Question:
A monoatomic ideal gas, initially at temperature $\text{T}_1$ is enclosed in a cylinder fitted with massless, frictionless piston. By releasing the piston suddenly, the gas is allowed to expand adiabatically to a temperature $\text{T}_2$. If $\text{L}_1$ and $\text{L}_2$ are the lengths of the gas columns before and after expansion respectively, then $(\text{T}_2/\text{T}_1)$ is given by}
A monoatomic ideal gas, initially at temperature $\text{T}_1$ is enclosed in a cylinder fitted with massless, frictionless piston. By releasing the piston suddenly, the gas is allowed to expand adiabatically to a temperature $\text{T}_2$. If $\text{L}_1$ and $\text{L}_2$ are the lengths of the gas columns before and after expansion respectively, then $(\text{T}_2/\text{T}_1)$ is given by}
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For monoatomic gases, $\gamma = 1.67$; for diatomic, $\gamma = 1.4$.
Updated On: Apr 26, 2026
- $\frac{\text{L}_1}{\text{L}_2}$
- $\frac{\text{L}_2}{\text{L}_1}$
- $(\frac{\text{L}_1}{\text{L}_2})^{2/3}$
- $(\frac{\text{L}_2}{\text{L}_1})^{2/3}$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
For an adiabatic process, $TV^{\gamma-1} = \text{constant}$.
Step 2: Analysis
Volume $V = \text{Area} \times \text{Length} = A \times L$.
So, $T_1 (AL_1)^{\gamma-1} = T_2 (AL_2)^{\gamma-1} \implies \frac{T_2}{T_1} = \left( \frac{L_1}{L_2} \right)^{\gamma-1}$.
Step 3: Calculation
For a monoatomic gas, $\gamma = 5/3$.
$\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$.
Thus, $\frac{T_2}{T_1} = \left( \frac{L_1}{L_2} \right)^{2/3}$.
Step 4: Conclusion
The correct ratio is $(\frac{L_1}{L_2})^{2/3}$.
Final Answer: (C)
For an adiabatic process, $TV^{\gamma-1} = \text{constant}$.
Step 2: Analysis
Volume $V = \text{Area} \times \text{Length} = A \times L$.
So, $T_1 (AL_1)^{\gamma-1} = T_2 (AL_2)^{\gamma-1} \implies \frac{T_2}{T_1} = \left( \frac{L_1}{L_2} \right)^{\gamma-1}$.
Step 3: Calculation
For a monoatomic gas, $\gamma = 5/3$.
$\gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3}$.
Thus, $\frac{T_2}{T_1} = \left( \frac{L_1}{L_2} \right)^{2/3}$.
Step 4: Conclusion
The correct ratio is $(\frac{L_1}{L_2})^{2/3}$.
Final Answer: (C)
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