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.
(1) Find the value of `W_4`.
(2) What is the efficiency of the cycle

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the First Law of Thermodynamics
The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W):
\[ \Delta U = Q - W \]
For a cyclic process, the change in internal energy is zero, hence:
\[ Q = W \]

### Step 2: Set Up the Equation for the Cycle
For the cyclic process, we can express the total heat and work done as:
\[ Q_1 + Q_2 + Q_3 + Q_4 = W_1 + W_2 + W_3 + W_4 \]

### Step 3: Substitute the Given Values
We substitute the values of heat (Q) and work (W) into the equation:
- \( Q_1 = 5960 \, J \)
- \( Q_2 = -5585 \, J \)
- \( Q_3 = -2980 \, J \)
- \( Q_4 = 3645 \, J \)
- \( W_1 = 2200 \, J \)
- \( W_2 = -825 \, J \)
- \( W_3 = -1100 \, J \)

The equation becomes:
\[ 5960 - 5585 - 2980 + 3645 = 2200 - 825 - 1100 + W_4 \]

### Step 4: Simplify the Equation
Now, we simplify both sides:
1. Left side:
\[ 5960 - 5585 - 2980 + 3645 = 5960 - 5585 - 2980 + 3645 = 5960 - 5585 = 375 + 3645 = 4020 \]

2. Right side:
\[ 2200 - 825 - 1100 + W_4 = 2200 - 825 - 1100 = 2200 - 1925 = 275 + W_4 \]

### Step 5: Solve for \( W_4 \)
Now we set the simplified left side equal to the right side:
\[ 4020 = 275 + W_4 \]
To find \( W_4 \), we rearrange the equation:
\[ W_4 = 4020 - 275 = 3745 \, J \]

### Step 6: Calculate the Efficiency of the Cycle
The efficiency (η) of the cycle is given by:
\[ \eta = \frac{\text{Total Work Done}}{\text{Heat Absorbed}} \times 100 \]

#### Step 6.1: Calculate Total Work Done
Total work done is:
\[ W_{\text{total}} = W_1 + W_2 + W_3 + W_4 \]
Substituting the values:
\[ W_{\text{total}} = 2200 - 825 - 1100 + 3745 \]
Calculating this gives:
\[ W_{\text{total}} = 2200 - 825 - 1100 + 3745 = 2200 - 1925 + 3745 = 2200 + 1820 = 4020 \, J \]

#### Step 6.2: Calculate Heat Absorbed
Heat absorbed is:
\[ Q_{\text{absorbed}} = Q_1 + Q_4 = 5960 + 3645 = 9605 \, J \]

#### Step 6.3: Calculate Efficiency
Now substituting into the efficiency formula:
\[ \eta = \frac{4020}{9605} \times 100 \]
Calculating this gives:
\[ \eta \approx 41.8\% \]

### Final Answers
1. The value of \( W_4 \) is \( 3745 \, J \).
2. The efficiency of the cycle is approximately \( 41.8\% \).