A system undergoes a cyclic process in which it absorbs `Q_(1)` heat and gives out `Q_(2)` heat. The efficiency of the process is `eta` and work done is `W.` Select correct statement:

Options

A. `W = Q_1 - Q_2`
B. `eta = W/Q_1`
C. `eta = Q_2/Q_1`
D. `eta = 1- Q_2/Q_1`

Explanation

To solve the problem, we need to analyze the cyclic process of the system and derive the relationships between the heat absorbed, heat given out, work done, and efficiency.

### Step-by-step Solution:

1. **Understanding the Efficiency of the Process:**
The efficiency (η) of a heat engine is defined as the ratio of the work done (W) to the heat absorbed (Q1). Mathematically, this can be expressed as:
\[ \eta = \frac{W}{Q_1} \]

2. **Relating Work Done to Heat Absorbed and Heat Given Out:**
In a cyclic process, the work done by the system can also be expressed in terms of the heat absorbed (Q1) and the heat rejected (Q2). The work done (W) is given by the difference between the heat absorbed and the heat rejected:
\[ W = Q_1 - Q_2 \]

3. **Substituting Work Done into the Efficiency Formula:**
We can substitute the expression for work done into the efficiency formula:
\[ \eta = \frac{Q_1 - Q_2}{Q_1} \]

4. **Simplifying the Efficiency Expression:**
This can be further simplified:
\[ \eta = 1 - \frac{Q_2}{Q_1} \]

5. **Conclusion:**
Based on the above derivations, we can conclude the following:
- The efficiency of the process is given by \( \eta = \frac{W}{Q_1} \).
- The work done can be expressed as \( W = Q_1 - Q_2 \).
- The efficiency can also be expressed as \( \eta = 1 - \frac{Q_2}{Q_1} \).

### Correct Statements:
From the analysis, we find that:
- The efficiency is correctly defined.
- The work done is correctly expressed.
- The alternative expression for efficiency is also correct.

Thus, the correct statements are A, B, and D.