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 find the heat absorbed by the system per cycle in a cyclic process involving one mole of an ideal monoatomic gas, we can use the first law of thermodynamics. Here’s the step-by-step solution:

### Step 1: Understand the First Law of Thermodynamics
The first law of thermodynamics states that:
\[ \Delta U = Q + W \]
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 2: Identify the Characteristics of a Cyclic Process
In a cyclic process, the system returns to its initial state, which means that the change in internal energy (\(\Delta U\)) is zero:
\[ \Delta U = 0 \]

### Step 3: Apply the First Law to the Cyclic Process
Since \(\Delta U = 0\), we can rewrite the first law as:
\[ 0 = Q + W \]
This implies:
\[ Q = -W \]

### Step 4: Substitute the Given Work Done
We are given that the work done by the system is:
\[ W = -50 \, \text{kJ} \]
Now, substituting this value into the equation for \(Q\):
\[ Q = -(-50 \, \text{kJ}) = 50 \, \text{kJ} \]

### Step 5: Conclusion
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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