Question

A mono-atomic 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. \(L_{1}\) and \(L_{2}\) are the lengths of gas columns before and after the expansion, respectively. The value of \(\frac{T_{1}}{T_{2}}\) is \(\left[\gamma =\frac{5}{3}\right]\)

• A. \(\left(\frac{L_{1}}{L_{2}}\right)^{\frac{2}{3}}\)
• B. \(\left(\frac{L_{2}}{L_{1}}\right)^{\frac{2}{3}}\)
• C. \(\left(\frac{L_{1}}{L_{2}}\right)^{\frac{5}{3}}\)
• D. \(\left(\frac{L_{2}}{L_{1}}\right)^{\frac{5}{3}}\)

Hint:

For an adiabatic process the product of temperature and volume raised to the power (gamma minus 1) stays constant, and volume is proportional to the gas column length.

Answer 1

The Correct Option is B

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

Step 2: Use length
The cross-section is fixed, so \(V\propto L\). Then \(T_{1}L_{1}^{\gamma-1}=T_{2}L_{2}^{\gamma-1}\).

Step 3: Calculation
\(\dfrac{T_{1}}{T_{2}}=\left(\dfrac{L_{2}}{L_{1}}\right)^{\gamma-1}=\left(\dfrac{L_{2}}{L_{1}}\right)^{2/3}\), since \(\gamma-1=\frac{2}{3}\).

Final Answer:
The ratio \(\frac{T_{1}}{T_{2}}\) is \(\left(\frac{L_{2}}{L_{1}}\right)^{2/3}\). \[ \boxed{\text{(B) }\left(\dfrac{L_{2}}{L_{1}}\right)^{\frac{2}{3}}} \]