An ideal gas is taken through a cyclic thermodynamical process through four steps. The amounts of heat involved in these steps are`Q_1=5960J,Q_2=-5585J,Q_3= - 2980J, Q_4 =3645 J,` respectively . The corresponding works involved are `W_1=2200J, W_2=-825 J, W_3= -1100J, W_4` respectively. The value of `W_4` is

Options

A. 1315 J
B. 275 J
C. 765 J
D. 675 J

Answer

To find the value of \( W_4 \) in the cyclic thermodynamic process involving an ideal gas, we can follow these steps:

### Step 1: Understand the First Law of Thermodynamics
The first law of thermodynamics states that:
\[ Q = \Delta U + W \]
where \( Q \) is the heat added to the system, \( \Delta U \) is the change in internal energy, and \( W \) is the work done by the system. In a cyclic process, the change in internal energy \( \Delta U \) is zero, so we have:
\[ Q = W \]

### Step 2: Calculate the Total Heat \( Q \)
We need to find the total heat absorbed by the system during the four steps:
\[ Q = Q_1 + Q_2 + Q_3 + Q_4 \]
Substituting the given values:
\[ Q = 5960\,J + (-5585\,J) + (-2980\,J) + 3645\,J \]
Calculating this step-by-step:
- \( 5960 - 5585 = 375 \)
- \( 375 - 2980 = -2605 \)
- \( -2605 + 3645 = 1040 \)

So, the total heat \( Q \) is:
\[ Q = 1040\,J \]

### Step 3: Calculate the Total Work \( W \)
Now, we can express the total work done by the system during the four steps:
\[ W = W_1 + W_2 + W_3 + W_4 \]
Substituting the known values:
\[ W = 2200\,J + (-825\,J) + (-1100\,J) + W_4 \]
Calculating this step-by-step:
- \( 2200 - 825 = 1375 \)
- \( 1375 - 1100 = 275 \)

So, we have:
\[ W = 275\,J + W_4 \]

### Step 4: Set Up the Equation
From the first law of thermodynamics in a cyclic process, we know:
\[ Q = W \]
Thus, we can set up the equation:
\[ 1040\,J = 275\,J + W_4 \]

### Step 5: Solve for \( W_4 \)
Rearranging the equation to solve for \( W_4 \):
\[ W_4 = 1040\,J - 275\,J \]
Calculating this gives:
\[ W_4 = 765\,J \]

### Final Answer
The value of \( W_4 \) is:
\[ \boxed{765\,J} \]