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. $L_1$ and $L_2$ are the lengths of the gas columns before and after the expansion respectively. Then $\frac{T_2}{T_1}$ is

• A. $\left(\frac{L_2}{L_1}\right)^{2/3}$
• B. $\left(\frac{L_1}{L_2}\right)^{2/3}$
• C. $\left(\frac{L_1}{L_2}\right)^{1/2}$
• D. $\left(\frac{L_2}{L_1}\right)^{1/2}$

Hint:

During gas expansions, volume increases ($L_2 > L_1$) which always causes temperature to drop ($T_2 < T_1$). Therefore, the ratio $\frac{T_2}{T_1}$ must evaluate to a value less than 1. This means the fractional core must put the smaller length on top, instantly filtering out options (A) and (D).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A collection of monoatomic ideal gas trapped within a cylindrical containment undergoes sudden adiabatic expansion.
As it pushes the boundaries from initial length $L_1$ to expanded length $L_2$, its internal thermal state shifts from temperature $T_1$ to $T_2$. We must find the mathematical ratio of these temperatures.

Step 2: Key Formula or Approach:
For a classic reversible adiabatic process, state variables temperature ($T$) and volume ($V$) obey Poisson's relations:
$$T V^{\gamma - 1} = \text{constant} \implies T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$$ The ratio can be isolated as:
$$\frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1}$$ For a uniform cylinder of cross-sectional area $A$, the volume scales linearly with column height: $V = A \cdot L \implies \frac{V_1}{V_2} = \frac{L_1}{L_2}$.
For a single-atom monoatomic ideal gas system, the ratio of specific heats is defined as $\gamma = \frac{5}{3}$.

Step 3: Detailed Explanation:
Let's substitute the direct geometrical relationship ($\frac{V_1}{V_2} = \frac{L_1}{L_2}$) into the temperature ratio equation:
$$\frac{T_2}{T_1} = \left(\frac{L_1}{L_2}\right)^{\gamma - 1}$$ Now substitute the thermodynamic constant $\gamma = \frac{5}{3}$ for the monoatomic gas:
$$\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$$ Combining these steps yields the final fractional exponent configuration:
$$\frac{T_2}{T_1} = \left(\frac{L_1}{L_2}\right)^{2/3}$$

Step 4: Final Answer:
The ratio $\frac{T_2}{T_1}$ is given by $\left(\frac{L_1}{L_2}\right)^{2/3}$, matching option (B).