An ideal gas is taken from the state A (pressure p, volume V) to the state B (pressure `p/2`, volume 2V) along a straight line path in the p-V diagram. Select the correct statement(s) from the following.

Options

A. The work done by the gas in the process A to B exceeds the work that would be done by it if the system were taken from A to B along an isotherm
B. In the T-V diagram, the path AB becomes a part of a parabola
C. In the p-T diageam, the AB becomes a part of a hyperbola
D. In going from A to B, the temperature T of the gas decreases

Answer

To solve the problem, we need to analyze the transition of an ideal gas from state A (pressure P, volume V) to state B (pressure P/2, volume 2V) along a straight line in the p-V diagram. We will evaluate the statements given in the question based on our analysis.

### Step-by-Step Solution:

1. **Identify the States**:
- State A: \( P = P \), \( V = V \)
- State B: \( P = \frac{P}{2} \), \( V = 2V \)

2. **Draw the p-V Diagram**:
- Plot point A at coordinates (V, P) and point B at (2V, P/2).
- Connect these two points with a straight line, representing the path taken by the gas.

3. **Calculate Work Done**:
- The work done by the gas during the transition from A to B can be calculated as the area under the p-V curve.
- For a straight line, the work done \( W \) can be expressed as:
\[ W = \text{Area of trapezoid} = \frac{1}{2} \times (P + \frac{P}{2}) \times (2V - V) = \frac{1}{2} \times \frac{3P}{2} \times V = \frac{3PV}{4} \]

4. **Compare with Isothermal Process**:
- For an isothermal process between the same points, the work done can be calculated using the formula:
\[ W_{\text{isothermal}} = nRT \ln\left(\frac{V_f}{V_i}\right) \]
- Since both states A and B have the same temperature (as derived from the ideal gas law), the work done in the isothermal process will be less than the work done along the straight line path.

5. **Evaluate Temperature Change**:
- The temperature at state A can be calculated as:
\[ T_A = \frac{PV}{nR} \]
- The temperature at state B is:
\[ T_B = \frac{\left(\frac{P}{2}\right)(2V)}{nR} = \frac{PV}{nR} \]
- Thus, \( T_A = T_B \), indicating that the temperatures at both states are equal.

6. **Determine the Shape of the T-V Graph**:
- The relationship between temperature and volume can be derived from the ideal gas law. The path taken is linear in the p-V diagram, which implies a parabolic relationship in the T-V diagram.

7. **Determine the Shape of the P-T Graph**:
- By manipulating the ideal gas law, we can derive the relationship between pressure and temperature, which also results in a parabolic equation.

### Conclusion:
Based on the analysis, we can conclude:
- The work done by the gas from A to B exceeds the work done during an isothermal process.
- The T-V graph is parabolic.
- The P-T graph is not a hyperbola but also follows a parabolic relationship.

### Correct Statements:
1. The work done by the gas in the process A to B exceeds the work done during an isothermal process.
2. The T-V graph is a parabola.