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. $\frac{\text{L}_1}{\text{L}_2}$
• B. $\frac{\text{L}_2}{\text{L}_1}$
• C. $(\frac{\text{L}_1}{\text{L}_2})^{2/3}$
• D. $(\frac{\text{L}_2}{\text{L}_1})^{2/3}$

Hint:

For monoatomic gases, $\gamma = 1.67$; for diatomic, $\gamma = 1.4$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A sudden expansion is an adiabatic process, meaning no heat is exchanged with the surroundings ($Q = 0$).
For an adiabatic process involving an ideal gas, the relationship between temperature and volume is governed by the adiabatic index (ratio of specific heats), $\gamma$.
Step 2: Key Formula or Approach:
The adiabatic relation between temperature $T$ and volume $V$ is: \[ T \cdot V^{\gamma - 1} = \text{constant} \] For two states 1 and 2, this is written as: \[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \] The volume of a cylinder is $V = A \cdot L$, where $A$ is the constant cross-sectional area and $L$ is the length of the gas column.
Step 3: Detailed Explanation:
First, express volumes in terms of lengths: \[ V_1 = A \cdot L_1 \] \[ V_2 = A \cdot L_2 \] Substitute these into the adiabatic relation: \[ T_1 (A \cdot L_1)^{\gamma - 1} = T_2 (A \cdot L_2)^{\gamma - 1} \] Since the area $A$ is constant, $A^{\gamma - 1}$ cancels out from both sides: \[ T_1 \cdot L_1^{\gamma - 1} = T_2 \cdot L_2^{\gamma - 1} \] We are looking for the ratio $\frac{T_2}{T_1}$: \[ \frac{T_2}{T_1} = \frac{L_1^{\gamma - 1}}{L_2^{\gamma - 1}} = \left( \frac{L_1}{L_2} \right)^{\gamma - 1} \] The gas is given as monoatomic. For a monoatomic ideal gas, the adiabatic index $\gamma$ is: \[ \gamma = \frac{5}{3} \] Calculate the exponent $\gamma - 1$: \[ \gamma - 1 = \frac{5}{3} - 1 = \frac{2}{3} \] Substitute this exponent back into the ratio equation: \[ \frac{T_2}{T_1} = \left( \frac{L_1}{L_2} \right)^{2/3} \] Step 4: Final Answer:
The ratio $(\text{T}_2/\text{T}_1)$ is given by $(\frac{\text{L}_1}{\text{L}_2})^{2/3}$.