One mole of an ideal gas at pressure `P_(0)` and temperature `T_(0)` is expanded isothermally to twice its volume and then compressed at constant pressure to `(V_(0) // 2)` and the gas is brought bac to original state by a process in which `P alpha V` (pressure is directly proportional to volume). The correct temperature of the process is

Options

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Explanation

To solve the problem step by step, we will analyze the three processes that the gas undergoes and determine the final temperature of the gas after it has returned to its original state.

### Step 1: Understand the Initial Conditions
We have 1 mole of an ideal gas at:
- Initial Pressure: \( P_0 \)
- Initial Temperature: \( T_0 \)
- Initial Volume: \( V_0 \)

### Step 2: Process 1 - Isothermal Expansion
The gas is expanded isothermally to twice its volume:
- Final Volume after expansion: \( V = 2V_0 \)

Since the process is isothermal, the temperature remains constant at \( T_0 \). According to Boyle's Law for isothermal processes:
\[ P_0 V_0 = P_1 (2V_0) \]
From this, we can find the new pressure \( P_1 \):
\[ P_1 = \frac{P_0 V_0}{2V_0} = \frac{P_0}{2} \]

### Step 3: Process 2 - Isobaric Compression
Next, the gas is compressed at constant pressure to \( V_0/2 \):
- Final Volume after compression: \( V = \frac{V_0}{2} \)
- Pressure remains constant: \( P_1 = P_0/2 \)

Using the ideal gas law \( PV = nRT \):
\[ \frac{P_0}{2} \cdot \frac{V_0}{2} = nRT_1 \]
Where \( T_1 \) is the temperature after this process. Substituting \( n = 1 \) (1 mole):
\[ \frac{P_0 V_0}{4} = RT_1 \]
Thus,
\[ T_1 = \frac{P_0 V_0}{4R} \]

### Step 4: Process 3 - P is Proportional to V
In the final process, the gas is brought back to the original state with \( P \propto V \). This means:
\[ P = kV \]
Using the ideal gas law again, we can express this as:
\[ PV = nRT \]
Substituting \( P = kV \):
\[ kV^2 = nRT \]
This implies:
\[ T = \frac{kV^2}{nR} \]

### Step 5: Returning to Original State
To find the temperature at the end of the third process, we need to know the values of \( k \) and \( V \). Since the process returns to the original state, we know:
- Final Volume: \( V = V_0 \)
- Final Pressure: \( P = P_0 \)

Substituting back into the equation:
\[ T = \frac{kV_0^2}{nR} \]

### Step 6: Relate \( k \) to Original Conditions
From the previous processes, we can relate \( k \) to the original conditions. Since we know the gas returns to its original state, we can set \( T \) equal to \( T_0 \):
\[ T_0 = \frac{kV_0^2}{R} \]

### Final Calculation
To find the correct temperature of the process, we need to relate it back to the initial temperature:
- After the first process, the temperature remains \( T_0 \).
- After the second process, the temperature is \( T_1 = \frac{P_0 V_0}{4R} \).
- The final temperature after the third process must also relate back to the original state.

Thus, the correct temperature of the process is:
\[ T = T_0 \]

### Conclusion
The correct temperature of the process is \( T_0 \).