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 \(L_{1}\) and \(L_{2}\) be the lengths of the gas column before and after expansion respectively, then \(\frac{T_{1}}{T_{2}}\) is

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

Hint:

Use TV^(gamma-1) = constant with gamma = 5/3 and volume proportional to the gas column length.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For an adiabatic process, \(TV^{\gamma-1}=\text{constant}\).

Step 2: Gamma
For a monoatomic gas \(\gamma=\dfrac{5}{3}\), so \(\gamma-1=\dfrac{2}{3}\).

Step 3: Volume and Length
The cross-section is uniform, so \(V\propto L\).

Step 4: Solve
\(T_1L_1^{2/3}=T_2L_2^{2/3}\), so \(\dfrac{T_1}{T_2}=\left(\dfrac{L_2}{L_1}\right)^{2/3}\).

Final Answer:
T1 over T2 equals L2 over L1 raised to the power two thirds. \[ \boxed{\text{(D) }\left(\dfrac{L_{2}}{L_{1}}\right)^{\frac{2}{3}}} \]