The work done by the system in a cyclic process involving one mole of an ideal monoatomic gas is `-50 kJ//cycle`. The heat absorbed by the system per cycle is

Options

A. Zero
B. 50 kJ
C. `-50 kJ`
D. 250 kJ

Explanation

To solve the problem, we will use the first law of thermodynamics, which states:

\[ \Delta U = Q - W \]

Where:
- \(\Delta U\) is the change in internal energy,
- \(Q\) is the heat absorbed by the system,
- \(W\) is the work done by the system.

### Step 1: Understand the cyclic process
In a cyclic process, the system returns to its initial state after completing the cycle. Therefore, the change in internal energy (\(\Delta U\)) for a complete cycle is zero.

\[ \Delta U = 0 \]

### Step 2: Apply the first law of thermodynamics
Since \(\Delta U = 0\), we can rewrite the first law of thermodynamics as:

\[ 0 = Q - W \]

This can be rearranged to find \(Q\):

\[ Q = W \]

### Step 3: Substitute the given values
We know from the problem that the work done by the system is \(-50 \, \text{kJ}\):

\[ W = -50 \, \text{kJ} \]

Substituting this value into the equation for \(Q\):

\[ Q = -(-50 \, \text{kJ}) = +50 \, \text{kJ} \]

### Step 4: Conclusion
Thus, the heat absorbed by the system per cycle is:

\[ Q = +50 \, \text{kJ} \]

### Final Answer:
The heat absorbed by the system per cycle is **50 kJ**.

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