Comprehension-3
An ideal gas initially at pressure `p_(0)` undergoes a free expansion (expansion against vacuum under adiabatic conditions) until its volume is `3` times its initial volume. The gas is next adiabatically compressed back to its original volume. The pressure after compression is `3^(2//3)p_(0)`.
What is the ratio of the average kinetic energy per molecule in the final state to that in the initial state?

Options

A. `1`
B. `3^(2//3)`
C. `3^(1//3)`
D. `3^(1//6)`

Answer

To solve the problem, we need to find the ratio of the average kinetic energy per molecule in the final state to that in the initial state of an ideal gas that undergoes a free expansion followed by adiabatic compression.

### Step 1: Understand the initial conditions
- We start with an ideal gas at initial pressure \( p_0 \), initial volume \( V_0 \), and initial temperature \( T_0 \).
- The average kinetic energy per molecule is given by the formula:
\[ E = \frac{3}{2} k_B T \]
where \( k_B \) is the Boltzmann constant.

### Step 2: Analyze the free expansion
- The gas undergoes free expansion to a volume \( 3V_0 \) against a vacuum.
- Since this is an adiabatic process (no heat exchange) and the work done is zero (expanding against vacuum), the internal energy change \( \Delta U \) is zero.
- Therefore, the temperature remains constant during this process:
\[ T_B = T_0 \]

### Step 3: Analyze the adiabatic compression
- The gas is then adiabatically compressed back to its original volume \( V_0 \).
- We know from the problem that the pressure after compression is \( p_C = 3^{2/3} p_0 \).
- For an adiabatic process, the relation between temperature and volume is given by:
\[ T V^{\gamma - 1} = \text{constant} \]
where \( \gamma \) is the heat capacity ratio.

### Step 4: Calculate the final temperature
- Using the relation for the adiabatic process from point B to C:
\[ T_B (3V_0)^{\gamma - 1} = T_C (V_0)^{\gamma - 1} \]
- Substituting \( T_B = T_0 \):
\[ T_0 (3V_0)^{\gamma - 1} = T_C (V_0)^{\gamma - 1} \]
- Simplifying gives:
\[ T_C = T_0 \cdot 3^{\gamma - 1} \]

### Step 5: Determine the value of \( \gamma \)
- From the information given, we know:
\[ p_B (3V_0)^{\gamma} = p_C V_0^{\gamma} \]
- Substituting \( p_B = \frac{p_0}{3} \) and \( p_C = 3^{2/3} p_0 \):
\[ \left(\frac{p_0}{3}\right)(3V_0)^{\gamma} = (3^{2/3} p_0)(V_0)^{\gamma} \]
- Canceling \( p_0 \) and \( V_0^{\gamma} \) leads to:
\[ \frac{1}{3} \cdot 3^{\gamma} = 3^{2/3} \]
- This simplifies to:
\[ 3^{\gamma - 1} = 3^{2/3} \]
- Therefore, \( \gamma - 1 = \frac{2}{3} \) leading to \( \gamma = \frac{5}{3} \).

### Step 6: Calculate the final temperature
- Now substituting \( \gamma \) back into the temperature equation:
\[ T_C = T_0 \cdot 3^{\frac{5}{3} - 1} = T_0 \cdot 3^{\frac{2}{3}} \]

### Step 7: Calculate the ratio of average kinetic energy
- The average kinetic energy per molecule in the final state is:
\[ E_C = \frac{3}{2} k_B T_C = \frac{3}{2} k_B (T_0 \cdot 3^{2/3}) \]
- The average kinetic energy per molecule in the initial state is:
\[ E_0 = \frac{3}{2} k_B T_0 \]
- The ratio of the average kinetic energies is:
\[ \frac{E_C}{E_0} = \frac{\frac{3}{2} k_B (T_0 \cdot 3^{2/3})}{\frac{3}{2} k_B T_0} = 3^{2/3} \]

### Final Answer
The ratio of the average kinetic energy per molecule in the final state to that in the initial state is:
\[ \frac{E_C}{E_0} = 3^{2/3} \]