Question

An ideal gas is taken through a cyclic thermodynamical process through four steps. The amounts of heat involved in these steps are \( Q_1 = 5960 \, \text{J} \), \( Q_2 = -5585 \, \text{J} \), \( Q_3 = -2980 \, \text{J} \) and \( Q_4 = 3645 \, \text{J} \), respectively. The value of \( W_4 \) is:

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

Hint:

In cyclic processes, the work done is equal to the net heat added or removed from the system, since the internal energy returns to its original value.

Answer 1

The Correct Option is C

The first law of thermodynamics states that the change in internal energy of the gas is the sum of the heat added to the system and the work done by the system: \[ \Delta U = Q - W \] For a cyclic process, the net change in internal energy \( \Delta U = 0 \).
Therefore, the net heat added equals the net work done: \[ W = Q_1 + Q_2 + Q_3 + Q_4 \] Substituting the given values: \[ W = 5960 + (-5585) + (-2980) + 3645 = 765 \, \text{J} \]
Thus, the correct answer is (c).