An ideal gas is taken through a cyclic thermodynamic process through four steps. The amounts of heat involved in these steps are `Q_1=5960J`, `Q_2=-5585J`, `Q_3=-2980J` and `Q_4=3645J` respectively. The corresponding quantities of work involved are `W_1=2200J`, `W_2=-825J`, `W_3=-1100J` and `W_4` respectively.
(a) Find the value of `W_4`.
(b) What are the efficiency of the cycle?

Explanation

To solve the problem step by step, we will address both parts of the question: finding the value of \( W_4 \) and calculating the efficiency of the cycle.

### Part (a): Finding the value of \( W_4 \)

1. **Understanding the First Law of Thermodynamics**:
The first law states that the change in internal energy (\( \Delta U \)) of a system is equal to the heat added to the system (\( Q \)) minus the work done by the system (\( W \)). For a cyclic process, \( \Delta U = 0 \), which gives us the equation:
\[ Q_{\text{total}} = W_{\text{total}} \]

2. **Writing the equation for the cyclic process**:
We can express this as:
\[ Q_1 + Q_2 + Q_3 + Q_4 = W_1 + W_2 + W_3 + W_4 \]

3. **Substituting the known values**:
Given:
- \( Q_1 = 5960 \, J \)
- \( Q_2 = -5585 \, J \)
- \( Q_3 = -2980 \, J \)
- \( Q_4 = 3645 \, J \)
- \( W_1 = 2200 \, J \)
- \( W_2 = -825 \, J \)
- \( W_3 = -1100 \, J \)
- \( W_4 = ? \)

Substitute these values into the equation:
\[ 5960 + (-5585) + (-2980) + 3645 = 2200 + (-825) + (-1100) + W_4 \]

4. **Calculating the left side**:
\[ 5960 - 5585 - 2980 + 3645 = 5960 - 5585 - 2980 + 3645 = 1040 \, J \]

5. **Calculating the right side**:
\[ 2200 - 825 - 1100 + W_4 = 2200 - 825 - 1100 + W_4 = 275 + W_4 \]

6. **Setting the left side equal to the right side**:
\[ 1040 = 275 + W_4 \]

7. **Solving for \( W_4 \)**:
\[ W_4 = 1040 - 275 = 765 \, J \]

### Part (b): Calculating the efficiency of the cycle

1. **Understanding Efficiency**:
The efficiency (\( \eta \)) of a thermodynamic cycle is given by the formula:
\[ \eta = \frac{W_{\text{net}}}{Q_{\text{in}}} \times 100 \]
where \( W_{\text{net}} \) is the total work done by the system and \( Q_{\text{in}} \) is the total heat absorbed by the system.

2. **Calculating \( W_{\text{net}} \)**:
\[ W_{\text{net}} = W_1 + W_2 + W_3 + W_4 = 2200 - 825 - 1100 + 765 = 1040 \, J \]

3. **Calculating \( Q_{\text{in}} \)**:
The heat absorbed by the system is the sum of the positive heat values:
\[ Q_{\text{in}} = Q_1 + Q_4 = 5960 + 3645 = 9605 \, J \]

4. **Calculating Efficiency**:
\[ \eta = \frac{1040}{9605} \times 100 \approx 10.82\% \]

### Final Answers:
(a) \( W_4 = 765 \, J \)
(b) Efficiency of the cycle \( \eta \approx 10.82\% \)