`550 kJ "cycle" ^(-1)` work is done by 1 mol of an ideal gas in a cyclic process. The amount of heat absorbed by the system in one cycle is

Options

A. 550 kJ
B. `-550 kJ`
C. 1100 kJ
D. Zero

Answer

To solve the problem, we will use the first law of thermodynamics, which 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).

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Work done by the system (W) = 550 kJ (since work is done by the system, it is considered negative in the context of the first law).
- The process is cyclic, which means the change in internal energy (ΔU) = 0.

2. **Apply the First Law of Thermodynamics:**
\[ \Delta U = Q - W \]
Since the process is cyclic, we have:
\[ 0 = Q - W \]

3. **Rearrange the Equation:**
\[ Q = W \]

4. **Substitute the Value of Work Done:**
Since W = -550 kJ (because work is done by the system),
\[ Q = -(-550 \text{ kJ}) = 550 \text{ kJ} \]

5. **Conclusion:**
The amount of heat absorbed by the system in one cycle is **550 kJ**.

### Final Answer:
The amount of heat absorbed by the system in one cycle is **550 kJ**.

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