An ideal gas is initially at `P_1,V_1` is expands to `P_2,V_2` and then compressed adiabatically to the same volume `V_1` and pressure `P_3.` If W is the net work done by the gas in complete process which of the following is true.

Options

A. `Wgt0, P_3gtP_1`
B. `Wlt0, P_3gtP_1`
C. `Wgt0, P_3ltP_1`
D. `Wlt0, P_3ltP_1`

Explanation

To solve the problem step by step, we will analyze the process of the ideal gas as it undergoes expansion and adiabatic compression, and then determine the net work done by the gas.

### Step 1: Understand the Initial State
The ideal gas starts at an initial state defined by pressure \( P_1 \) and volume \( V_1 \).

### Step 2: Expansion Process
The gas expands to a new state defined by pressure \( P_2 \) and volume \( V_2 \). According to Boyle's law, if the volume increases (from \( V_1 \) to \( V_2 \)), the pressure decreases. Thus, we can conclude that:
\[ P_2 < P_1 \]

### Step 3: Work Done During Expansion
The work done by the gas during this expansion can be represented as the area under the curve in a PV diagram. Since the gas is expanding, the work done \( W_{exp} \) is positive.

### Step 4: Compression Process
Next, the gas is compressed adiabatically back to the original volume \( V_1 \) but now at a new pressure \( P_3 \). During this compression, the volume decreases, which leads to an increase in pressure:
\[ P_3 > P_1 \]

### Step 5: Work Done During Compression
The work done on the gas during this adiabatic compression is negative, as the gas is being compressed. We can denote this work as \( W_{comp} \), which is negative.

### Step 6: Net Work Done
The net work done \( W \) by the gas over the entire process can be calculated as:
\[ W = W_{exp} + W_{comp} \]
Since \( W_{exp} \) is positive and \( W_{comp} \) is negative, the net work done will depend on the magnitudes of these two quantities.

### Step 7: Direction of the Process
The overall process can be visualized in a PV diagram as an anti-clockwise loop, indicating that the net work done by the gas is negative. This is because the area enclosed by the loop represents the net work done, and in an anti-clockwise direction, the work done is conventionally considered negative.

### Conclusion
From the analysis:
1. The net work done \( W \) is negative.
2. The pressure \( P_3 \) after adiabatic compression is greater than the initial pressure \( P_1 \).

Thus, the correct conclusion is:
- **The net work done \( W < 0 \) and \( P_3 > P_1 \)**.