Question

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 \( L \) and \( 2L \) are the lengths of the gas column before and after expansion respectively, then \( \frac{T_1}{T_2} \) is

• A. \( 2^{3/2} \)
• B. \( 2^{2/3} \)
• C. \( \left(\frac{1}{2}\right)^{2/3} \)
• D. \( \left(\frac{1}{2}\right)^{3/2} \)

Hint:

In adiabatic process, use \( TV^{\gamma-1} = \text{constant} \). For monoatomic gas, \( \gamma = \frac{5}{3} \).

Answer 1

The Correct Option is B

Step 1: Use adiabatic relation.
\[ TV^{\gamma-1} = \text{constant} \]

Step 2: Write relation between initial and final states.
\[ T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \]

Step 3: Write ratio.
\[ \frac{T_1}{T_2} = \left(\frac{V_2}{V_1}\right)^{\gamma-1} \]

Step 4: Relate volume with length.
Since cross-section constant:
\[ V \propto L \]

Step 5: Substitute length ratio.
\[ \frac{V_2}{V_1} = \frac{2L}{L} = 2 \]

Step 6: Use value of \( \gamma \).
For monoatomic gas:
\[ \gamma = \frac{5}{3} \]
\[ \gamma - 1 = \frac{2}{3} \]

Step 7: Final calculation.
\[ \frac{T_1}{T_2} = 2^{2/3} \]