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. (a) The work done by the gas in the process A to B is negative
B. (b) In the T-V diagram, the path AB becomes a part of parabola
C. (c) In the p-T diagram, the path AB becomes a part of a hyperbola
D. (d) In going from A to B, the temperature T of the gas first increases to a maximum value and then decreases.

Explanation

To solve the problem, we need to analyze the transition of an ideal gas from state A to state B in a p-V diagram, where state A has pressure \( p \) and volume \( V \), and state B has pressure \( \frac{p}{2} \) and volume \( 2V \). We will examine the work done, the relationship between pressure and volume, and the behavior of temperature during this process.

### Step-by-Step Solution:

1. **Identify the Initial and Final States:**
- Initial state (A): \( P_A = p \), \( V_A = V \)
- Final state (B): \( P_B = \frac{p}{2} \), \( V_B = 2V \)

2. **Calculate the Work Done:**
- The work done by the gas during an expansion from state A to state B along a straight line in the p-V diagram can be calculated using the formula for work done in a p-V process:
\[ W = \int_{V_A}^{V_B} P \, dV \]
- Since the path is a straight line, we can express pressure \( P \) as a linear function of volume \( V \). The equation of the line connecting points A and B can be derived from the two points:
\[ P = mV + c \]
- The slope \( m \) can be calculated as:
\[ m = \frac{P_B - P_A}{V_B - V_A} = \frac{\frac{p}{2} - p}{2V - V} = \frac{-\frac{p}{2}}{V} = -\frac{p}{2V} \]
- The y-intercept \( c \) is equal to the pressure at point A, which is \( p \).
- Thus, the equation of the line is:
\[ P = -\frac{p}{2V} V + p \]

3. **Integrate to Find Work Done:**
- Substitute the equation of the line into the work done formula:
\[ W = \int_{V}^{2V} \left(-\frac{p}{2V} V + p\right) dV \]
- Simplifying gives:
\[ W = \int_{V}^{2V} \left(-\frac{p}{2} + p\right) dV = \int_{V}^{2V} \left(\frac{p}{2}\right) dV \]
- Evaluating the integral:
\[ W = \frac{p}{2} \left[ V \right]_{V}^{2V} = \frac{p}{2} (2V - V) = \frac{pV}{2} \]

4. **Analyze Temperature Change:**
- The ideal gas law states that \( PV = nRT \). We can express temperature in terms of pressure and volume:
\[ T = \frac{PV}{nR} \]
- At state A:
\[ T_A = \frac{pV}{nR} \]
- At state B:
\[ T_B = \frac{\left(\frac{p}{2}\right)(2V)}{nR} = \frac{pV}{nR} \]
- As the gas moves from A to B, the pressure decreases while the volume increases. The relationship between pressure and volume will create a parabolic path in the T-V diagram, indicating that temperature will first increase to a maximum and then decrease.

5. **Conclusion:**
- The correct statements based on the analysis are:
- The work done by the gas is positive, as the volume increases.
- The relationship between temperature and volume follows a parabolic path, indicating that temperature first increases to a maximum value and then decreases.

### Correct Options:
- Option B: The relationship between T and V is a parabola.
- Option D: In going from A to B, the temperature T of the gas first increases to a maximum value and then decreases.