`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 need to apply the first law of thermodynamics, which relates the change in internal energy (ΔU) of a system to the heat added to the system (Q) and the work done by the system (W). The equation is given by:

\[ \Delta U = Q - W \]

In a cyclic process, the system returns to its initial state, which means that the change in internal energy (ΔU) is zero. Therefore, we can write:

\[ 0 = Q - W \]

From this, we can rearrange the equation to find the heat absorbed (Q):

\[ Q = W \]

Given that the work done by the gas in one cycle is 550 kJ, we can substitute this value into the equation:

\[ Q = 550 \text{ kJ} \]

Thus, the amount of heat absorbed by the system in one cycle is:

\[ Q = 550 \text{ kJ} \]

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

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