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= and -2980J Q_4 =3645 J,` respectively . The corresponding works involved are `W_1=2200J, W_2=-825 J, W_3= and -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, we can use the first law of thermodynamics, which states:

\[ \Delta U = Q - W \]

In a cyclic process, the change in internal energy \( \Delta U \) is zero. Therefore, we can write:

\[ Q = W \]

This means that the total heat added to the system is equal to the total work done by the system over the entire cycle.

### Step-by-step Solution:

1. **Write the equation for total heat and total work:**
\[ Q_1 + Q_2 + Q_3 + Q_4 = W_1 + W_2 + W_3 + W_4 \]

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

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

3. **Calculate the left-hand side:**
\[ 5960 - 5585 = 375 \\ 375 - 2980 = -2605 \\ -2605 + 3645 = 1040 \]

So, the left-hand side equals \( 1040 \, J \).

4. **Calculate the right-hand side:**
\[ 2200 - 825 = 1375 \\ 1375 - 1100 = 275 \\ 275 + W_4 \]

5. **Set the left-hand side equal to the right-hand side:**
\[ 1040 = 275 + W_4 \]

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

### Final Answer:
\[ W_4 = 765 \, J \]