L_1/L_2`
C. (c) `L_2/L_1`
D. (d) `((L_2)/(L_1))^(2/3)`

Explanation

To solve the problem, we will use the principles of thermodynamics, specifically focusing on the adiabatic process for a monoatomic ideal gas. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Adiabatic Process
In an adiabatic process, there is no heat exchange with the surroundings. For an ideal gas undergoing an adiabatic expansion, the relationship between temperature and volume can be expressed as:
\[ T V^{\gamma - 1} = \text{constant} \]
where \( \gamma \) (gamma) is the heat capacity ratio.

### Step 2: Write the Relationship for Initial and Final States
For the initial state (before expansion) and the final state (after expansion), we can write:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]

### Step 3: Express Volume in Terms of Length
The volume \( V \) of the gas can be expressed in terms of the cross-sectional area \( A \) and the length \( L \) of the gas column:
\[ V = A \cdot L \]
Thus, we can rewrite the volumes as:
\[ V_1 = A \cdot L_1 \quad \text{and} \quad V_2 = A \cdot L_2 \]

### Step 4: Substitute Volumes into the Equation
Substituting the expressions for \( V_1 \) and \( V_2 \) into the adiabatic condition gives:
\[ T_1 (A \cdot L_1)^{\gamma - 1} = T_2 (A \cdot L_2)^{\gamma - 1} \]
Since the area \( A \) is the same in both cases, it cancels out:
\[ T_1 L_1^{\gamma - 1} = T_2 L_2^{\gamma - 1} \]

### Step 5: Rearranging the Equation
Rearranging the equation to find the ratio \( \frac{T_1}{T_2} \):
\[ \frac{T_1}{T_2} = \frac{L_2^{\gamma - 1}}{L_1^{\gamma - 1}} = \left( \frac{L_2}{L_1} \right)^{\gamma - 1} \]

### Step 6: Substitute the Value of Gamma
For a monoatomic ideal gas, \( \gamma = \frac{5}{3} \). Therefore:
\[ \gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3} \]
Substituting this value into the equation gives:
\[ \frac{T_1}{T_2} = \left( \frac{L_2}{L_1} \right)^{\frac{2}{3}} \]

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