In the figure shown below, an ideal gas is carried around the cyclic process. How much work is done in one cycle if `P_(0) = 8 atm and V_(0) = 7.00` litre?

Options

A. 5656 J
B. `-5656 J`
C. `10600 J`
D. `11300 J`

Answer

To solve the problem of calculating the work done in one cycle of an ideal gas carried around a cyclic process, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Initial Pressure, \( P_0 = 8 \, \text{atm} \)
- Initial Volume, \( V_0 = 7.00 \, \text{litre} \)

2. **Convert Units**:
- Convert pressure from atm to Pascal:
\[ P_0 = 8 \, \text{atm} \times 1.01 \times 10^5 \, \text{Pa/atm} = 8.08 \times 10^5 \, \text{Pa} \]
- Convert volume from litres to cubic meters:
\[ V_0 = 7.00 \, \text{litre} = 7.00 \times 10^{-3} \, \text{m}^3 \]

3. **Understand the Work Done in a Cyclic Process**:
- The work done by the gas during a cyclic process is equal to the area enclosed by the process on the PV diagram.
- Since the process is anticlockwise, the work done will be negative.

4. **Calculate the Work Done**:
- For a rectangular area on the PV diagram, the work done \( W \) can be calculated as:
\[ W = -P_0 \times V_0 \]
- Substitute the values:
\[ W = - (8.08 \times 10^5 \, \text{Pa}) \times (7.00 \times 10^{-3} \, \text{m}^3) \]
- Calculate:
\[ W = - (8.08 \times 10^5) \times (7.00 \times 10^{-3}) = -5656 \, \text{J} \]

5. **Final Result**:
- The work done in one cycle is:
\[ W = -5656 \, \text{J} \]

### Summary:
The work done in one cycle of the gas is \(-5656 \, \text{J}\).