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, then the value of $T_{1}/T_{2}$ will be}
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, then the value of $T_{1}/T_{2}$ will be}
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For monoatomic gases, $\gamma = 5/3$; for diatomic, $\gamma = 7/5$. Always use the relation $TV^{\gamma - 1}$ for adiabatic temperature changes.
Updated On: Jul 4, 2026
- $(L_{1}/L_{2})^{2/3}$
- $(L_{2}/L_{1})^{2/3}$
- $L_{2}/L_{1}$
- $L_{1}/L_{2}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
For an adiabatic process, the relationship between temperature ($T$) and volume ($V$) is $TV^{\gamma - 1} = \text{constant}$.
Step 2: Meaning
Since the cylinder has a constant cross-sectional area, volume is directly proportional to the length ($L$) of the gas column ($V \propto L$). Thus, $TL^{\gamma - 1} = \text{constant}$.
Step 3: Analysis
For a monoatomic gas, the ratio of specific heats $\gamma = 5/3$. Substituting this value: $T_{1}L_{1}^{(5/3)-1} = T_{2}L_{2}^{(5/3)-1}$. This simplifies to $T_{1}L_{1}^{2/3} = T_{2}L_{2}^{2/3}$.
Step 4: Conclusion
Rearranging the equation gives $T_{1}/T_{2} = (L_{2}/L_{1})^{2/3}$. Final Answer: (B)
For an adiabatic process, the relationship between temperature ($T$) and volume ($V$) is $TV^{\gamma - 1} = \text{constant}$.
Step 2: Meaning
Since the cylinder has a constant cross-sectional area, volume is directly proportional to the length ($L$) of the gas column ($V \propto L$). Thus, $TL^{\gamma - 1} = \text{constant}$.
Step 3: Analysis
For a monoatomic gas, the ratio of specific heats $\gamma = 5/3$. Substituting this value: $T_{1}L_{1}^{(5/3)-1} = T_{2}L_{2}^{(5/3)-1}$. This simplifies to $T_{1}L_{1}^{2/3} = T_{2}L_{2}^{2/3}$.
Step 4: Conclusion
Rearranging the equation gives $T_{1}/T_{2} = (L_{2}/L_{1})^{2/3}$. Final Answer: (B)
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