In given figure, let `Delta W` and `Delta W_(2)` be the work done by the gas in process `A` and `B` respectively then (given change in volume is same in both process)

Options

A. `Delta W_(1) gt Delta W_(2)`
B. `Delta W_(1) = Delta W_(2)`
C. `Delta W_(1) lt Delta W_(2)`
D. Nothing can be said about the relation between `Delta W_(1)` and `Delta W_(2)`

Explanation

To solve the problem regarding the work done by the gas in processes A and B, we will follow these steps:

### Step 1: Understand the Work Done by a Gas
The work done by a gas during expansion or compression can be calculated using the formula:
\[ \Delta W = P \Delta V \]
where \( P \) is the pressure and \( \Delta V \) is the change in volume.

### Step 2: Identify the Processes
In the given question, we have two processes:
- Process A (at constant pressure \( P_A \))
- Process B (at constant pressure \( P_B \))

### Step 3: Write the Expressions for Work Done
For process A, the work done \( \Delta W_1 \) can be expressed as:
\[ \Delta W_1 = P_A \Delta V \]

For process B, the work done \( \Delta W_2 \) can be expressed as:
\[ \Delta W_2 = P_B \Delta V \]

### Step 4: Compare the Work Done
Since the change in volume \( \Delta V \) is the same for both processes, we can compare the work done by taking the ratio:
\[ \frac{\Delta W_1}{\Delta W_2} = \frac{P_A \Delta V}{P_B \Delta V} = \frac{P_A}{P_B} \]

### Step 5: Analyze the Pressure Values
From the problem statement, we know that the pressure \( P_B \) is greater than \( P_A \) (i.e., \( P_B > P_A \)). Therefore, the ratio \( \frac{P_A}{P_B} \) will be less than 1:
\[ \frac{P_A}{P_B} < 1 \]

### Step 6: Conclude the Comparison of Work Done
Since \( \frac{\Delta W_1}{\Delta W_2} < 1 \), we can conclude that:
\[ \Delta W_1 < \Delta W_2 \]

### Final Answer
Thus, the work done by the gas in process A is less than the work done in process B:
\[ \Delta W_1 < \Delta W_2 \]