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)`.
The pressure of the gas after the free expansion is:

Options

A. `(p_(0))/(3)`
B. `p_(0)^(1//3)`
C. `p_(0)`
D. `3p_(0)`

Answer

To solve the problem, we will analyze the situation step by step.

### Step 1: Understand the Initial Conditions
Initially, we have an ideal gas at pressure \( p_0 \) and volume \( V_0 \). The temperature is \( T_0 \).

### Step 2: Free Expansion
The gas undergoes a free expansion against a vacuum until its volume becomes three times its initial volume. Therefore, the final volume after the free expansion is:
\[ V_B = 3V_0 \]

### Step 3: Analyze the Free Expansion Process
In a free expansion:
- There is no external pressure opposing the expansion (since it is against a vacuum).
- The process is adiabatic, meaning there is no heat exchange with the surroundings (\( Q = 0 \)).
- The work done by the gas is also zero (\( W = 0 \)) because there is no opposing force.

### Step 4: Apply the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ Q = \Delta U + W \]
Since \( Q = 0 \) and \( W = 0 \), we have:
\[ 0 = \Delta U + 0 \implies \Delta U = 0 \]
This indicates that the internal energy of the gas does not change during the free expansion.

### Step 5: Relate Internal Energy to Temperature
For an ideal gas, the change in internal energy (\( \Delta U \)) is given by:
\[ \Delta U = nC_v \Delta T \]
Since \( \Delta U = 0 \), it follows that:
\[ \Delta T = 0 \]
This means the temperature of the gas remains constant during the free expansion.

### Step 6: Use the Ideal Gas Law
Since the temperature remains constant, we can use the ideal gas law:
\[ PV = nRT \]
At the initial state (point A):
\[ p_0 V_0 = nRT_0 \]
At the final state after free expansion (point B):
\[ P_B \cdot (3V_0) = nRT_0 \]
Since \( nRT_0 \) is constant, we can equate the two states:
\[ p_0 V_0 = P_B \cdot (3V_0) \]

### Step 7: Solve for \( P_B \)
Dividing both sides by \( V_0 \) (assuming \( V_0 \neq 0 \)):
\[ p_0 = P_B \cdot 3 \]
Rearranging gives:
\[ P_B = \frac{p_0}{3} \]

### Conclusion
The pressure of the gas after the free expansion is:
\[ \boxed{\frac{p_0}{3}} \]