Q150 / 180 · COMEDK 2025 (10 May Shift 2) · Physics
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 $L$ and $2 L$ are the lengths of the gas column before and after expansion respectively, then $\dfrac{T_1}{T_2}$ is
- A$\left(\dfrac{1}{2}\right)^{2 / 3}$
- B$2^{3 / 2}$
- Correct answerC$2^{2 / 3}$
- D$\left(\dfrac{1}{2}\right)^{3 / 2}$
Solution
For an adiabatic process of an ideal gas, the relationship between temperature and volume is given by $T V^{\gamma - 1} = \text{constant}$.
For a monoatomic gas, the adiabatic index $\gamma = \dfrac{5}{3}$. Thus, $\gamma - 1 = \dfrac{5}{3} - 1 = \dfrac{2}{3}$.
The volume of the gas is proportional to the length of the gas column, $V \propto L$. Therefore, $T_1 L_1^{2/3} = T_2 L_2^{2/3}$.
Given $L_1 = L$ and $L_2 = 2L$, we have $T_1 L^{2/3} = T_2 (2L)^{2/3}$.
Rearranging to find the ratio $\dfrac{T_1}{T_2}$:
$\dfrac{T_1}{T_2} = \dfrac{(2L)^{2/3}}{L^{2/3}} = \left(\dfrac{2L}{L}\right)^{2/3} = 2^{2/3}$.
Answer: $2^{2 / 3}$
For a monoatomic gas, the adiabatic index $\gamma = \dfrac{5}{3}$. Thus, $\gamma - 1 = \dfrac{5}{3} - 1 = \dfrac{2}{3}$.
The volume of the gas is proportional to the length of the gas column, $V \propto L$. Therefore, $T_1 L_1^{2/3} = T_2 L_2^{2/3}$.
Given $L_1 = L$ and $L_2 = 2L$, we have $T_1 L^{2/3} = T_2 (2L)^{2/3}$.
Rearranging to find the ratio $\dfrac{T_1}{T_2}$:
$\dfrac{T_1}{T_2} = \dfrac{(2L)^{2/3}}{L^{2/3}} = \left(\dfrac{2L}{L}\right)^{2/3} = 2^{2/3}$.
Answer: $2^{2 / 3}$
Hence, the correct answer is option: 'C'.