A fixed mass of a gas is first heated isobarically to double the volume and then cooled isochorically

Options

A. (a)2
B. (b)`1//2`
C. (c)1n 2
D. (d)1n 3

Explanation

To solve the problem step by step, we will analyze the processes involved and calculate the work done during each process.

### Step 1: Initial Conditions
Let the initial pressure of the gas be \( P \) and the initial volume be \( V \).

### Step 2: Isobaric Expansion
The gas is heated isobarically (at constant pressure) to double its volume. Therefore, the final volume after this process is:
\[ V_f = 2V \]
Since the process is isobaric, the pressure remains constant at \( P \).

### Step 3: Work Done During Isobaric Expansion
The work done during an isobaric process is given by:
\[ W = P \Delta V \]
Where \( \Delta V \) is the change in volume. Here, the change in volume is:
\[ \Delta V = V_f - V_i = 2V - V = V \]
Thus, the work done during the isobaric expansion becomes:
\[ W_{isobaric} = P \cdot V \]

### Step 4: Final Conditions After Isobaric Expansion
After the isobaric expansion, the gas has:
- Volume: \( 2V \)
- Pressure: \( P \)

### Step 5: Isochoric Cooling
Next, the gas is cooled isochorically (at constant volume). Since the volume does not change during this process, the work done during the isochoric process is:
\[ W_{isochoric} = P \Delta V = P \cdot 0 = 0 \]

### Step 6: Total Work Done
The total work done in the two processes is the sum of the work done during the isobaric and isochoric processes:
\[ W_{total} = W_{isobaric} + W_{isochoric} = PV + 0 = PV \]

### Step 7: Conclusion
The total work done during the entire process is \( PV \).

### Final Answer
The total work done during the two successive processes is \( PV \).

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