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 $l_2$ are the lengths of the gas column, before and after the expansion respectively, then the value of $\frac{T_1}{T_2}$ will be :

• A. $\frac{l_1}{l_2}$
• B. $\frac{l_2}{l_1}$
• C. $(\frac{l_1}{l_2})^{2/3}$
• D. $(\frac{l_2}{l_1})^{2/3}$

Hint:

For adiabatic processes, remember that expansion always leads to cooling (\(T_2<T_1\)). Since \(l_2>l_1\), the ratio \(T_1/T_2\) must be greater than 1, which fits the mathematical result.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
In an adiabatic process, there is no heat exchange with the surroundings (\(\Delta Q = 0\)). The state variables satisfy the adiabatic equation involving temperature and volume.
Step 2: Key Formula or Approach:
1. Adiabatic relation: \(T V^{\gamma - 1} = \text{constant}\).
2. Volume of cylinder: \(V = \text{Area} \times \text{Length} = A \cdot l\).
3. For monoatomic gas: \(\gamma = \frac{5}{3}\).
Step 3: Detailed Explanation:
From the adiabatic relation:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \implies \frac{T_1}{T_2} = \left( \frac{V_2}{V_1} \right)^{\gamma - 1} \]
Since area \(A\) is constant:
\[ \frac{V_2}{V_1} = \frac{A \cdot l_2}{A \cdot l_1} = \frac{l_2}{l_1} \]
Substituting \(\gamma = 5/3\):
\[ \gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3} \]
Therefore:
\[ \frac{T_1}{T_2} = \left( \frac{l_2}{l_1} \right)^{2/3} \]
Step 4: Final Answer:
The ratio of temperatures is \((\frac{l_2}{l_1})^{2/3}\).