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 is the efficiency of the cycle?

Explanation

To solve the problem step-by-step, we will follow the principles of thermodynamics, specifically applying the first law of thermodynamics to a cyclic process.

### Step-by-Step Solution:

**Step 1: Understand the First Law of Thermodynamics**
The first law of thermodynamics states that:
\[ \Delta Q = \Delta U + \Delta W \]
For a cyclic process, the change in internal energy (\(\Delta U\)) is zero, hence:
\[ Q_{\text{net}} = W_{\text{net}} \]

**Step 2: Calculate the Net Heat Transfer**
The net heat transfer (\(Q_{\text{net}}\)) for the cycle can be calculated by summing the heat transfers for each step:
\[ Q_{\text{net}} = Q_1 + Q_2 + Q_3 + Q_4 \]
Substituting the given values:
\[ Q_{\text{net}} = 5960\,J - 5585\,J - 2980\,J + 3645\,J \]
Calculating this gives:
\[ Q_{\text{net}} = 5960 - 5585 - 2980 + 3645 = 1040\,J \]

**Step 3: Calculate the Net Work Done**
Similarly, the net work done (\(W_{\text{net}}\)) can be expressed as:
\[ W_{\text{net}} = W_1 + W_2 + W_3 + W_4 \]
Substituting the known values:
\[ W_{\text{net}} = 2200\,J - 825\,J - 1100\,J + W_4 \]
This simplifies to:
\[ W_{\text{net}} = 2200 - 825 - 1100 + W_4 = 275 + W_4 \]

**Step 4: Set Up the Equation**
From the first law, we know:
\[ Q_{\text{net}} = W_{\text{net}} \]
Substituting the expressions from Steps 2 and 3:
\[ 1040\,J = 275\,J + W_4 \]

**Step 5: Solve for \(W_4\)**
Rearranging the equation to find \(W_4\):
\[ W_4 = 1040\,J - 275\,J = 765\,J \]

**Step 6: Calculate the Efficiency of the Cycle**
The efficiency (\(\eta\)) of the cycle is given by:
\[ \eta = \frac{W_{\text{net}}}{Q_{\text{absorbed}}} \times 100 \]
Where \(Q_{\text{absorbed}}\) is the sum of the heat absorbed during the process:
\[ Q_{\text{absorbed}} = Q_1 + Q_4 = 5960\,J + 3645\,J = 9605\,J \]
Now substituting the values:
\[ \eta = \frac{1040\,J}{9605\,J} \times 100 \]
Calculating this gives:
\[ \eta \approx 10.82\% \]

### Final Answers:
(a) The value of \(W_4\) is \(765\,J\).
(b) The efficiency of the cycle is approximately \(10.82\%\).