During an experiment, an ideal gas is found to obey a condition `Vp^2 =` constant. The gas is initially at a temperature (T), pressure (p) and volume (V). The gas expands to volume (4V).

Options

A. The pressure of gas changes to `(p)/(2)`
B. The temperature of the gas changes to `4 T`
C. The graph of the above process on (p - T) diagram is parabola.
D. The graph of the above process on (p - T) diagram is hyperbola.

Explanation

To solve the problem, we start with the given condition for the ideal gas:

**Step 1: Understand the given condition.**
We have \( V P^2 = \text{constant} \). This means that for two states of the gas, we can write:
\[ V_1 P_1^2 = V_2 P_2^2 \]

**Step 2: Identify the initial and final states.**
Let the initial state be denoted by subscript 1:
- Initial volume, \( V_1 = V \)
- Initial pressure, \( P_1 = P \)

The final state (after expansion) is denoted by subscript 2:
- Final volume, \( V_2 = 4V \)
- Final pressure, \( P_2 \) (unknown)

**Step 3: Set up the equation using the condition.**
Substituting the values into the equation:
\[ V \cdot P^2 = 4V \cdot P_2^2 \]

**Step 4: Simplify the equation.**
We can cancel \( V \) from both sides (assuming \( V \neq 0 \)):
\[ P^2 = 4 P_2^2 \]

**Step 5: Solve for \( P_2 \).**
Taking the square root of both sides:
\[ P = 2 P_2 \implies P_2 = \frac{P}{2} \]

**Step 6: Relate pressure and temperature using the ideal gas law.**
The ideal gas law states:
\[ PV = nRT \]
From this, we can express pressure in terms of temperature and volume:
\[ P = \frac{nRT}{V} \]

**Step 7: Substitute the final volume into the ideal gas law.**
For the final state, substituting \( V_2 = 4V \):
\[ P_2 = \frac{nRT_2}{4V} \]

**Step 8: Set up the equation using the final pressure.**
We already found \( P_2 = \frac{P}{2} \). Now substituting this into the equation:
\[ \frac{P}{2} = \frac{nRT_2}{4V} \]

**Step 9: Substitute for \( P \) using the initial state.**
From the initial state:
\[ P = \frac{nRT}{V} \]
Substituting this into the equation gives:
\[ \frac{1}{2} \cdot \frac{nRT}{V} = \frac{nRT_2}{4V} \]

**Step 10: Simplify to find \( T_2 \).**
Cancelling \( nR/V \) from both sides:
\[ \frac{T}{2} = \frac{T_2}{4} \]
Cross-multiplying gives:
\[ 4T = 2T_2 \implies T_2 = 2T \]

**Final Results:**
- The final pressure \( P_2 = \frac{P}{2} \)
- The final temperature \( T_2 = 2T \)

**Step 11: Graph of \( P \) vs \( T \).**
Since we established that \( PT = \text{constant} \), the graph of \( P \) vs \( T \) will be a hyperbola.

### Summary of Results:
1. The pressure of the gas after expansion is \( P_2 = \frac{P}{2} \).
2. The temperature of the gas after expansion is \( T_2 = 2T \).
3. The graph of \( P \) vs \( T \) is a hyperbola.